Atomic Structure – Subjective Questions and Answers

The heat capacity of a solid according to the Debye model is given by: $C_V = 9 N k_B \left(\frac{T}{\Theta_D}\right)^3 \int_0^{\Theta_D/T} \frac{x^4 e^x}{(e^x – 1)^2} dx$ where $N$ is the number of atoms, $k_B$ is the Boltzmann constant, $T$ is the temperature, and $\Theta_D$ is the Debye temperature. For one mole, $N k_B = R$, so $3 N k_B = 3R$. The heat capacity relative to the classical value is $C_V / (3R)$.

Case 1: T = 20 K $\frac{T}{\Theta_D} = \frac{20}{400} = 0.05$ $\frac{\Theta_D}{T} = \frac{400}{20} = 20$ Since $T \ll \Theta_D$, we can use the low-temperature approximation: $C_V \approx \frac{12 \pi^4}{5} N k_B \left(\frac{T}{\Theta_D}\right)^3$ $\frac{C_V}{3R} \approx \frac{1}{3} \frac{12 \pi^4}{5} \left(\frac{T}{\Theta_D}\right)^3 = \frac{4 \pi^4}{5} \left(\frac{T}{\Theta_D}\right)^3$ $\frac{C_V}{3R} \approx \frac{4 \times (3.14159)^4}{5} \times (0.05)^3 = \frac{4 \times 97.409}{5} \times 0.000125 = 77.9272 \times 0.000125 \approx 0.00974$

Case 2: T = 800 K $\frac{T}{\Theta_D} = \frac{800}{400} = 2$ $\frac{\Theta_D}{T} = \frac{400}{800} = 0.5$ Since $T > \Theta_D$, we can use the high-temperature expansion of the Debye function for heat capacity. The Debye function for heat capacity is $D_C(y) = \frac{3}{y^3} \int_0^y \frac{x^4 e^x}{(e^x – 1)^2} dx$, where $y = \Theta_D/T$. Then $C_V = 3R D_C(\Theta_D/T)$. For high temperatures ($T \gg \Theta_D$, or small $y$), $D_C(y) \approx 1 – \frac{y^2}{40}$. $\frac{C_V}{3R} \approx 1 – \frac{1}{40} \left(\frac{\Theta_D}{T}\right)^2$ $\frac{C_V}{3R} \approx 1 – \frac{1}{40} (0.5)^2 = 1 – \frac{1}{40} \times 0.25 = 1 – 0.00625 = 0.99375$

Final Answer: The final answer is $\boxed{0.00974, 0.994}$

The molar heat capacity at constant volume according to the Einstein model is given by: $C_V = 3R \left(\frac{\theta_E}{T}\right)^2 \frac{e^{\theta_E/T}}{(e^{\theta_E/T} – 1)^2}$ where $\theta_E$ is the Einstein temperature and $T$ is the ambient temperature.

The Dulong-Petit law states that the molar heat capacity of a solid is approximately $3R$.

Given that the Einstein temperature is much larger than the ambient temperature, we have $\theta_E \gg T$. This means that $\frac{\theta_E}{T}$ is a large number.

Let $x = \frac{\theta_E}{T}$. Since $\theta_E \gg T$, we have $x \gg 1$.

Now, let’s examine the Einstein formula in this limit: $C_V = 3R x^2 \frac{e^x}{(e^x – 1)^2}$

Since $x$ is large, $e^x \gg 1$, so we can approximate $(e^x – 1)^2 \approx (e^x)^2 = e^{2x}$.

Therefore, the Einstein formula becomes approximately: $C_V \approx 3R x^2 \frac{e^x}{e^{2x}} = 3R x^2 e^{-x}$

Substitute back $x = \frac{\theta_E}{T}$: $C_V \approx 3R \left(\frac{\theta_E}{T}\right)^2 e^{-\theta_E/T}$

Since $\theta_E \gg T$, the term $e^{-\theta_E/T}$ is very small (approaching zero). Although $\left(\frac{\theta_E}{T}\right)^2$ is large, the exponential term decays much faster. Therefore, the product $\left(\frac{\theta_E}{T}\right)^2 e^{-\theta_E/T}$ will be a very small number.

Thus, the heat capacity $C_V$ will be much smaller than $3R$.

Final Answer: The final answer is $\boxed{\text{The heat capacity is much smaller than the Dulong-Petit law prediction.}}$

The heat capacity of a diatomic gas is higher than that of a monatomic gas because diatomic molecules possess additional modes of energy storage that monatomic molecules do not. These additional modes are related to the internal structure of the molecule.

Monatomic Gases:

• Translational Motion: Monatomic gases (like Helium, Neon, Argon) can only store energy through the kinetic energy of their translational motion. This means they can move in three independent directions (x, y, and z).
• Degrees of Freedom: According to the equipartition theorem, each degree of freedom contributes $\frac{1}{2}kT$ of energy per molecule (or $\frac{1}{2}RT$ per mole), where $k$ is the Boltzmann constant and $R$ is the ideal gas constant. Monatomic gases have 3 translational degrees of freedom.
• Heat Capacity: For a monatomic ideal gas, the molar heat capacity at constant volume ($C_V$) is determined by the translational degrees of freedom: $C_V = \frac{3}{2}R$

Diatomic Gases:

• Translational Motion: Like monatomic gases, diatomic gases (like Oxygen, Nitrogen, Hydrogen) can also store energy through translational motion in three directions.
• Rotational Motion: Diatomic molecules can rotate around two axes perpendicular to the bond axis. Rotation around the bond axis itself doesn’t contribute significantly to the energy at typical temperatures because the moment of inertia around that axis is very small. This adds 2 rotational degrees of freedom.
• Vibrational Motion: The two atoms in a diatomic molecule can vibrate back and forth along the bond axis. This vibrational motion involves both kinetic and potential energy. It contributes 2 degrees of freedom (one for kinetic and one for potential energy).
• Degrees of Freedom: In total, a diatomic molecule has 3 translational + 2 rotational + 2 vibrational = 7 degrees of freedom.
• Heat Capacity: According to the equipartition theorem:
  • At low temperatures: Only translational degrees of freedom are significantly active. $C_V \approx \frac{3}{2}R$.
  • At moderate temperatures: Translational and rotational degrees of freedom are active. $C_V \approx \frac{3}{2}R + \frac{2}{2}R = \frac{5}{2}R$.
  • At high temperatures: Translational, rotational, and vibrational degrees of freedom are active. $C_V \approx \frac{3}{2}R + \frac{2}{2}R + \frac{2}{2}R = \frac{7}{2}R$.

Comparison:

As you can see, the molar heat capacity at constant volume ($C_V$) for a diatomic gas is higher than that of a monatomic gas because diatomic molecules can absorb energy into rotational and vibrational modes in addition to translational motion. This means that for the same increase in temperature, a diatomic gas needs more energy input than a monatomic gas.

In summary: Diatomic molecules have more ways to store energy (more degrees of freedom) than monatomic molecules. This leads to a higher heat capacity because more energy is required to increase the temperature by a certain amount. The additional energy is absorbed by the rotational and vibrational motion of the diatomic molecules.

Estimating the Molar Heat Capacity of Diamond:

Given that the Debye temperature ($\Theta_D$) of diamond is very high ($\approx 2230$ K) and the ambient temperature ($T$) is $25^\circ$C ($298$ K), we are in the regime where $T \ll \Theta_D$. In this low-temperature regime, the Debye model for the molar heat capacity at constant volume ($C_V$) is approximately given by:

$C_V \approx \frac{12 \pi^4}{5} R \left(\frac{T}{\Theta_D}\right)^3$

where $R$ is the ideal gas constant ($8.314 , \text{J/mol K}$).

Let’s plug in the values:

$C_V \approx \frac{12 \pi^4}{5} (8.314 , \text{J/mol K}) \left(\frac{298 , \text{K}}{2230 , \text{K}}\right)^3$

$C_V \approx 233.725 \times 8.314 \times (0.13363)^3$

$C_V \approx 233.725 \times 8.314 \times 0.002381$

$C_V \approx 1943.15 \times 0.002381$

$C_V \approx 4.62 , \text{J/mol K}$

Therefore, the estimated molar heat capacity of diamond at $25^\circ$C is approximately $4.62 , \text{J/mol K}$.

Why it is Different from the Dulong-Petit Law:

The Dulong-Petit law states that the molar heat capacity of a solid is approximately $3R$, which is:

$C_V \approx 3 \times 8.314 , \text{J/mol K} = 24.942 , \text{J/mol K}$

The value we calculated using the Debye model ($4.62 , \text{J/mol K}$) is significantly lower than the prediction by the Dulong-Petit law ($24.942 , \text{J/mol K}$). This difference arises from the underlying assumptions of the two models:

• Dulong-Petit Law: This law is a classical result based on the equipartition theorem. It assumes that each atom in the solid behaves as a harmonic oscillator with 6 degrees of freedom (3 kinetic and 3 potential, corresponding to vibrations in three dimensions). At sufficiently high temperatures, each degree of freedom contributes $\frac{1}{2}kT$ of energy per atom (or $\frac{1}{2}RT$ per mole). Therefore, the total molar internal energy is $3RT$, and the molar heat capacity at constant volume is its temperature derivative, $C_V = 3R$. This law works well at high temperatures where all vibrational modes are fully excited.

• Debye Model: This model is a quantum mechanical approach that considers the quantized nature of lattice vibrations (phonons). It introduces the Debye temperature ($\Theta_D$), which is a measure of the highest frequency of these vibrations. At temperatures much lower than the Debye temperature ($T \ll \Theta_D$), many of the high-frequency vibrational modes are “frozen out” because there is not enough thermal energy available to excite them. Only the low-frequency modes are significantly excited, contributing to the heat capacity. This leads to a heat capacity that is much lower than the classical prediction.

In summary: The Dulong-Petit law fails at low temperatures because it does not account for the quantization of vibrational energy. For diamond, with its very high Debye temperature, room temperature ($25^\circ$C) is effectively a low temperature. Therefore, only a small fraction of the vibrational modes are excited, resulting in a significantly lower heat capacity than predicted by the classical Dulong-Petit law.

Final Answer: The final answer is $\boxed{4.62 , \text{J/mol K}}$ and the explanation for the difference is due to the quantization of vibrational energy levels which is not considered in the Dulong-Petit law, causing it to overestimate the heat capacity at temperatures much lower than the Debye temperature.

The molar heat capacity of a solid according to the Einstein model is given by: $C_V = 3R \left(\frac{\theta_E}{T}\right)^2 \frac{e^{\theta_E/T}}{(e^{\theta_E/T} – 1)^2}$ where $\theta_E$ is the Einstein temperature and $T$ is the temperature of the metal. Given $\theta_E = 300$ K and $T = 30$ K.

Substitute the values into the formula: $C_V = 3R \left(\frac{300}{30}\right)^2 \frac{e^{300/30}}{(e^{300/30} – 1)^2}$ $C_V = 3R (10)^2 \frac{e^{10}}{(e^{10} – 1)^2}$ $C_V = 300R \frac{e^{10}}{(e^{10} – 1)^2}$

Since $e^{10}$ is a large number ($e^{10} \approx 22026$), we can approximate $e^{10} – 1 \approx e^{10}$. $C_V \approx 300R \frac{e^{10}}{(e^{10})^2}$ $C_V \approx 300R \frac{1}{e^{10}}$ $C_V \approx 300R e^{-10}$

Using $R \approx 8.314 , \text{J/mol K}$: $C_V \approx 300 \times 8.314 \times e^{-10}$ $C_V \approx 2494.2 \times e^{-10}$

Now we calculate $e^{-10}$: $e^{-10} \approx 4.53999 \times 10^{-5}$

$C_V \approx 2494.2 \times 4.53999 \times 10^{-5}$ $C_V \approx 0.11324 , \text{J/mol K}$

Rounding to two significant figures (as in the given temperatures), we get: $C_V \approx 0.11 , \text{J/mol K}$

Alternatively, we can use the low-temperature approximation for the Einstein model where $T \ll \theta_E$: $C_V \approx 3R \left(\frac{\theta_E}{T}\right)^2 e^{-\theta_E/T}$ $C_V \approx 3 \times 8.314 , \text{J/mol K} \times \left(\frac{300 , \text{K}}{30 , \text{K}}\right)^2 \times e^{-300 , \text{K}/30 , \text{K}}$ $C_V \approx 24.942 , \text{J/mol K} \times (10)^2 \times e^{-10}$ $C_V \approx 2494.2 , \text{J/mol K} \times e^{-10}$ $C_V \approx 2494.2 , \text{J/mol K} \times 4.53999 \times 10^{-5}$ $C_V \approx 0.11324 , \text{J/mol K}$ $C_V \approx 0.11 , \text{J/mol K}$

Final Answer: The final answer is $\boxed{0.11 , \text{J/mol K}}$

The Planck radiation formula describes the spectral radiance of a blackbody as a function of wavelength ($\lambda$) and temperature ($T$):

$B(\lambda, T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_B T} – 1}$

where:

• $B(\lambda, T)$ is the spectral radiance (power emitted per unit solid angle per unit projected area per unit wavelength).
• $h$ is Planck’s constant.
• $c$ is the speed of light.
• $k_B$ is the Boltzmann constant.

To find the wavelength at which the spectral radiance is maximum, we need to take the derivative of $B(\lambda, T)$ with respect to $\lambda$ and set it equal to zero:

$\frac{\partial B(\lambda, T)}{\partial \lambda} = 0$

Let $x = \frac{hc}{\lambda k_B T}$. Then the Planck formula can be written as:

$B(\lambda, T) = 2hc^2 \lambda^{-5} (e^x – 1)^{-1}$

Now, we need to find the derivative with respect to $\lambda$. Using the chain rule:

$\frac{\partial B}{\partial \lambda} = 2hc^2 \left[ -5\lambda^{-6} (e^x – 1)^{-1} + \lambda^{-5} (-1) (e^x – 1)^{-2} \frac{\partial (e^x – 1)}{\partial \lambda} \right]$

We need to find $\frac{\partial x}{\partial \lambda}$:

$\frac{\partial x}{\partial \lambda} = \frac{\partial}{\partial \lambda} \left(\frac{hc}{\lambda k_B T}\right) = -\frac{hc}{\lambda^2 k_B T}$

And $\frac{\partial (e^x – 1)}{\partial \lambda} = e^x \frac{\partial x}{\partial \lambda} = e^x \left(-\frac{hc}{\lambda^2 k_B T}\right)$

Substitute this back into the derivative of $B$:

$\frac{\partial B}{\partial \lambda} = 2hc^2 \left[ -\frac{5}{\lambda^6 (e^x – 1)} + \frac{1}{\lambda^5 (e^x – 1)^2} e^x \frac{hc}{\lambda^2 k_B T} \right]$

Set the derivative to zero:

$-\frac{5}{\lambda^6 (e^x – 1)} + \frac{hc e^x}{\lambda^7 k_B T (e^x – 1)^2} = 0$

Multiply by $\lambda^7 (e^x – 1)^2$:

$-5\lambda (e^x – 1) + hc e^x / k_B T = 0$

Substitute back $x = \frac{hc}{\lambda k_B T}$:

$-5\lambda \left(e^{hc/\lambda k_B T} – 1\right) + hc e^{hc/\lambda k_B T} / k_B T = 0$

Divide by $\lambda$:

$-5 \left(e^{hc/\lambda k_B T} – 1\right) + \frac{hc}{\lambda k_B T} e^{hc/\lambda k_B T} = 0$

Let $y = \frac{hc}{\lambda k_B T}$. The equation becomes:

$-5(e^y – 1) + y e^y = 0$

$y e^y – 5e^y + 5 = 0$

This is a transcendental equation. We need to find the value of $y$ that satisfies this equation. Numerically, the solution for $y$ is approximately:

$y \approx 4.9651$

Now, substitute back the expression for $y$:

$\frac{hc}{\lambda_{max} k_B T} = 4.9651$

Rearrange to find the relationship between $\lambda_{max}$ and $T$:

$\lambda_{max} T = \frac{hc}{4.9651 k_B}$

The right-hand side is a constant. This constant is known as Wien’s displacement constant, $b$:

$b = \frac{hc}{4.9651 k_B}$

Therefore, we arrive at Wien’s displacement law:

$\boxed{\lambda_{max} T = b}$

where $b \approx 2.898 \times 10^{-3} , \text{m K}$.

Let $N_1$ and $N_2$ be the number of photons emitted at frequencies $\nu_1$ and $\nu_2$, respectively. The energy of a single photon with frequency $\nu$ is $E = h\nu$.

The spectral energy density of a blackbody at temperature $T$ is given by Planck’s law: $u(\nu, T) = \frac{8\pi h \nu^3}{c^3} \frac{1}{e^{h\nu/k_B T} – 1}$

The number of photons per unit volume per unit frequency interval is given by the energy density divided by the energy of a single photon: $n(\nu, T) = \frac{u(\nu, T)}{h\nu} = \frac{8\pi \nu^2}{c^3} \frac{1}{e^{h\nu/k_B T} – 1}$

The number of photons emitted by a blackbody at a given frequency is proportional to the number density of photons at that frequency. Therefore, the ratio of the number of photons emitted at two different frequencies is the ratio of their number densities:

$\frac{N_1}{N_2} = \frac{n(\nu_1, T)}{n(\nu_2, T)}$

Substituting the expression for $n(\nu, T)$:

$\frac{N_1}{N_2} = \frac{\frac{8\pi \nu_1^2}{c^3} \frac{1}{e^{h\nu_1/k_B T} – 1}}{\frac{8\pi \nu_2^2}{c^3} \frac{1}{e^{h\nu_2/k_B T} – 1}}$

Simplifying the expression, we get:

$\frac{N_1}{N_2} = \frac{\nu_1^2 / (e^{h\nu_1/k_B T} – 1)}{\nu_2^2 / (e^{h\nu_2/k_B T} – 1)}$

$\boxed{\frac{N_1}{N_2} = \frac{\nu_1^2}{\nu_2^2} \frac{e^{h\nu_2/k_B T} – 1}{e^{h\nu_1/k_B T} – 1}}$

To calculate the equilibrium temperature, we assume that the system is isolated, and thus the heat lost by the iron is equal to the heat gained by the water.

The heat lost or gained by a substance is given by the formula: $Q = mc\Delta T$ where: $Q$ is the heat transferred $m$ is the mass of the substance $c$ is the specific heat capacity of the substance $\Delta T$ is the change in temperature ($T_{final} – T_{initial}$)

Let $T_f$ be the final equilibrium temperature.

For iron: Mass of iron, $m_{Fe} = 100 , \text{g} = 0.1 , \text{kg}$ Initial temperature of iron, $T_{i,Fe} = 500 , \text{K}$ Final temperature of iron, $T_{f,Fe} = T_f$ Specific heat capacity of iron, $c_{Fe} \approx 450 , \text{J/kg K}$ (This value can vary slightly, but we’ll use this approximation)

Heat lost by iron, $Q_{lost} = m_{Fe} c_{Fe} (T_{i,Fe} – T_f) = 0.1 , \text{kg} \times 450 , \text{J/kg K} \times (500 , \text{K} – T_f)$

For water: Mass of water, $m_w = 100 , \text{g} = 0.1 , \text{kg}$ Initial temperature of water, $T_{i,w} = 300 , \text{K}$ Final temperature of water, $T_{f,w} = T_f$ Specific heat capacity of water, $c_w \approx 4186 , \text{J/kg K}$

Heat gained by water, $Q_{gained} = m_w c_w (T_f – T_{i,w}) = 0.1 , \text{kg} \times 4186 , \text{J/kg K} \times (T_f – 300 , \text{K})$

According to the principle of calorimetry, the heat lost by the iron is equal to the heat gained by the water: $Q_{lost} = Q_{gained}$ $0.1 \times 450 \times (500 – T_f) = 0.1 \times 4186 \times (T_f – 300)$ $45 \times (500 – T_f) = 418.6 \times (T_f – 300)$ $22500 – 45 T_f = 418.6 T_f – 125580$ $22500 + 125580 = 418.6 T_f + 45 T_f$ $148080 = 463.6 T_f$ $T_f = \frac{148080}{463.6}$ $T_f \approx 319.41 , \text{K}$

The equilibrium temperature is approximately 319.41 K.

Final Answer: The final answer is $\boxed{319.41 , \text{K}}$

To estimate the vibrational contribution to the heat capacity of a solid, we can use the Einstein model. The Einstein model treats the vibrational modes of a solid as a collection of independent quantum harmonic oscillators, all with the same frequency.

The molar heat capacity at constant volume due to vibrations according to the Einstein model is given by: $C_V = 3R \left(\frac{\theta_E}{T}\right)^2 \frac{e^{\theta_E/T}}{(e^{\theta_E/T} – 1)^2}$ where: $R$ is the ideal gas constant ($8.314 , \text{J/mol K}$) $T$ is the temperature in Kelvin (room temperature is approximately 298 K) $\theta_E$ is the Einstein temperature, given by $\theta_E = \frac{h\nu}{k_B}$ $h$ is Planck’s constant ($6.626 \times 10^{-34} , \text{J s}$) $\nu$ is the vibrational frequency ($10^{13} , \text{Hz}$) $k_B$ is the Boltzmann constant ($1.381 \times 10^{-23} , \text{J/K}$)

First, calculate the Einstein temperature: $\theta_E = \frac{(6.626 \times 10^{-34} , \text{J s})(10^{13} , \text{s}^{-1})}{1.381 \times 10^{-23} , \text{J/K}}$ $\theta_E = \frac{6.626 \times 10^{-21}}{1.381 \times 10^{-23}} , \text{K}$ $\theta_E \approx 479.8 , \text{K}$

Now, we can calculate the heat capacity at room temperature ($T = 298 , \text{K}$): $\frac{\theta_E}{T} = \frac{479.8 , \text{K}}{298 , \text{K}} \approx 1.610$

Substitute this into the heat capacity formula: $C_V = 3(8.314 , \text{J/mol K}) \left(1.610\right)^2 \frac{e^{1.610}}{(e^{1.610} – 1)^2}$

Calculate the exponential terms: $e^{1.610} \approx 5.005$

Now substitute back: $C_V = 24.942 , \text{J/mol K} \times (2.592) \frac{5.005}{(5.005 – 1)^2}$ $C_V = 24.942 , \text{J/mol K} \times 2.592 \times \frac{5.005}{(4.005)^2}$ $C_V = 64.64 , \text{J/mol K} \times \frac{5.005}{16.040}$ $C_V = 64.64 , \text{J/mol K} \times 0.3120$ $C_V \approx 20.17 , \text{J/mol K}$

The vibrational contribution to the heat capacity of the solid at room temperature is approximately $20.17 , \text{J/mol K}$.

Final Answer: The final answer is $\boxed{20.17 , \text{J/mol K}}$

The power emitted by a blackbody is given by the Stefan-Boltzmann law: $P = \sigma A T^4$ where: $P$ is the power emitted (in Watts) $\sigma$ is the Stefan-Boltzmann constant ($5.67 \times 10^{-8} , \text{W m}^{-2} \text{K}^{-4}$) $A$ is the surface area of the blackbody (in m²) $T$ is the temperature of the blackbody (in Kelvin)

Given: $P = 1000 , \text{W}$ $A = 0.1 , \text{m}^2$

We need to find the temperature $T$. Rearrange the Stefan-Boltzmann law to solve for $T$: $T^4 = \frac{P}{\sigma A}$ $T = \sqrt[4]{\frac{P}{\sigma A}}$

Substitute the given values: $T = \sqrt[4]{\frac{1000 , \text{W}}{(5.67 \times 10^{-8} , \text{W m}^{-2} \text{K}^{-4})(0.1 , \text{m}^2)}}$ $T = \sqrt[4]{\frac{1000}{5.67 \times 10^{-9}} , \text{K}^4}$ $T = \sqrt[4]{1.76366843 \times 10^{11}} , \text{K}$

To calculate the fourth root, we can take the square root twice: $\sqrt{1.76366843 \times 10^{11}} = \sqrt{176.366843 \times 10^{9}} = 13.2803 \times 10^{4.5} = 13.2803 \times 10^4 \sqrt{10} \approx 13.2803 \times 10^4 \times 3.162 \approx 4.1996 \times 10^5$ $\sqrt{4.1996 \times 10^5} = \sqrt{41.996 \times 10^4} = 6.4804 \times 10^2 = 648.04$

Alternatively, using direct calculation: $T = (1.76366843 \times 10^{11})^{1/4}$ $T \approx 648 , \text{K}$

Final Answer: The final answer is $\boxed{648 , \text{K}}$

The energy density of a blackbody at a given wavelength $\lambda$ and temperature $T$ is given by Planck’s formula: $u(\lambda, T) = \frac{8\pi hc}{\lambda^5} \frac{1}{e^{hc/\lambda k_B T} – 1}$

We are asked to find the ratio of the energy density at a wavelength $\lambda/2$ when the blackbody is at temperature $T_1$, to the energy density at a wavelength $\lambda$ when the blackbody is at temperature $T_2 = 2T_1$.

Let $u_1$ be the energy density at wavelength $\lambda/2$ and temperature $T_1$: $u_1 = u(\lambda/2, T_1) = \frac{8\pi hc}{(\lambda/2)^5} \frac{1}{e^{hc/(\lambda/2) k_B T_1} – 1} = \frac{8\pi hc}{\lambda^5/32} \frac{1}{e^{2hc/\lambda k_B T_1} – 1} = \frac{32 \times 8\pi hc}{\lambda^5} \frac{1}{e^{2hc/\lambda k_B T_1} – 1}$

Let $u_2$ be the energy density at wavelength $\lambda$ and temperature $T_2$: $u_2 = u(\lambda, T_2) = \frac{8\pi hc}{\lambda^5} \frac{1}{e^{hc/\lambda k_B T_2} – 1}$ Given that $T_2 = 2T_1$: $u_2 = \frac{8\pi hc}{\lambda^5} \frac{1}{e^{hc/\lambda k_B (2T_1)} – 1} = \frac{8\pi hc}{\lambda^5} \frac{1}{e^{hc/2\lambda k_B T_1} – 1}$

Now we find the ratio $\frac{u_1}{u_2}$: $\frac{u_1}{u_2} = \frac{\frac{32 \times 8\pi hc}{\lambda^5} \frac{1}{e^{2hc/\lambda k_B T_1} – 1}}{\frac{8\pi hc}{\lambda^5} \frac{1}{e^{hc/2\lambda k_B T_1} – 1}} = 32 \times \frac{e^{hc/2\lambda k_B T_1} – 1}{e^{2hc/\lambda k_B T_1} – 1}$

Let $x = \frac{hc}{2\lambda k_B T_1}$. Then the ratio becomes: $\frac{u_1}{u_2} = 32 \times \frac{e^x – 1}{e^{2x} – 1}$ Using the identity $e^{2x} – 1 = (e^x – 1)(e^x + 1)$: $\frac{u_1}{u_2} = 32 \times \frac{e^x – 1}{(e^x – 1)(e^x + 1)} = \frac{32}{e^x + 1}$

Substitute back the value of $x$: $\frac{u_1}{u_2} = \frac{32}{e^{hc/2\lambda k_B T_1} + 1}$

Final Answer: The final answer is $\boxed{\frac{32}{e^{hc/2\lambda k_B T_1} + 1}}$

The average energy per mode for the electromagnetic radiation in a blackbody cavity at a given temperature $T$ and wavelength $\lambda$ is given by: $E_{avg}(\lambda, T) = \frac{hc}{\lambda} \frac{1}{e^{hc/\lambda k_B T} – 1}$ where: $h$ is Planck’s constant ($6.626 \times 10^{-34} , \text{J s}$) $c$ is the speed of light ($3 \times 10^8 , \text{m/s}$) $\lambda$ is the wavelength ($10 \times 10^{-6} , \text{m}$) $k_B$ is the Boltzmann constant ($1.381 \times 10^{-23} , \text{J/K}$) $T$ is the temperature ($500 , \text{K}$)

First, calculate the value of the exponent term $\frac{hc}{\lambda k_B T}$: $\frac{hc}{\lambda k_B T} = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{(10 \times 10^{-6} , \text{m})(1.381 \times 10^{-23} , \text{J/K})(500 , \text{K})}$ $\frac{hc}{\lambda k_B T} = \frac{19.878 \times 10^{-26} , \text{J m}}{6.905 \times 10^{-26} , \text{J m}}$ $\frac{hc}{\lambda k_B T} \approx 2.8788$

Now calculate the term $e^{hc/\lambda k_B T}$: $e^{2.8788} \approx 17.78$

Next, calculate the term $\frac{hc}{\lambda}$: $\frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{10 \times 10^{-6} , \text{m}}$ $\frac{hc}{\lambda} = \frac{19.878 \times 10^{-26} , \text{J m}}{10^{-5} , \text{m}}$ $\frac{hc}{\lambda} = 19.878 \times 10^{-21} , \text{J}$

Finally, calculate the average energy per mode: $E_{avg} = (19.878 \times 10^{-21} , \text{J}) \times \frac{1}{17.78 – 1}$ $E_{avg} = (19.878 \times 10^{-21} , \text{J}) \times \frac{1}{16.78}$ $E_{avg} \approx 1.1846 \times 10^{-21} , \text{J}$

Final Answer: The final answer is $\boxed{1.18 \times 10^{-21} , \text{J}}$

To calculate the wavelength at which the intensity of radiation is maximized for a blackbody at a given temperature, we use Wien’s displacement law. Wien’s displacement law states that the wavelength of maximum emission ($\lambda_{max}$) of a blackbody is inversely proportional to its temperature ($T$):

$\lambda_{max} T = b$

where $b$ is Wien’s displacement constant, approximately $2.898 \times 10^{-3} , \text{m K}$.

First, we need to convert the temperature of the human body from Celsius to Kelvin: $T(K) = T(^\circ C) + 273.15$ $T = 37^\circ C + 273.15 = 310.15 , \text{K}$

Now, we can use Wien’s displacement law to find the wavelength at which the intensity is maximized: $\lambda_{max} = \frac{b}{T}$ $\lambda_{max} = \frac{2.898 \times 10^{-3} , \text{m K}}{310.15 , \text{K}}$

Calculate the value of $\lambda_{max}$: $\lambda_{max} = 9.3439 \times 10^{-6} , \text{m}$

We can express this wavelength in micrometers ($\mu m$): $\lambda_{max} = 9.3439 , \mu m$

Rounding to a reasonable number of significant figures (since the temperature was given as $37^\circ C$, which has two significant figures), we can round the wavelength to two significant figures as well.

$\lambda_{max} \approx 9.3 \times 10^{-6} , \text{m}$ or $9.3 , \mu m$.

Final Answer: The final answer is $\boxed{9.34 \times 10^{-6} , \text{m}}$

The Einstein model simplifies the complex vibrational behavior of atoms in a solid by making a key assumption: all atoms vibrate independently at the same single frequency, often denoted as $\nu_E$ (related to the Einstein temperature $\theta_E$). This is akin to imagining all the atoms are connected by identical springs and oscillate with the same frequency.

How this differs from reality:

In reality, the atoms in a solid are not independent but are coupled to their neighbors through interatomic forces. This coupling leads to collective vibrational modes, which are quantized as phonons. These phonons have a distribution of frequencies, ranging from very low frequencies (long wavelengths, like sound waves traversing the material) to a maximum frequency.

Here’s a breakdown of the discrepancies:

• Single Frequency vs. a Spectrum of Frequencies: The most significant difference is that the Einstein model predicts a single vibrational frequency, while in a real solid, there exists a continuous spectrum of vibrational frequencies.
• Ignoring Interactions: The Einstein model treats each atom as an independent oscillator, neglecting the interactions and coupling between neighboring atoms. These interactions are crucial in determining the actual vibrational modes and their frequencies.
• Low-Temperature Behavior: The Einstein model predicts the heat capacity to decrease exponentially as the temperature approaches zero ($C_V \propto e^{-\theta_E/T}$). While this shows a decrease, the actual behavior of real solids at very low temperatures follows a $T^3$ dependence, which the Einstein model fails to capture accurately. This discrepancy arises because the low-frequency modes are the ones excited at low temperatures, and the Einstein model lacks these low-frequency modes.

How the Debye model attempts to improve on this:

The Debye model provides a significant improvement over the Einstein model by introducing a distribution of vibrational frequencies instead of a single frequency. The Debye model makes the following key assumptions:

• Continuum Approximation: For long wavelengths (low frequencies), the solid can be treated as a continuous elastic medium.
• Maximum Frequency (Debye Frequency): The model assumes that the distribution of vibrational frequencies extends from zero up to a maximum frequency, known as the Debye frequency ($\nu_D$), or equivalently, the Debye temperature ($\theta_D = h\nu_D/k_B$). This maximum frequency is related to the speed of sound in the solid and the density of atoms.
• Density of States: The Debye model introduces a density of states function, $g(\nu)$, which describes the number of vibrational modes per unit frequency interval. For low frequencies, this function is proportional to $\nu^2$.

Key improvements of the Debye model:

• Frequency Distribution: By considering a distribution of frequencies, the Debye model more accurately reflects the collective nature of atomic vibrations in a solid.
• Accurate Low-Temperature Behavior: The Debye model correctly predicts the $T^3$ dependence of the heat capacity at low temperatures. This is because the density of states function includes low-frequency modes, which are the dominant contributors to the heat capacity at low temperatures.
• Better Agreement with Experiment: The Debye model provides a much better fit to the experimental heat capacity data for many solids over a wider range of temperatures compared to the Einstein model.

In summary: The Einstein model’s assumption of a single vibrational frequency is a significant oversimplification. The Debye model improves upon this by introducing a distribution of frequencies up to a maximum frequency, better reflecting the collective nature of atomic vibrations and leading to a more accurate description of the heat capacity of solids, especially at low temperatures. While the Debye model is still an approximation (it simplifies the actual phonon dispersion relations), it represents a crucial step forward from the Einstein model.

To calculate the equilibrium temperature, we assume that the system is isolated and that the heat lost by the aluminum block is equal to the heat gained by the water.

The heat lost or gained by a substance is given by the formula: $Q = mc\Delta T$ where: $Q$ is the heat transferred $m$ is the mass of the substance $c$ is the specific heat capacity of the substance $\Delta T$ is the change in temperature ($T_{final} – T_{initial}$)

Let $T_f$ be the final equilibrium temperature.

For the aluminum block: Mass of aluminum, $m_{Al} = 50 , \text{g} = 0.05 , \text{kg}$ Initial temperature of aluminum, $T_{i,Al} = 80^\circ \text{C}$ Final temperature of aluminum, $T_{f,Al} = T_f$ Specific heat capacity of aluminum, $c_{Al} \approx 900 , \text{J/kg}^\circ \text{C}$

Heat lost by the aluminum block: $Q_{lost} = m_{Al} c_{Al} (T_{i,Al} – T_f) = 0.05 , \text{kg} \times 900 , \text{J/kg}^\circ \text{C} \times (80^\circ \text{C} – T_f)$

For the water: Mass of water, $m_w = 100 , \text{g} = 0.1 , \text{kg}$ Initial temperature of water, $T_{i,w} = 20^\circ \text{C}$ Final temperature of water, $T_{f,w} = T_f$ Specific heat capacity of water, $c_w \approx 4186 , \text{J/kg}^\circ \text{C}$

Heat gained by the water: $Q_{gained} = m_w c_w (T_f – T_{i,w}) = 0.1 , \text{kg} \times 4186 , \text{J/kg}^\circ \text{C} \times (T_f – 20^\circ \text{C})$

According to the principle of conservation of energy, the heat lost by the aluminum is equal to the heat gained by the water: $Q_{lost} = Q_{gained}$ $0.05 \times 900 \times (80 – T_f) = 0.1 \times 4186 \times (T_f – 20)$ $45 \times (80 – T_f) = 418.6 \times (T_f – 20)$ $3600 – 45 T_f = 418.6 T_f – 8372$ $3600 + 8372 = 418.6 T_f + 45 T_f$ $11972 = 463.6 T_f$ $T_f = \frac{11972}{463.6}$ $T_f \approx 25.82 , ^\circ\text{C}$

The equilibrium temperature is approximately $25.82^\circ \text{C}$.

Final Answer: The final answer is $\boxed{25.82 , ^\circ\text{C}}$

The amount of heat required to change the temperature of a substance can be calculated using the formula: $Q = mc\Delta T$ where: $Q$ is the heat energy (in Joules) $m$ is the mass of the substance (in kg) $c$ is the specific heat capacity of the substance (in J/kg.K) $\Delta T$ is the change in temperature (in K or °C)

Given the following values: Mass of the metal block, $m = 1 , \text{kg}$ Specific heat of the metal, $c = 500 , \text{J/kg.K}$ Initial temperature, $T_i = 20^\circ\text{C}$ Final temperature, $T_f = 100^\circ\text{C}$

First, calculate the change in temperature: $\Delta T = T_f – T_i = 100^\circ\text{C} – 20^\circ\text{C} = 80^\circ\text{C}$ Since a change in temperature in Celsius is equivalent to a change in temperature in Kelvin, $\Delta T = 80 , \text{K}$.

Now, substitute the values into the heat formula: $Q = (1 , \text{kg}) \times (500 , \text{J/kg.K}) \times (80 , \text{K})$ $Q = 500 \times 80 , \text{J}$ $Q = 40000 , \text{J}$

Therefore, the amount of heat required to heat the metal block from 20°C to 100°C is 40000 Joules.

Final Answer: The final answer is $\boxed{40000 , \text{J}}$

To find out how long it will take to raise the temperature of the copper block, we need to calculate the total heat energy required and then use the power of the heater to determine the time.

1. Calculate the heat energy required:

The heat energy ($Q$) required to raise the temperature of a substance is given by the formula: $Q = mc\Delta T$ where:

• $m$ is the mass of the substance (in kg)
• $c$ is the specific heat capacity of the substance (in J/kg°C)
• $\Delta T$ is the change in temperature (°C)

Given:

• Mass of the copper block, $m = 5 , \text{kg}$
• Initial temperature, $T_i = 20 , ^\circ\text{C}$
• Final temperature, $T_f = 100 , ^\circ\text{C}$
• Specific heat capacity of copper, $c \approx 385 , \text{J/kg}^\circ\text{C}$ (This is a common value for copper, you might need to look it up for more precise calculations)

First, calculate the change in temperature: $\Delta T = T_f – T_i = 100 , ^\circ\text{C} – 20 , ^\circ\text{C} = 80 , ^\circ\text{C}$

Now, calculate the heat energy required: $Q = (5 , \text{kg}) \times (385 , \text{J/kg}^\circ\text{C}) \times (80 , ^\circ\text{C})$ $Q = 1925 \times 80 , \text{J}$ $Q = 154000 , \text{J}$

2. Calculate the time required:

The power of the heater ($P$) is the rate at which it delivers energy, and it’s given by: $P = \frac{Q}{t}$ where:

• $P$ is the power (in Watts)
• $Q$ is the energy (in Joules)
• $t$ is the time (in seconds)

Given:

• Power of the heater, $P = 1 , \text{kW} = 1000 , \text{W}$
• Heat energy required, $Q = 154000 , \text{J}$

Rearrange the formula to solve for time: $t = \frac{Q}{P}$ $t = \frac{154000 , \text{J}}{1000 , \text{W}}$ $t = 154 , \text{s}$

To convert seconds to minutes: $t = \frac{154}{60} , \text{minutes} \approx 2.57 , \text{minutes}$

Final Answer: The final answer is $\boxed{154 , \text{s}}$

Here’s how to calculate the temperature rise of the lead object:

1. Calculate the potential energy of the object before it’s dropped:

• Potential Energy (PE) = $mgh$

• where:

  • $m$ = mass = 10 g = 0.01 kg (convert grams to kilograms)
  • $g$ = acceleration due to gravity ≈ 9.8 m/s²
  • $h$ = height = 10 m

• PE = $(0.01 , \text{kg})(9.8 , \text{m/s}^2)(10 , \text{m}) = 0.98 , \text{J}$

2. Assume all potential energy is converted to kinetic energy just before impact:

• Kinetic Energy (KE) = PE = 0.98 J

3. Assume all kinetic energy is converted to heat upon impact:

• Heat (Q) = KE = 0.98 J

4. Use the formula for heat to calculate the temperature rise:

• $Q = mc\Delta T$

• where:

  • $Q$ = heat energy = 0.98 J
  • $m$ = mass = 0.01 kg
  • $c$ = specific heat capacity of lead ≈ 128 J/kg°C (This value can vary slightly, but this is a common approximation)
  • $\Delta T$ = change in temperature (what we want to find)

• Rearrange the formula to solve for $\Delta T$:

  • $\Delta T = \frac{Q}{mc}$

• Substitute the values:

  • $\Delta T = \frac{0.98 , \text{J}}{(0.01 , \text{kg})(128 , \text{J/kg}^\circ\text{C})}$
  • $\Delta T = \frac{0.98}{1.28} , ^\circ\text{C}$
  • $\Delta T \approx 0.765625 , ^\circ\text{C}$

5. Round to a reasonable number of significant figures:

Since the given values have two significant figures (10 g and 10 m), we should round our answer accordingly.

• $\Delta T \approx 0.77 , ^\circ\text{C}$

Therefore, the temperature of the lead object will rise by approximately 0.77 degrees Celsius.

Final Answer: The final answer is $\boxed{0.77 , ^\circ\text{C}}$

The Planck distribution law, which accurately describes the spectral radiance of electromagnetic radiation emitted by a blackbody, has its physical basis in the revolutionary idea of energy quantization. Here’s a breakdown of the key physical concepts:

1. Quantization of Energy:

• Classical Physics Fails: Classical physics treated electromagnetic radiation as a continuous wave and predicted the “ultraviolet catastrophe”—an infinite amount of energy at high frequencies, which was clearly not observed experimentally.
• Planck’s Postulate: Max Planck’s groundbreaking contribution was to propose that the energy of electromagnetic radiation is not continuous but is emitted and absorbed in discrete packets called quanta or photons. The energy of a single photon is directly proportional to its frequency:
  • $E = h\nu$
  • where $E$ is the energy of the photon, $h$ is Planck’s constant, and $\nu$ is the frequency of the radiation.

2. Quantized Oscillators in the Cavity Walls:

• Blackbody as a Cavity: A blackbody can be conceptually modeled as a cavity with perfectly absorbing walls. The electromagnetic radiation within the cavity is in thermal equilibrium with the oscillating charges (atoms and electrons) in the cavity walls.
• Quantized Oscillations: Planck further proposed that the energy of these oscillators in the cavity walls is also quantized. An oscillator vibrating at a frequency $\nu$ can only possess energies that are integer multiples of $h\nu$:
  • $E_n = nh\nu$, where $n = 0, 1, 2, 3, …$ is an integer called the quantum number.
• Discrete Energy Levels: This means the oscillators can only exist in specific, discrete energy levels, like steps on a ladder, rather than a continuous ramp.

3. Statistical Mechanics and the Boltzmann Distribution:

• Thermal Equilibrium: The distribution of energy among the oscillators is governed by the principles of statistical mechanics, specifically the Boltzmann distribution.
• Probability of Energy Levels: The Boltzmann distribution states that the probability of an oscillator being in a particular energy state $E_n$ at a temperature $T$ is proportional to the Boltzmann factor:
  • $P(E_n) \propto e^{-E_n/k_BT} = e^{-nh\nu/k_BT}$
  • where $k_B$ is the Boltzmann constant.
• Decreasing Probability with Increasing Energy: This means that at a given temperature, it’s more probable for an oscillator to be in a lower energy state than a higher one. The probability decreases exponentially with increasing energy.

4. Calculating the Average Energy of an Oscillator:

• Weighted Average: The average energy of an oscillator at a given frequency is calculated by taking the weighted average of the energies of all possible states, where the weights are the probabilities of those states.
• Mathematical Derivation: This involves summing over all possible energy levels:
  • $\langle E \rangle = \frac{\sum_{n=0}^{\infty} E_n P(E_n)}{\sum_{n=0}^{\infty} P(E_n)} = \frac{\sum_{n=0}^{\infty} nh\nu e^{-nh\nu/k_BT}}{\sum_{n=0}^{\infty} e^{-nh\nu/k_BT}}$
• Resulting Average Energy: Performing this summation leads to the expression for the average energy of an oscillator at frequency $\nu$:
  • $\langle E \rangle = \frac{h\nu}{e^{h\nu/k_BT} – 1}$

5. Connecting Average Energy to Energy Density:

• Density of Modes: To find the energy density of the radiation within the cavity, we need to consider the number of electromagnetic modes (standing waves) per unit volume per unit frequency interval. This is given by the density of states:
  • $g(\nu) d\nu = \frac{8\pi \nu^2}{c^3} d\nu$
  • where $c$ is the speed of light.
• Planck Distribution Formula: The energy density $u(\nu, T) d\nu$ is then the product of the average energy per mode and the number of modes in that frequency interval:
  • $u(\nu, T) d\nu = \langle E \rangle g(\nu) d\nu = \frac{h\nu}{e^{h\nu/k_BT} – 1} \frac{8\pi \nu^2}{c^3} d\nu$
  • $u(\nu, T) = \frac{8\pi h\nu^3}{c^3} \frac{1}{e^{h\nu/k_BT} – 1}$

1. Calculate the total heat lost:

• The piece of iron loses 100 J of energy every second.
• Over 10 seconds, the total energy lost is: $Q = \text{rate of heat loss} \times \text{time}$ $Q = 100 , \text{J/s} \times 10 , \text{s} = 1000 , \text{J}$

2. Relate the heat lost to the temperature change using the heat capacity:

• The heat capacity ($C$) of an object is the amount of heat required to raise its temperature by 1 Kelvin (or 1 degree Celsius).
• The relationship between heat lost ($Q$), heat capacity ($C$), and temperature change ($\Delta T$) is: $Q = C \Delta T$
• Where $\Delta T = T_{initial} – T_{final}$ (since the iron is cooling, the initial temperature is higher).

3. Solve for the temperature change:

• Rearrange the formula to solve for $\Delta T$: $\Delta T = \frac{Q}{C}$
• Substitute the values: $\Delta T = \frac{1000 , \text{J}}{200 , \text{J/K}} = 5 , \text{K}$

4. Calculate the final temperature:

• We know the initial temperature and the change in temperature: $\Delta T = T_{initial} – T_{final}$
• Rearrange to solve for the final temperature ($T_{final}$): $T_{final} = T_{initial} – \Delta T$
• Substitute the values: $T_{final} = 400 , \text{K} – 5 , \text{K} = 395 , \text{K}$

Therefore, the temperature of the piece of iron after 10 seconds will be 395 K.

Final Answer: The final answer is $\boxed{395 , \text{K}}$

The Planck distribution for the spectral energy density of blackbody radiation is given by: $u(\nu, T) = \frac{8 \pi h \nu^3}{c^3} \frac{1}{e^{h\nu/k_B T} – 1}$

The power radiated per unit area per unit frequency interval is given by: $M_\nu(\nu, T) = \frac{c}{4} u(\nu, T) = \frac{2 \pi h \nu^3}{c^2} \frac{1}{e^{h\nu/k_B T} – 1}$

To find the total power radiated per unit area over all frequencies, we integrate $M_\nu(\nu, T)$ from $0$ to $\infty$: $P = \int_0^\infty M_\nu(\nu, T) , d\nu = \int_0^\infty \frac{2 \pi h \nu^3}{c^2} \frac{1}{e^{h\nu/k_B T} – 1} , d\nu$

Let $x = \frac{h\nu}{k_B T}$. Then $\nu = \frac{k_B T x}{h}$, and $d\nu = \frac{k_B T}{h} dx$. When $\nu = 0$, $x = 0$. When $\nu \to \infty$, $x \to \infty$.

Substitute these into the integral: $P = \int_0^\infty \frac{2 \pi h}{c^2} \left( \frac{k_B T x}{h} \right)^3 \frac{1}{e^x – 1} \left( \frac{k_B T}{h} dx \right)$ $P = \int_0^\infty \frac{2 \pi h}{c^2} \frac{(k_B T)^3 x^3}{h^3} \frac{1}{e^x – 1} \frac{k_B T}{h} dx$ $P = \frac{2 \pi h (k_B T)^4}{c^2 h^4} \int_0^\infty \frac{x^3}{e^x – 1} dx$ $P = \frac{2 \pi k_B^4 T^4}{c^2 h^3} \int_0^\infty \frac{x^3}{e^x – 1} dx$

The integral $\int_0^\infty \frac{x^3}{e^x – 1} dx$ is a known definite integral, and its value is $\frac{\pi^4}{15}$.

Substituting this value back into the equation for P: $P = \frac{2 \pi k_B^4 T^4}{c^2 h^3} \left( \frac{\pi^4}{15} \right)$ $P = \left( \frac{2 \pi^5 k_B^4}{15 c^2 h^3} \right) T^4$

Let $\sigma = \frac{2 \pi^5 k_B^4}{15 c^2 h^3}$. Then $P = \sigma T^4$.

Therefore, the total power radiated by a blackbody is proportional to $T^4$.

$\boxed{P = \sigma T^4}$

Classical physics predicted that the energy density of blackbody radiation would increase indefinitely with increasing frequency. Specifically, the Rayleigh-Jeans law, derived from classical equipartition of energy, stated that the energy density $u(\nu, T)$ is proportional to $\nu^2$:

$u(\nu, T) = \frac{8 \pi \nu^2}{c^3} k_B T$

This implies that as frequency $\nu$ increases (especially in the ultraviolet region), the energy density would become infinitely large. This was in stark contradiction with experimental observations which showed that the energy density actually peaked at a certain frequency and then decreased at higher frequencies. This discrepancy was termed the “ultraviolet catastrophe”.

The quantum nature of energy, as proposed by Planck, resolves this issue by stating that energy is not continuous but exists in discrete packets called quanta, with the energy of each quantum being proportional to its frequency: $E = h\nu$.

At high frequencies (in the ultraviolet region), the energy of each quantum ($h\nu$) becomes large. According to classical physics, each mode of oscillation would receive an equal amount of energy ($k_B T$). However, with quantized energy, to excite a high-frequency mode, you need to provide at least one quantum of energy $h\nu$.

If $h\nu \gg k_B T$, the probability of having enough energy ($h\nu$) available to excite these high-frequency modes becomes exponentially small, following the Boltzmann distribution. Essentially, there isn’t enough thermal energy available to populate these high-energy, high-frequency modes.

This means that while classical physics allowed for an unlimited number of high-frequency modes to be excited, quantum mechanics restricts this. The number of modes with significant energy falls off rapidly at high frequencies because the energy required to excite them is much larger than the available thermal energy.

Therefore, the energy density does not continue to increase indefinitely with frequency, but instead decreases at high frequencies, avoiding the ultraviolet catastrophe and aligning with experimental results. Planck’s distribution, which incorporates this quantization of energy, correctly predicts the observed blackbody spectrum.

Final Answer: The classical Rayleigh-Jeans law predicted that the energy density of blackbody radiation would increase indefinitely with frequency, leading to an “ultraviolet catastrophe”. However, Planck’s quantum theory states that energy is quantized in packets $E=h\nu$. At high frequencies, the energy of each quantum is large. The probability of exciting these high-energy modes is low because the thermal energy $k_B T$ is insufficient to provide the required energy $h\nu$. This limits the number of excited high-frequency modes, causing the energy density to decrease at high frequencies, thus resolving the ultraviolet catastrophe.

Solution: The Einstein model considers a solid of $N$ atoms, where each atom can oscillate independently in 3 dimensions. Each atom is treated as a 3D quantum harmonic oscillator. This is equivalent to $3N$ independent 1D quantum harmonic oscillators.

Let the energy of each oscillator be quantized in units of $\hbar\omega$. Let $q$ be the total number of energy quanta in the system. The number of ways to distribute these $q$ indistinguishable quanta among $3N$ distinguishable oscillators is given by the number of combinations with repetition, which can be calculated using the stars and bars method: $$ \Omega = \binom{q + 3N – 1}{3N – 1} = \binom{q + 3N – 1}{q} $$ where $\Omega$ is the number of microstates for a given total energy.

The entropy $S$ of the system is given by Boltzmann’s formula: $$ S = k_B \ln \Omega = k_B \ln \binom{q + 3N – 1}{3N – 1} $$ Using Stirling’s approximation for large numbers, $\ln n! \approx n \ln n – n$, we can approximate the logarithm of the binomial coefficient: $\begin{align*} \ln \binom{q + 3N – 1}{3N – 1} &= \ln \frac{(q + 3N – 1)!}{(3N – 1)! q!} \ &\approx (q + 3N – 1) \ln(q + 3N – 1) – (q + 3N – 1) \ &- [(3N – 1) \ln(3N – 1) – (3N – 1)] \ &- [q \ln q – q] \ &\approx (q + 3N) \ln(q + 3N) – 3N \ln(3N) – q \ln q \end{align*}$ where we have ignored the $-1$ terms as $q$ and $N$ are large.

So, the entropy is approximately: $$ S = k_B [(q + 3N) \ln(q + 3N) – 3N \ln(3N) – q \ln q] $$ Rearranging the terms: $\begin{align*} S &= k_B [q \ln(q + 3N) + 3N \ln(q + 3N) – 3N \ln(3N) – q \ln q] \ &= k_B \left[ q (\ln(q + 3N) – \ln q) + 3N (\ln(q + 3N) – \ln(3N)) \right] \ &= k_B \left[ q \ln\left(\frac{q + 3N}{q}\right) + 3N \ln\left(\frac{q + 3N}{3N}\right) \right] \ &= k_B \left[ q \ln\left(1 + \frac{3N}{q}\right) + 3N \ln\left(1 + \frac{q}{3N}\right) \right] \end{align*}$

Final Answer: The final answer is $\boxed{S = k_B \left[ q \ln\left(1 + \frac{3N}{q}\right) + 3N \ln\left(1 + \frac{q}{3N}\right) \right]}$

To calculate the power generated by the solar panel, we need to determine the power incident on the panel from the sun and then apply the panel’s efficiency.

First, we calculate the power radiated per unit area by the sun using the Stefan-Boltzmann law: $P/A = \sigma T^4$ where $\sigma$ is the Stefan-Boltzmann constant ($5.67 \times 10^{-8} , W m^{-2} K^{-4}$) and $T$ is the temperature of the sun (5800 K).

$P/A = (5.67 \times 10^{-8} , W m^{-2} K^{-4}) \times (5800 , K)^4$ $P/A = 5.67 \times 10^{-8} \times 1.1316 \times 10^{15} , W m^{-2}$ $P/A = 6.415972 \times 10^7 , W m^{-2}$

This is the power radiated per unit area at the surface of the sun. To find the power incident on the solar panel, we need to consider the solar irradiance at the Earth’s distance. However, since the problem states the sun is directly overhead and ignores atmospheric effects, we can use the concept of solar constant or solar irradiance, which is the power per unit area received from the sun at the Earth’s distance. The standard value for solar irradiance ($I$) is approximately $1361 , W/m^2$.

The total power incident on the solar panel is the solar irradiance multiplied by the area of the panel: $P_{incident} = I \times Area$ $P_{incident} = 1361 , W/m^2 \times 2 , m^2$ $P_{incident} = 2722 , W$

Now, we apply the efficiency of the solar panel to find the power it can generate: $P_{generated} = P_{incident} \times Efficiency$ $P_{generated} = 2722 , W \times 0.15$ $P_{generated} = 408.3 , W$

$\boxed{408.3 , W}$

The number of photons emitted per unit time by a blackbody source can be determined by calculating the power emitted in the specified wavelength range and then dividing by the energy of a single photon at that wavelength.

The Planck’s law for spectral radiance $B(\lambda, T)$ is given by: $B(\lambda, T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_B T} – 1}$

Given values: Area $A = 100 , cm^2 = 0.01 , m^2$ Temperature $T = 1500 , K$ Wavelength range $\Delta \lambda = 505 , nm – 500 , nm = 5 \times 10^{-9} , m$

We can approximate the spectral radiance within this small range using the value at the center wavelength $\lambda_c = \frac{500 + 505}{2} , nm = 502.5 \times 10^{-9} , m$.

Constants: $h = 6.626 \times 10^{-34} , Js$ $c = 3 \times 10^8 , m/s$ $k_B = 1.38 \times 10^{-23} , J/K$

First, calculate the term $\frac{hc}{\lambda_c k_B T}$: $\frac{hc}{\lambda_c k_B T} = \frac{(6.626 \times 10^{-34} , Js)(3 \times 10^8 , m/s)}{(502.5 \times 10^{-9} , m)(1.38 \times 10^{-23} , J/K)(1500 , K)} \approx 19.137$

Now calculate the spectral radiance at $\lambda_c$: $B(\lambda_c, T) = \frac{2(6.626 \times 10^{-34})(3 \times 10^8)^2}{(502.5 \times 10^{-9})^5} \frac{1}{e^{19.137} – 1}$ $B(\lambda_c, T) \approx 2.453 \times 10^6 , W m^{-3} sr^{-1}$

The power radiated per unit area per unit solid angle in this wavelength range is approximately $B(\lambda_c, T) \Delta \lambda$. $P’ = B(\lambda_c, T) \Delta \lambda = 2.453 \times 10^6 , W m^{-3} sr^{-1} \times 5 \times 10^{-9} , m = 0.012265 , W m^{-2} sr^{-1}$

The total power emitted by the area $A$ into a hemisphere ($2\pi$ steradians) in this wavelength range is: $P_{total} = A \times \int B(\lambda, T) d\Omega d\lambda \approx A \times 2\pi \times B(\lambda_c, T) \Delta \lambda$ $P_{total} = 0.01 , m^2 \times 2\pi \times 0.012265 , W m^{-2} sr^{-1} = 7.718 \times 10^{-4} , W$

The energy of a single photon at the center wavelength $\lambda_c$ is: $E_{photon} = \frac{hc}{\lambda_c} = \frac{(6.626 \times 10^{-34} , Js)(3 \times 10^8 , m/s)}{502.5 \times 10^{-9} , m} \approx 3.956 \times 10^{-19} , J$

The number of photons emitted per unit time is the total power divided by the energy per photon: Number of photons per second $= \frac{P_{total}}{E_{photon}} = \frac{7.718 \times 10^{-4} , W}{3.956 \times 10^{-19} , J/photon} \approx 1.951 \times 10^{15} , photons/s$

$\boxed{1.95 \times 10^{15}}$

Consider a cubic cavity of side length $L$ and volume $V = L^3$. The electromagnetic field inside the cavity must satisfy certain boundary conditions. For simplicity, we use periodic boundary conditions, which require the field to be the same at opposite walls of the cavity.

The wavevector $\vec{k}$ of the electromagnetic waves can be written as $\vec{k} = (k_x, k_y, k_z)$. The components of the wavevector must satisfy: $e^{ik_x L} = 1 \implies k_x L = 2 \pi n_x$ $e^{ik_y L} = 1 \implies k_y L = 2 \pi n_y$ $e^{ik_z L} = 1 \implies k_z L = 2 \pi n_z$ where $n_x$, $n_y$, and $n_z$ are integers (0, ±1, ±2, …).

The allowed values of the wavevector components are thus: $k_x = \frac{2 \pi n_x}{L}$ $k_y = \frac{2 \pi n_y}{L}$ $k_z = \frac{2 \pi n_z}{L}$

Each unique set of integers $(n_x, n_y, n_z)$ corresponds to a distinct mode of oscillation. In k-space, the separation between adjacent allowed k-vectors in each direction is $\Delta k_x = \frac{2 \pi}{L}$, $\Delta k_y = \frac{2 \pi}{L}$, and $\Delta k_z = \frac{2 \pi}{L}$. The volume occupied by a single mode in k-space is $(\Delta k_x)(\Delta k_y)(\Delta k_z) = \left(\frac{2 \pi}{L}\right)^3 = \frac{8 \pi^3}{L^3}$.

To find the number of modes with wavevectors between $k$ and $k+dk$, consider a spherical shell in k-space with radius $k$ and thickness $dk$. The volume of this shell is $4 \pi k^2 dk$. The number of modes $dN$ within this shell is the volume of the shell divided by the volume per mode: $dN = \frac{4 \pi k^2 dk}{8 \pi^3 / L^3} = \frac{L^3 k^2 dk}{2 \pi^2}$

Electromagnetic waves are transverse waves, which means for each wavevector $\vec{k}$, there are two independent polarization states. Therefore, we multiply the number of modes by 2: $dN = 2 \times \frac{L^3 k^2 dk}{2 \pi^2} = \frac{L^3 k^2 dk}{\pi^2}$

To find the number of modes per unit volume, we divide by the volume of the cavity $V = L^3$: $\frac{dN}{V} = \frac{k^2 dk}{\pi^2}$

Now, we want to express this in terms of frequency $\nu$. The relationship between the magnitude of the wavevector $k$ and the frequency $\nu$ is $k = \frac{\omega}{c} = \frac{2 \pi \nu}{c}$. Differentiating with respect to $\nu$, we get $dk = \frac{2 \pi}{c} d\nu$.

Substitute these into the expression for the number of modes per unit volume: $\frac{dN}{V} = \frac{\left(\frac{2 \pi \nu}{c}\right)^2 \left(\frac{2 \pi}{c} d\nu\right)}{\pi^2} = \frac{\frac{4 \pi^2 \nu^2}{c^2} \frac{2 \pi}{c} d\nu}{\pi^2} = \frac{8 \pi^3 \nu^2 d\nu}{c^3 \pi^2} = \frac{8 \pi \nu^2}{c^3} d\nu$

The number of modes of electromagnetic radiation per unit volume per unit frequency, denoted by $n(\nu)$, is: $n(\nu) = \frac{1}{V} \frac{dN}{d\nu} = \frac{8 \pi \nu^2}{c^3}$

Final Answer: The final answer is $\boxed{\frac{8 \pi \nu^2}{c^3}}$ Note: We have now derived an equation similar to Rayleigh-Jeans law.

To determine how many times greater the total power output of the star is compared to the sun, we can use Wien’s Law and the Stefan-Boltzmann Law.

First, we use Wien’s Law to find the relationship between the temperatures of the star and the sun: $\lambda_{max} T = b$ where $\lambda_{max}$ is the wavelength at which the emission is strongest, $T$ is the temperature, and $b$ is Wien’s displacement constant.

For the star: $\lambda_{max, star} T_{star} = b$ For the sun: $\lambda_{max, sun} T_{sun} = b$

Therefore, $\lambda_{max, star} T_{star} = \lambda_{max, sun} T_{sun}$. We are given $\lambda_{max, star} = 400 , nm$ and $\lambda_{max, sun} = 500 , nm$. $\frac{T_{star}}{T_{sun}} = \frac{\lambda_{max, sun}}{\lambda_{max, star}} = \frac{500 , nm}{400 , nm} = \frac{5}{4}$

Next, we use the Stefan-Boltzmann Law, which states that the total power output of a blackbody is proportional to the fourth power of its temperature and its surface area: $P = \sigma A T^4$ where $P$ is the power output, $\sigma$ is the Stefan-Boltzmann constant, and $A$ is the surface area. For a sphere, the surface area is $A = 4 \pi R^2$.

For the star: $P_{star} = \sigma (4 \pi R_{star}^2) T_{star}^4$ For the sun: $P_{sun} = \sigma (4 \pi R_{sun}^2) T_{sun}^4$

We are given that the radius of the star is 5 times that of the sun, so $R_{star} = 5 R_{sun}$. Now we can find the ratio of the power outputs: $\frac{P_{star}}{P_{sun}} = \frac{\sigma (4 \pi R_{star}^2) T_{star}^4}{\sigma (4 \pi R_{sun}^2) T_{sun}^4} = \left(\frac{R_{star}}{R_{sun}}\right)^2 \left(\frac{T_{star}}{T_{sun}}\right)^4$

Substitute the ratios we found: $\frac{P_{star}}{P_{sun}} = \left(\frac{5 R_{sun}}{R_{sun}}\right)^2 \left(\frac{5}{4}\right)^4 = (5)^2 \left(\frac{5^4}{4^4}\right) = 25 \times \frac{625}{256}$

Calculate the final value: $\frac{P_{star}}{P_{sun}} = \frac{25 \times 625}{256} = \frac{15625}{256} \approx 61.04$

The total power output of the star is approximately 61.04 times greater than that of the sun.

Final Answer: The final answer is $\boxed{61.04}$

Let the initial temperature of the blackbody sphere be $T_1$. The temperature is increased by 10%, so the final temperature $T_2 = T_1 + 0.10 T_1 = 1.10 T_1$.

The total power emitted by a blackbody sphere is given by the Stefan-Boltzmann Law: $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant and $A$ is the surface area of the sphere.

The initial power emitted is $P_1 = \sigma A T_1^4$. The final power emitted is $P_2 = \sigma A T_2^4 = \sigma A (1.10 T_1)^4 = \sigma A (1.10)^4 T_1^4 = 1.4641 \sigma A T_1^4$.

The change in the total power emitted is $\Delta P = P_2 – P_1 = 1.4641 \sigma A T_1^4 – \sigma A T_1^4 = 0.4641 \sigma A T_1^4$. The percentage change in the total power emitted is $\frac{\Delta P}{P_1} \times 100% = \frac{0.4641 \sigma A T_1^4}{\sigma A T_1^4} \times 100% = 0.4641 \times 100% = 46.41%$.

The wavelength at which most of the power is emitted is given by Wien’s Displacement Law: $\lambda_{max} T = b$, where $b$ is Wien’s displacement constant.

Let the initial wavelength of maximum emission be $\lambda_{max,1}$ and the final wavelength of maximum emission be $\lambda_{max,2}$. $\lambda_{max,1} T_1 = b$ $\lambda_{max,2} T_2 = b$

Therefore, $\lambda_{max,1} T_1 = \lambda_{max,2} T_2$. Substituting $T_2 = 1.10 T_1$: $\lambda_{max,1} T_1 = \lambda_{max,2} (1.10 T_1)$ $\lambda_{max,1} = 1.10 \lambda_{max,2}$ $\lambda_{max,2} = \frac{\lambda_{max,1}}{1.10}$

The change in the wavelength at which most of the power is emitted is $\Delta \lambda_{max} = \lambda_{max,2} – \lambda_{max,1} = \frac{\lambda_{max,1}}{1.10} – \lambda_{max,1} = \lambda_{max,1} \left( \frac{1}{1.10} – 1 \right) = \lambda_{max,1} \left( \frac{1 – 1.10}{1.10} \right) = -\frac{0.10}{1.10} \lambda_{max,1} = -\frac{1}{11} \lambda_{max,1}$. The change is a decrease in wavelength by a factor of $\frac{1}{11}$ of the original wavelength. The percentage change in wavelength is $\frac{\Delta \lambda_{max}}{\lambda_{max,1}} \times 100% = -\frac{1}{11} \times 100% \approx -9.09%$.

Final Answer: The final answer is $\boxed{46.41% \text{ increase in power, and a decrease in wavelength by } \frac{1}{11} \text{ of the original wavelength.}}$

The heat capacity at constant volume according to the Debye model is given by: $C_V = 9 N k_B \left( \frac{T}{\Theta_D} \right)^3 \int_0^{\Theta_D/T} \frac{x^4 e^x}{(e^x – 1)^2} dx$

The internal energy $U$ is related to the heat capacity by: $C_V = \left( \frac{\partial U}{\partial T} \right)_V$

To find the temperature dependence of the internal energy, we integrate the heat capacity with respect to temperature: $U(T) = \int C_V(T) dT$

Consider the low-temperature limit ($T \ll \Theta_D$). In this limit, the upper limit of the integral in the Debye model approaches infinity, and the integral becomes a constant. $C_V \approx 9 N k_B \left( \frac{T}{\Theta_D} \right)^3 \int_0^\infty \frac{x^4 e^x}{(e^x – 1)^2} dx$ The integral $\int_0^\infty \frac{x^4 e^x}{(e^x – 1)^2} dx = \frac{4 \pi^4}{15}$. So, $C_V \approx \frac{12 \pi^4 N k_B}{5 \Theta_D^3} T^3$.

Integrating $C_V$ with respect to $T$ to find the internal energy at low temperatures: $U(T) = \int \frac{12 \pi^4 N k_B}{5 \Theta_D^3} T^3 dT = \frac{12 \pi^4 N k_B}{5 \Theta_D^3} \frac{T^4}{4} + U_0 = \frac{3 \pi^4 N k_B}{5 \Theta_D^3} T^4 + U_0$ where $U_0$ is the integration constant, representing the zero-point energy. Thus, at low temperatures, $U \propto T^4$.

Consider the high-temperature limit ($T \gg \Theta_D$). In this limit, we can approximate the integral. Let $y = \frac{hc}{\lambda k_B T}$. For high temperatures, the integrand can be approximated by using the Dulong-Petit law, where $C_V \approx 3 N k_B$. Integrating $C_V$ with respect to $T$ to find the internal energy at high temperatures: $U(T) = \int 3 N k_B dT = 3 N k_B T + U’_0$ where $U’_0$ is another integration constant. Thus, at high temperatures, $U \propto T$.

Final Answer: The temperature dependence of the internal energy of the solid according to the Debye Model is $U \propto T^4$ at low temperatures and $U \propto T$ at high temperatures.

The Planck formula for the spectral energy density of blackbody radiation is given by: $u(\nu, T) = \frac{8 \pi h \nu^3}{c^3} \frac{1}{e^{h\nu/k_B T} – 1}$

We want to show that in the limit of very low frequencies ($\nu \to 0$) or very high temperatures ($T \to \infty$), this formula reduces to the Rayleigh-Jeans formula: $u(\nu, T)_{RJ} = \frac{8 \pi \nu^2}{c^3} k_B T$

Consider the exponent in the Planck formula: $\frac{h\nu}{k_B T}$.

For very low frequencies ($\nu \to 0$) or very high temperatures ($T \to \infty$), the value of the exponent becomes very small: $\frac{h\nu}{k_B T} \ll 1$

We can use the Taylor series expansion for the exponential function $e^x$ around $x=0$, which is $e^x \approx 1 + x$ for small $x$. In our case, $x = \frac{h\nu}{k_B T}$. Thus, we have: $e^{h\nu/k_B T} \approx 1 + \frac{h\nu}{k_B T}$

Substituting this approximation into the denominator of the Planck formula: $e^{h\nu/k_B T} – 1 \approx \left(1 + \frac{h\nu}{k_B T}\right) – 1 = \frac{h\nu}{k_B T}$

Now, substitute this approximation back into the Planck formula: $u(\nu, T) \approx \frac{8 \pi h \nu^3}{c^3} \frac{1}{\frac{h\nu}{k_B T}}$ $u(\nu, T) \approx \frac{8 \pi h \nu^3}{c^3} \times \frac{k_B T}{h\nu}$

Simplify the expression by cancelling out the common terms $h$ and one factor of $\nu$: $u(\nu, T) \approx \frac{8 \pi \nu^2 k_B T}{c^3}$

This is the Rayleigh-Jeans formula. Thus, we have shown that in the limit of very low frequencies or very high temperatures, the Planck formula reduces to the Rayleigh-Jeans formula.

The average energy of a single molecule in thermal equilibrium can be determined using the equipartition theorem. For a diatomic molecule, we consider the contributions from translational, rotational, and vibrational degrees of freedom.

At a temperature of 200 K, we assume that the translational and rotational degrees of freedom are fully excited. A diatomic molecule has: $\begin{itemize} \item 3 translational degrees of freedom, each contributing $\frac{1}{2} k_B T$ to the average energy. \item 2 rotational degrees of freedom, each contributing $\frac{1}{2} k_B T$ to the average energy. \end{itemize}$ The vibrational degrees of freedom are typically not fully excited at such low temperatures. The characteristic vibrational temperature for most diatomic molecules is much higher than 200 K. Therefore, we will neglect the contribution from vibrational energy.

The average energy due to translational motion is: $E_{trans} = 3 \times \frac{1}{2} k_B T = \frac{3}{2} k_B T$

The average energy due to rotational motion is: $E_{rot} = 2 \times \frac{1}{2} k_B T = k_B T$

The total average energy of a single diatomic molecule is the sum of the translational and rotational energies: $E_{avg} = E_{trans} + E_{rot} = \frac{3}{2} k_B T + k_B T = \frac{5}{2} k_B T$

Now, we can substitute the values for $k_B$ and $T$: $k_B = 1.38 \times 10^{-23} , J/K$ $T = 200 , K$

$E_{avg} = \frac{5}{2} \times (1.38 \times 10^{-23} , J/K) \times (200 , K)$ $E_{avg} = 2.5 \times 1.38 \times 10^{-23} \times 200 , J$ $E_{avg} = 3.45 \times 10^{-23} \times 200 , J$ $E_{avg} = 6.90 \times 10^{-21} , J$

Final Answer: The final answer is $\boxed{6.90 \times 10^{-21} , J}$

The energy of a single photon is given by the equation $E = hf$, where $h$ is Planck’s constant and $f$ is the frequency of the light. The frequency $f$ is related to the wavelength $\lambda$ by the speed of light $c$, such that $c = \lambda f$, so $f = \frac{c}{\lambda}$. Therefore, the energy of a single photon can be expressed as: $E = h \frac{c}{\lambda}$

Given values: Wavelength $\lambda = 650 , nm = 650 \times 10^{-9} , m$ Power output $P = 5 , mW = 5 \times 10^{-3} , W$ Planck’s constant $h = 6.626 \times 10^{-34} , Js$ Speed of light $c = 3 \times 10^8 , m/s$

Calculate the energy of a single photon: $E = (6.626 \times 10^{-34} , Js) \times \frac{3 \times 10^8 , m/s}{650 \times 10^{-9} , m}$ $E = \frac{6.626 \times 3 \times 10^{-34+8-(-9)}}{650} , J$ $E = \frac{19.878 \times 10^{-34+8+9}}{650} , J$ $E = \frac{19.878 \times 10^{-17}}{650} , J$ $E \approx 0.0305815 \times 10^{-17} , J$ $E \approx 3.058 \times 10^{-19} , J$

The power output of the laser is the total energy emitted per unit time. If $N$ is the number of photons emitted per second, then the total power $P$ is given by the energy of a single photon multiplied by the number of photons emitted per second: $P = N \times E$

To find the number of photons emitted per second, we rearrange the formula: $N = \frac{P}{E}$

Substitute the given power output and the calculated energy of a single photon: $N = \frac{5 \times 10^{-3} , W}{3.058 \times 10^{-19} , J/photon}$ $N = \frac{5}{3.058} \times \frac{10^{-3}}{10^{-19}} , photons/s$ $N \approx 1.635055 \times 10^{-3 – (-19)} , photons/s$ $N \approx 1.635 \times 10^{-3 + 19} , photons/s$ $N \approx 1.635 \times 10^{16} , photons/s$

Final Answer: The final answer is $\boxed{1.64 \times 10^{16}}$

The power hitting the solar panel is given by the product of the intensity of sunlight and the area of the panel: $P = I \times A$ $P = 1000 , \text{W/m}^2 \times 1 , \text{m}^2$ $P = 1000 , \text{W}$

The energy of a single photon is given by the equation: $E = hf$ where $h$ is Planck’s constant ($6.626 \times 10^{-34} , \text{J s}$) and $f$ is the frequency of the photon. We know that the speed of light $c = \lambda f$, so $f = c/\lambda$. Given the average wavelength $\lambda = 550 , \text{nm} = 550 \times 10^{-9} , \text{m}$, and the speed of light $c = 3 \times 10^8 , \text{m/s}$, we can find the frequency: $f = \frac{3 \times 10^8 , \text{m/s}}{550 \times 10^{-9} , \text{m}}$ $f = \frac{3}{550} \times 10^{17} , \text{Hz}$ $f \approx 0.0054545 \times 10^{17} , \text{Hz}$ $f \approx 5.4545 \times 10^{14} , \text{Hz}$

Now, calculate the energy of a single photon: $E = hf = 6.626 \times 10^{-34} , \text{J s} \times 5.4545 \times 10^{14} , \text{Hz}$ $E \approx 36.14 \times 10^{-20} , \text{J}$ $E \approx 3.614 \times 10^{-19} , \text{J}$

Alternatively, we can use the formula $E = \frac{hc}{\lambda}$: $E = \frac{6.626 \times 10^{-34} , \text{J s} \times 3 \times 10^8 , \text{m/s}}{550 \times 10^{-9} , \text{m}}$ $E = \frac{19.878 \times 10^{-26}}{550 \times 10^{-9}} , \text{J}$ $E \approx 0.03614 \times 10^{-17} , \text{J}$ $E \approx 3.614 \times 10^{-19} , \text{J}$

Let $N$ be the number of photons hitting the solar panel per second. The total power is the energy of each photon multiplied by the number of photons per second: $P = N \times E$ $N = \frac{P}{E}$ $N = \frac{1000 , \text{W}}{3.614 \times 10^{-19} , \text{J}}$ $N \approx 276.70 \times 10^{19}$ $N \approx 2.77 \times 10^{21}$

Final Answer: The final answer is $\boxed{2.77 \times 10^{21}}$

The energy of a single photon is given by the equation: $E = hf$ where $h$ is Planck’s constant ($6.626 \times 10^{-34} , \text{J s}$) and $f$ is the frequency of the photon.

Given the frequency of the radio transmission $f = 100 , \text{MHz}$. First, convert the frequency to Hz: $f = 100 \times 10^6 , \text{Hz} = 1 \times 10^8 , \text{Hz}$

Now, calculate the energy of a single photon: $E = 6.626 \times 10^{-34} , \text{J s} \times 1 \times 10^8 , \text{Hz}$ $E = 6.626 \times 10^{-26} , \text{J}$

The power of the radio transmission is given as $P = 50 , \text{kW}$. First, convert the power to Watts: $P = 50 \times 10^3 , \text{W}$

Let $N$ be the number of photons emitted per second. The total power is the energy of each photon multiplied by the number of photons per second: $P = N \times E$

To find the number of photons emitted per second, rearrange the equation: $N = \frac{P}{E}$ $N = \frac{50 \times 10^3 , \text{W}}{6.626 \times 10^{-26} , \text{J}}$ $N = \frac{5 \times 10^4}{6.626 \times 10^{-26}}$ $N = \frac{5}{6.626} \times 10^{4 – (-26)}$ $N \approx 0.7546 \times 10^{30}$ $N \approx 7.546 \times 10^{29}$

Final Answer: The final answer is $\boxed{7.55 \times 10^{29}}$

The energy of a single photon emitted by the LED is given by the equation: $E = hf = \frac{hc}{\lambda}$ where $h = 6.626 \times 10^{-34} , \text{J s}$ is Planck’s constant, $c = 3 \times 10^8 , \text{m/s}$ is the speed of light, and $\lambda = 520 , \text{nm} = 520 \times 10^{-9} , \text{m}$ is the wavelength of the emitted light.

$E = \frac{(6.626 \times 10^{-34} , \text{J s}) \times (3 \times 10^8 , \text{m/s})}{520 \times 10^{-9} , \text{m}}$ $E = \frac{19.878 \times 10^{-26}}{520 \times 10^{-9}} , \text{J}$ $E \approx 3.823 \times 10^{-19} , \text{J}$

The electrical power input to the LED is $P_{in} = 100 , \text{mW} = 100 \times 10^{-3} , \text{W} = 0.1 , \text{W}$. The efficiency of the LED is 20%, so the optical power output $P_{out}$ is: $P_{out} = \text{Efficiency} \times P_{in}$ $P_{out} = 0.20 \times 0.1 , \text{W}$ $P_{out} = 0.02 , \text{W}$

Let $N$ be the number of photons emitted per second. The total optical power output is the energy of each photon multiplied by the number of photons emitted per second: $P_{out} = N \times E$

Solving for $N$: $N = \frac{P_{out}}{E}$ $N = \frac{0.02 , \text{W}}{3.823 \times 10^{-19} , \text{J}}$ $N = \frac{2 \times 10^{-2}}{3.823 \times 10^{-19}}$ $N \approx 0.52315 \times 10^{17}$ $N \approx 5.23 \times 10^{16}$

Final Answer: The final answer is $\boxed{5.23 \times 10^{16}}$

The energy of a single photon is given by the equation: $E = hf = \frac{hc}{\lambda}$ where $h = 6.626 \times 10^{-34} , \text{J s}$ is Planck’s constant, $c = 3 \times 10^8 , \text{m/s}$ is the speed of light, and $\lambda$ is the wavelength of the photon.

Given the minimum wavelength $\lambda = 0.1 , \text{nm} = 0.1 \times 10^{-9} , \text{m} = 1 \times 10^{-10} , \text{m}$.

Calculate the energy of a single photon at this minimum wavelength: $E = \frac{(6.626 \times 10^{-34} , \text{J s}) \times (3 \times 10^8 , \text{m/s})}{1 \times 10^{-10} , \text{m}}$ $E = \frac{19.878 \times 10^{-26}}{1 \times 10^{-10}} , \text{J}$ $E = 19.878 \times 10^{-16} , \text{J}$

The power generated by the X-ray tube is $P = 10 , \text{W}$. The power is the energy emitted per unit time. If $N$ is the number of photons emitted per second, then the total power is the energy of each photon multiplied by the number of photons per second: $P = N \times E$

To find the rate at which photons are emitted ($N$), rearrange the equation: $N = \frac{P}{E}$ $N = \frac{10 , \text{W}}{19.878 \times 10^{-16} , \text{J}}$ $N = \frac{10}{19.878} \times 10^{16}$ $N \approx 0.50306 \times 10^{16}$ $N \approx 5.03 \times 10^{15}$

Final Answer: The final answer is $\boxed{5.03 \times 10^{15}}$

The energy of a photon is given by $E = hf = \frac{hc}{\lambda}$. For the first case, $\lambda_1 = 400 , \text{nm} = 400 \times 10^{-9} , \text{m}$. The energy of the photon is $E_1 = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{400 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{400 \times 10^{-9}} , \text{J} = 4.9695 \times 10^{-19} , \text{J}$. Convert this energy to electron volts: $E_1 = \frac{4.9695 \times 10^{-19}}{1.602 \times 10^{-19}} , \text{eV} \approx 3.10 , \text{eV}$.

The photoelectric equation is $E = \phi + KE_{max}$. Given $KE_{max} = 1.8 , \text{eV}$, we can find the work function $\phi$: $\phi = E_1 – KE_{max} = 3.10 , \text{eV} – 1.8 , \text{eV} = 1.30 , \text{eV}$.

For the second case, $\lambda_2 = 300 , \text{nm} = 300 \times 10^{-9} , \text{m}$. The energy of the photon is $E_2 = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{300 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{300 \times 10^{-9}} , \text{J} = 6.626 \times 10^{-19} , \text{J}$. Convert this energy to electron volts: $E_2 = \frac{6.626 \times 10^{-19}}{1.602 \times 10^{-19}} , \text{eV} \approx 4.136 , \text{eV}$.

Now, use the photoelectric equation to find the maximum kinetic energy for the second case: $KE_{max, 2} = E_2 – \phi = 4.136 , \text{eV} – 1.30 , \text{eV} = 2.836 , \text{eV}$.

Final Answer: The final answer is $\boxed{\text{Work function: } 1.30 , \text{eV, Maximum KE: } 2.84 , \text{eV}}$

The work function of potassium is given as $\phi = 2.3 , \text{eV}$. First, convert the work function to Joules: $\phi = 2.3 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} = 3.6846 \times 10^{-19} , \text{J}$.

The wavelength of the incident light is $\lambda = 350 , \text{nm} = 350 \times 10^{-9} , \text{m}$. The energy of the incident photon is: $E = hf = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{350 \times 10^{-9} , \text{m}}$ $E = \frac{19.878 \times 10^{-26}}{350 \times 10^{-9}} , \text{J} = 5.6794 \times 10^{-19} , \text{J}$.

Convert the photon energy to eV: $E (\text{eV}) = \frac{5.6794 \times 10^{-19} , \text{J}}{1.602 \times 10^{-19} , \text{J/eV}} \approx 3.545 , \text{eV}$.

The maximum kinetic energy of the photoelectrons is given by the photoelectric equation: $KE_{max} = E – \phi = 3.545 , \text{eV} – 2.3 , \text{eV} = 1.245 , \text{eV}$.

Convert the maximum kinetic energy to Joules: $KE_{max} = 1.245 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} = 1.99449 \times 10^{-19} , \text{J}$.

The maximum kinetic energy is also given by $KE_{max} = \frac{1}{2}mv_{max}^2$, where $m$ is the mass of an electron ($9.109 \times 10^{-31} , \text{kg}$) and $v_{max}$ is the maximum velocity. $v_{max}^2 = \frac{2 \times KE_{max}}{m} = \frac{2 \times 1.99449 \times 10^{-19} , \text{J}}{9.109 \times 10^{-31} , \text{kg}} = \frac{3.98898 \times 10^{-19}}{9.109 \times 10^{-31}} , \text{m}^2/\text{s}^2$ $v_{max}^2 \approx 4.37916 \times 10^{11} , \text{m}^2/\text{s}^2$ $v_{max} = \sqrt{4.37916 \times 10^{11}} , \text{m/s} \approx 6.6175 \times 10^5 , \text{m/s}$.

The cutoff wavelength $\lambda_c$ is the wavelength for which the photon energy is equal to the work function ($E = \phi$). $\frac{hc}{\lambda_c} = \phi$ $\lambda_c = \frac{hc}{\phi} = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{3.6846 \times 10^{-19} , \text{J}}$ $\lambda_c = \frac{19.878 \times 10^{-26}}{3.6846 \times 10^{-19}} , \text{m} = 5.4 \times 10^{-7} , \text{m}$. Convert to nm: $\lambda_c = 540 , \text{nm}$.

Final Answer: The final answer is $\boxed{\text{Maximum velocity: } 6.62 \times 10^5 , \text{m/s, Cutoff wavelength: } 540 , \text{nm}}$

The energy of the incident photon is given by $E = \frac{hc}{\lambda}$. Given the wavelength $\lambda = 200 , \text{nm} = 200 \times 10^{-9} , \text{m}$. $E = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{200 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{2 \times 10^{-7}} , \text{J} = 9.939 \times 10^{-19} , \text{J}$.

Convert the energy to electron volts: $E (\text{eV}) = \frac{9.939 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 6.204 , \text{eV}$.

For metal surface A, the work function $\phi_A = 3.5 , \text{eV}$. The maximum kinetic energy of the photoelectrons from surface A is $KE_A = E – \phi_A$. $KE_A = 6.204 , \text{eV} – 3.5 , \text{eV} = 2.704 , \text{eV}$. We are given that the maximum velocity of these electrons is $v$, so $KE_A = \frac{1}{2}mv^2$.

For metal surface B, the maximum kinetic energy of the photoelectrons $KE_B$ is twice that of surface A. $KE_B = 2 \times KE_A = 2 \times 2.704 , \text{eV} = 5.408 , \text{eV}$.

The photoelectric equation for metal B is $KE_B = E – \phi_B$, where $\phi_B$ is the work function of metal B. Rearranging the equation to solve for $\phi_B$: $\phi_B = E – KE_B$. $\phi_B = 6.204 , \text{eV} – 5.408 , \text{eV} = 0.796 , \text{eV}$.

Final Answer: The final answer is $\boxed{0.796 , \text{eV}}$

The energy of a photon is given by $E = \frac{hc}{\lambda}$, and the maximum kinetic energy of a photoelectron is $KE_{max} = \frac{1}{2}mv_{max}^2$. The photoelectric equation is $E = \phi + KE_{max}$.

For the first case, with $\lambda_1 = 250 , \text{nm} = 250 \times 10^{-9} , \text{m}$ and $v_{max1} = 8.0 \times 10^5 , \text{m/s}$: The energy of the photon is $E_1 = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{250 \times 10^{-9} , \text{m}} = 7.9512 \times 10^{-19} , \text{J}$. The maximum kinetic energy of the photoelectrons is $KE_{max1} = \frac{1}{2}(9.109 \times 10^{-31} , \text{kg})(8.0 \times 10^5 , \text{m/s})^2 = 2.91488 \times 10^{-19} , \text{J}$. Using the photoelectric equation, $\phi = E_1 – KE_{max1} = 7.9512 \times 10^{-19} , \text{J} – 2.91488 \times 10^{-19} , \text{J} = 5.03632 \times 10^{-19} , \text{J}$.

For the second case, with $\lambda_2 = 300 , \text{nm} = 300 \times 10^{-9} , \text{m}$ and $v_{max2} = 5.8 \times 10^5 , \text{m/s}$: The energy of the photon is $E_2 = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{300 \times 10^{-9} , \text{m}} = 6.626 \times 10^{-19} , \text{J}$. The maximum kinetic energy of the photoelectrons is $KE_{max2} = \frac{1}{2}(9.109 \times 10^{-31} , \text{kg})(5.8 \times 10^5 , \text{m/s})^2 = 1.53397 \times 10^{-19} , \text{J}$. Using the photoelectric equation, $\phi = E_2 – KE_{max2} = 6.626 \times 10^{-19} , \text{J} – 1.53397 \times 10^{-19} , \text{J} = 5.09203 \times 10^{-19} , \text{J}$.

The values of the work function calculated from the two cases are slightly different due to precision. We can take the average value or use one of them. Let’s use the value from the first case. $\phi = 5.03632 \times 10^{-19} , \text{J}$.

To find the threshold wavelength $\lambda_0$, we use the relation $E = \phi$, so $\frac{hc}{\lambda_0} = \phi$. $\lambda_0 = \frac{hc}{\phi} = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{5.03632 \times 10^{-19} , \text{J}} = \frac{19.878 \times 10^{-26}}{5.03632 \times 10^{-19}} , \text{m} = 3.9468 \times 10^{-7} , \text{m}$. Converting to nanometers, $\lambda_0 = 394.68 , \text{nm}$.

Final Answer: The final answer is $\boxed{\text{Work function: } 5.04 \times 10^{-19} , \text{J} \text{ or } 3.15 , \text{eV, Threshold wavelength: } 395 , \text{nm}}$

The energy of a photon is given by $E = \frac{hc}{\lambda}$. The work function of the metal is $\phi = 3.0 , \text{eV}$. The maximum kinetic energy of the photoelectrons is $KE_{max} = E – \phi$. The reverse potential (stopping potential) $V_s$ is related to the maximum kinetic energy by $eV_s = KE_{max}$.

Case 1: Wavelength $\lambda_1 = 220 , \text{nm} = 220 \times 10^{-9} , \text{m}$. Energy of the photon $E_1 = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{220 \times 10^{-9} , \text{m}}$ $E_1 = \frac{19.878 \times 10^{-26}}{2.2 \times 10^{-7}} , \text{J} = 9.035 \times 10^{-19} , \text{J}$ Convert to eV: $E_1 (\text{eV}) = \frac{9.035 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 5.64 , \text{eV}$ Maximum kinetic energy $KE_{max1} = E_1 – \phi = 5.64 , \text{eV} – 3.0 , \text{eV} = 2.64 , \text{eV}$ Reverse potential $V_{s1} = \frac{KE_{max1}}{e} = \frac{2.64 , \text{eV}}{e} = 2.64 , \text{V}$

Case 2: Wavelength $\lambda_2 = 280 , \text{nm} = 280 \times 10^{-9} , \text{m}$. Energy of the photon $E_2 = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{280 \times 10^{-9} , \text{m}}$ $E_2 = \frac{19.878 \times 10^{-26}}{2.8 \times 10^{-7}} , \text{J} = 7.10 \times 10^{-19} , \text{J}$ Convert to eV: $E_2 (\text{eV}) = \frac{7.10 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.43 , \text{eV}$ Maximum kinetic energy $KE_{max2} = E_2 – \phi = 4.43 , \text{eV} – 3.0 , \text{eV} = 1.43 , \text{eV}$ Reverse potential $V_{s2} = \frac{KE_{max2}}{e} = \frac{1.43 , \text{eV}}{e} = 1.43 , \text{V}$

Final Answer: The final answer is $\boxed{\text{Reverse potential for 220 nm: } 2.64 , \text{V, Reverse potential for 280 nm: } 1.43 , \text{V}}$

A green laser pointer emits 532 nm light with 3 mW of power. How many photons are emitted per second?

The energy of a single photon is given by $E = \frac{hc}{\lambda}$. Given $\lambda = 532 , \text{nm} = 532 \times 10^{-9} , \text{m}$, $h = 6.626 \times 10^{-34} , \text{J s}$, and $c = 3 \times 10^8 , \text{m/s}$. $E = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{532 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{532 \times 10^{-9}} , \text{J} \approx 3.736 \times 10^{-19} , \text{J}$.

The power of the laser is $P = 3 , \text{mW} = 3 \times 10^{-3} , \text{W}$. The number of photons emitted per second, $N$, is given by $N = \frac{P}{E}$. $N = \frac{3 \times 10^{-3} , \text{W}}{3.736 \times 10^{-19} , \text{J}} \approx 8.029 \times 10^{15}$.

A UV lamp emits 250 nm light with a power of 5 W. What is the energy of each photon, and how many photons are emitted per second?

The energy of a single photon is given by $E = \frac{hc}{\lambda}$. Given $\lambda = 250 , \text{nm} = 250 \times 10^{-9} , \text{m}$, $h = 6.626 \times 10^{-34} , \text{J s}$, and $c = 3 \times 10^8 , \text{m/s}$. $E = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{250 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{250 \times 10^{-9}} , \text{J} \approx 7.951 \times 10^{-19} , \text{J}$.

The power of the UV lamp is $P = 5 , \text{W}$. The number of photons emitted per second, $N$, is given by $N = \frac{P}{E}$. $N = \frac{5 , \text{W}}{7.951 \times 10^{-19} , \text{J}} \approx 6.289 \times 10^{18}$.

A radio transmitter broadcasts at 98 MHz with a power of 10 kW. How many photons does it emit per second?

The energy of a single photon is given by $E = hf$. Given $f = 98 , \text{MHz} = 98 \times 10^6 , \text{Hz}$ and $h = 6.626 \times 10^{-34} , \text{J s}$. $E = (6.626 \times 10^{-34} , \text{J s})(98 \times 10^6 , \text{Hz}) = 6.49348 \times 10^{-26} , \text{J}$.

The power of the transmitter is $P = 10 , \text{kW} = 10 \times 10^3 , \text{W} = 10^4 , \text{W}$. The number of photons emitted per second, $N$, is given by $N = \frac{P}{E}$. $N = \frac{10^4 , \text{W}}{6.49348 \times 10^{-26} , \text{J}} \approx 1.540 \times 10^{29}$.

The human eye can detect light with an intensity as low as 10^-10 W/m². If the light has a wavelength of 500 nm, how many photons per second are incident on a 1 mm² area of the retina?

The area of the retina is $A = 1 , \text{mm}^2 = (1 \times 10^{-3} , \text{m})^2 = 1 \times 10^{-6} , \text{m}^2$. The power incident on the retina is $P = I \times A = (10^{-10} , \text{W/m}^2)(1 \times 10^{-6} , \text{m}^2) = 10^{-16} , \text{W}$.

The energy of a single photon is given by $E = \frac{hc}{\lambda}$. Given $\lambda = 500 , \text{nm} = 500 \times 10^{-9} , \text{m}$, $h = 6.626 \times 10^{-34} , \text{J s}$, and $c = 3 \times 10^8 , \text{m/s}$. $E = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{500 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{500 \times 10^{-9}} , \text{J} \approx 3.9756 \times 10^{-19} , \text{J}$.

The number of photons per second, $N$, is given by $N = \frac{P}{E}$. $N = \frac{10^{-16} , \text{W}}{3.9756 \times 10^{-19} , \text{J}} \approx 251.5$.

A 60 W incandescent lightbulb emits light of average wavelength 600 nm. Estimate the number of photons emitted per second, assuming all the electrical power is converted to visible light.

The energy of a single photon is given by $E = \frac{hc}{\lambda}$. Given $\lambda = 600 , \text{nm} = 600 \times 10^{-9} , \text{m}$, $h = 6.626 \times 10^{-34} , \text{J s}$, and $c = 3 \times 10^8 , \text{m/s}$. $E = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{600 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{600 \times 10^{-9}} , \text{J} \approx 3.313 \times 10^{-19} , \text{J}$.

The power of the lightbulb is $P = 60 , \text{W}$. The number of photons emitted per second, $N$, is given by $N = \frac{P}{E}$. $N = \frac{60 , \text{W}}{3.313 \times 10^{-19} , \text{J}} \approx 1.811 \times 10^{20}$.

The work function for a particular metal is 3.0 eV. What is the longest wavelength of light that will cause photoemission?

The longest wavelength of light that will cause photoemission corresponds to the threshold frequency, where the photon energy is equal to the work function. The work function $\phi = 3.0 , \text{eV}$. Convert this to Joules: $\phi = 3.0 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} = 4.806 \times 10^{-19} , \text{J}$.

The energy of the photon is $E = \frac{hc}{\lambda}$. At the threshold wavelength $\lambda_0$, $E = \phi$. So, $\phi = \frac{hc}{\lambda_0}$. Rearrange to solve for $\lambda_0$: $\lambda_0 = \frac{hc}{\phi} = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{4.806 \times 10^{-19} , \text{J}} = \frac{19.878 \times 10^{-26}}{4.806 \times 10^{-19}} , \text{m} \approx 4.136 \times 10^{-7} , \text{m}$. Convert to nanometers: $\lambda_0 = 4.136 \times 10^{-7} , \text{m} \times \frac{10^9 , \text{nm}}{1 , \text{m}} = 413.6 , \text{nm}$.

Light with a wavelength of 450 nm shines on a metal, and the fastest emitted electrons have a kinetic energy of 1.5 eV. What is the metal’s work function?

The energy of the incident photon is given by $E = \frac{hc}{\lambda}$. Given $\lambda = 450 , \text{nm} = 450 \times 10^{-9} , \text{m}$, $h = 6.626 \times 10^{-34} , \text{J s}$, and $c = 3 \times 10^8 , \text{m/s}$. $E = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{450 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{450 \times 10^{-9}} , \text{J} \approx 4.417 \times 10^{-19} , \text{J}$.

Convert the photon energy to electron volts: $E (\text{eV}) = \frac{4.417 \times 10^{-19} , \text{J}}{1.602 \times 10^{-19} , \text{J/eV}} \approx 2.757 , \text{eV}$.

The photoelectric equation is $E = \phi + KE_{max}$, where $\phi$ is the work function and $KE_{max}$ is the maximum kinetic energy of the emitted electrons. Given $KE_{max} = 1.5 , \text{eV}$. $\phi = E – KE_{max} = 2.757 , \text{eV} – 1.5 , \text{eV} = 1.257 , \text{eV}$.

If the frequency of incident light on a photoelectric material is doubled, how does the maximum kinetic energy of the photoelectrons change?

The photoelectric equation is given by $KE_{max} = hf – \phi$, where $KE_{max}$ is the maximum kinetic energy of the photoelectrons, $h$ is Planck’s constant, $f$ is the frequency of the incident light, and $\phi$ is the work function of the material.

Let the initial frequency be $f_1$ and the initial maximum kinetic energy be $KE_{max1}$. Then, $KE_{max1} = hf_1 – \phi$.

If the frequency is doubled, the new frequency is $f_2 = 2f_1$. Let the new maximum kinetic energy be $KE_{max2}$. Then, $KE_{max2} = hf_2 – \phi = h(2f_1) – \phi = 2hf_1 – \phi$.

We want to find the relationship between $KE_{max2}$ and $KE_{max1}$. From the first equation, $hf_1 = KE_{max1} + \phi$. Substitute this into the equation for $KE_{max2}$: $KE_{max2} = 2(KE_{max1} + \phi) – \phi = 2KE_{max1} + 2\phi – \phi = 2KE_{max1} + \phi$.

So, the new maximum kinetic energy is the original maximum kinetic energy doubled, plus the work function. The maximum kinetic energy does not simply double. It increases by an amount equal to the original kinetic energy plus the work function.

What is the stopping potential required to halt the most energetic photoelectrons emitted from a surface with a work function of 2.5 eV when light of wavelength 400 nm is incident on it?

First, calculate the energy of the incident photon using the formula $E = \frac{hc}{\lambda}$. Given $\lambda = 400 , \text{nm} = 400 \times 10^{-9} , \text{m}$, $h = 6.626 \times 10^{-34} , \text{J s}$, and $c = 3 \times 10^8 , \text{m/s}$. $E = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{400 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{400 \times 10^{-9}} , \text{J} \approx 4.9695 \times 10^{-19} , \text{J}$.

Convert the photon energy to electron volts: $E (\text{eV}) = \frac{4.9695 \times 10^{-19} , \text{J}}{1.602 \times 10^{-19} , \text{J/eV}} \approx 3.102 , \text{eV}$.

The maximum kinetic energy of the photoelectrons is given by the photoelectric equation: $KE_{max} = E – \phi$, where $\phi$ is the work function. Given $\phi = 2.5 , \text{eV}$. $KE_{max} = 3.102 , \text{eV} – 2.5 , \text{eV} = 0.602 , \text{eV}$.

The stopping potential $V_s$ is the potential difference required to stop the most energetic photoelectrons. The kinetic energy of the fastest electrons is equal to the work done by the stopping potential to halt them, i.e., $KE_{max} = eV_s$. Therefore, $V_s = \frac{KE_{max}}{e} = \frac{0.602 , \text{eV}}{e} = 0.602 , \text{V}$.

How does the number of photoelectrons change if the intensity of the incident light is doubled, while keeping the frequency constant?

The intensity of light is proportional to the number of photons incident on the surface per unit time. If the intensity of the incident light is doubled, it means that the number of photons incident on the surface per unit time is doubled.

Since each photon with sufficient energy can liberate one electron, doubling the number of incident photons (while keeping the frequency constant, ensuring each photon still has enough energy to overcome the work function) will double the number of photoelectrons emitted per unit time.

Therefore, if the intensity of the incident light is doubled while keeping the frequency constant, the number of photoelectrons emitted per second is also doubled.

What is the threshold frequency for photoelectric emission for a metal with a work function of 4.0 eV?

The work function $\phi$ is the minimum energy required to remove an electron from the surface of a metal. At the threshold frequency $f_0$, the energy of the photon is equal to the work function: $E = hf_0 = \phi$.

First, convert the work function from eV to Joules: $\phi = 4.0 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} = 6.408 \times 10^{-19} , \text{J}$.

Now, use the equation $hf_0 = \phi$ to solve for the threshold frequency $f_0$: $f_0 = \frac{\phi}{h} = \frac{6.408 \times 10^{-19} , \text{J}}{6.626 \times 10^{-34} , \text{J s}} \approx 9.66 \times 10^{14} , \text{Hz}$.

Light of wavelength 300 nm is shone on a metal surface with a work function of 2 eV. Calculate the maximum kinetic energy and velocity of emitted electrons.

First, calculate the energy of the incident photon using $E = \frac{hc}{\lambda}$. Given $\lambda = 300 , \text{nm} = 300 \times 10^{-9} , \text{m}$, $h = 6.626 \times 10^{-34} , \text{J s}$, and $c = 3 \times 10^8 , \text{m/s}$. $E = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{300 \times 10^{-9} , \text{m}} = \frac{19.878 \times 10^{-26}}{300 \times 10^{-9}} , \text{J} \approx 6.626 \times 10^{-19} , \text{J}$.

Convert the photon energy to electron volts: $E (\text{eV}) = \frac{6.626 \times 10^{-19} , \text{J}}{1.602 \times 10^{-19} , \text{J/eV}} \approx 4.136 , \text{eV}$.

The maximum kinetic energy of the emitted electrons is given by the photoelectric equation: $KE_{max} = E – \phi$. Given the work function $\phi = 2 , \text{eV}$. $KE_{max} = 4.136 , \text{eV} – 2 , \text{eV} = 2.136 , \text{eV}$.

Convert the maximum kinetic energy to Joules: $KE_{max} = 2.136 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} \approx 3.422 \times 10^{-19} , \text{J}$.

To find the maximum velocity $v_{max}$, use the kinetic energy formula $KE_{max} = \frac{1}{2}mv_{max}^2$, where $m$ is the mass of an electron ($9.109 \times 10^{-31} , \text{kg}$). $v_{max}^2 = \frac{2 \times KE_{max}}{m} = \frac{2 \times 3.422 \times 10^{-19} , \text{J}}{9.109 \times 10^{-31} , \text{kg}} = \frac{6.844 \times 10^{-19}}{9.109 \times 10^{-31}} , \text{m}^2/\text{s}^2 \approx 7.514 \times 10^{11} , \text{m}^2/\text{s}^2$. $v_{max} = \sqrt{7.514 \times 10^{11}} , \text{m/s} \approx 8.668 \times 10^5 , \text{m/s}$.

How does the maximum kinetic energy of photoelectrons depend on the wavelength of the incident light and the work function of the metal? Explain qualitatively.

The maximum kinetic energy of photoelectrons is determined by the energy of the incident photons and the work function of the metal, as described by the photoelectric equation: $KE_{max} = E_{photon} – \phi$.

The energy of a photon is inversely proportional to its wavelength ($E = hc/\lambda$). Therefore, as the wavelength of the incident light decreases (meaning the frequency increases), the energy of the photons increases, leading to an increase in the maximum kinetic energy of the photoelectrons. Shorter wavelengths (like blue or ultraviolet light) carry more energy per photon than longer wavelengths (like red light). This extra energy is transferred to the photoelectrons, giving them more kinetic energy.

The work function ($\phi$) is a property of the metal and represents the minimum energy required to liberate an electron from its surface. It’s like an energy barrier the photon needs to overcome. According to the photoelectric equation, the maximum kinetic energy of the photoelectrons is obtained after this work function energy is “used up” to free the electron. Therefore, a higher work function means that more energy is needed to release the electron, and for the same incident light, the maximum kinetic energy of the emitted electron will be lower. Different metals have different work functions.

In summary:

• Shorter wavelength (higher frequency) light leads to higher maximum kinetic energy.
• Higher work function of the metal leads to lower maximum kinetic energy.

What is the relationship between the stopping potential and the maximum kinetic energy of the photoelectrons?

The stopping potential ($V_s$) is the minimum negative potential applied to the anode (collector) relative to the cathode (emitter) that is just sufficient to stop the most energetic photoelectrons from reaching the anode, making the photocurrent zero.

The relationship between the stopping potential and the maximum kinetic energy of the photoelectrons is that the work done by the stopping potential in halting the fastest electrons is equal to their maximum kinetic energy.

Mathematically, this relationship is expressed as:

$eV_s = KE_{max}$

where:

• $e$ is the elementary charge (the magnitude of the charge of an electron).
• $V_s$ is the stopping potential.
• $KE_{max}$ is the maximum kinetic energy of the photoelectrons.

This equation shows a direct proportionality between the stopping potential and the maximum kinetic energy. A higher maximum kinetic energy requires a larger stopping potential to bring those electrons to a halt.

Calculate the momentum of a photon with a wavelength of 600nm.

The momentum of a photon is given by the equation $p = \frac{h}{\lambda}$, where $h$ is Planck’s constant and $\lambda$ is the wavelength of the photon. Given $\lambda = 600 , \text{nm} = 600 \times 10^{-9} , \text{m}$ and $h = 6.626 \times 10^{-34} , \text{J s}$. $p = \frac{6.626 \times 10^{-34} , \text{J s}}{600 \times 10^{-9} , \text{m}} = \frac{6.626}{600} \times 10^{-34+9} , \text{kg m/s}$ $p \approx 0.01104 \times 10^{-25} , \text{kg m/s}$ $p \approx 1.104 \times 10^{-27} , \text{kg m/s}$

Calculate the de Broglie wavelength of an electron moving at 1% the speed of light.

The de Broglie wavelength of a particle is given by the equation $\lambda = \frac{h}{mv}$, where $h$ is Planck’s constant, $m$ is the mass of the particle, and $v$ is its velocity. The speed of light $c = 3 \times 10^8 , \text{m/s}$. The velocity of the electron $v = 0.01 \times c = 0.01 \times 3 \times 10^8 , \text{m/s} = 3 \times 10^6 , \text{m/s}$. The mass of an electron $m_e = 9.109 \times 10^{-31} , \text{kg}$. $\lambda = \frac{6.626 \times 10^{-34} , \text{J s}}{(9.109 \times 10^{-31} , \text{kg})(3 \times 10^6 , \text{m/s})} = \frac{6.626}{9.109 \times 3} \times 10^{-34+31-6} , \text{m}$ $\lambda \approx \frac{6.626}{27.327} \times 10^{-9} , \text{m}$ $\lambda \approx 0.2425 \times 10^{-9} , \text{m}$ $\lambda \approx 2.425 \times 10^{-10} , \text{m}$

What is the de Broglie wavelength of a proton moving at 10^6 m/s?

The de Broglie wavelength of a particle is given by the equation $\lambda = \frac{h}{mv}$. The velocity of the proton $v = 10^6 , \text{m/s}$. The mass of a proton $m_p = 1.672 \times 10^{-27} , \text{kg}$. $\lambda = \frac{6.626 \times 10^{-34} , \text{J s}}{(1.672 \times 10^{-27} , \text{kg})(10^6 , \text{m/s})} = \frac{6.626}{1.672} \times 10^{-34+27-6} , \text{m}$ $\lambda \approx 3.963 \times 10^{-13} , \text{m}$

A baseball of mass 0.15 kg is moving at 30 m/s. What is its de Broglie wavelength? Is this wavelength significant?

$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{(0.15)(30)} \approx \boxed{1.47 \times 10^{-34} , \text{m}}$. This wavelength is not significant.

An electron is accelerated through a potential difference of 500 V. What is its de Broglie wavelength?

$KE = eV = (1.602 \times 10^{-19})(500) = 8.01 \times 10^{-17} , \text{J}$. $p = \sqrt{2mKE} = \sqrt{2(9.109 \times 10^{-31})(8.01 \times 10^{-17})} \approx 1.21 \times 10^{-23} , \text{kg m/s}$. $\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{1.21 \times 10^{-23}} \approx \boxed{5.48 \times 10^{-11} , \text{m}}$.

If the de Broglie wavelength of an electron is 0.1 nm, what is its kinetic energy?

$p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{1 \times 10^{-10}} = 6.63 \times 10^{-24} , \text{kg m/s}$. $KE = \frac{p^2}{2m} = \frac{(6.63 \times 10^{-24})^2}{2(9.109 \times 10^{-31})} \approx \boxed{2.41 \times 10^{-17} , \text{J}}$.

What potential difference must be applied to accelerate an electron so that it has a de Broglie wavelength equal to the diameter of an atom (approx. 0.1 nm)?

From the previous question, $KE \approx 2.41 \times 10^{-17} , \text{J}$. $eV = KE \implies V = \frac{KE}{e} = \frac{2.41 \times 10^{-17}}{1.602 \times 10^{-19}} \approx \boxed{150 , \text{V}}$.

If the de Broglie wavelength of a particle is halved, how does its momentum change?

Since $p = \frac{h}{\lambda}$, if $\lambda$ is halved, the momentum is $\boxed{\text{doubled}}$.

Explain the concept of wave-particle duality and how it applies to electrons and photons.

Wave-particle duality is a fundamental concept in quantum mechanics stating that all matter exhibits both wave-like and particle-like properties. Classically, waves and particles are considered distinct entities. Waves, like light or water waves, exhibit phenomena like diffraction and interference, while particles, like marbles, have definite positions and momentum.

However, at the quantum level, this distinction blurs. Photons, which were initially understood as packets of energy or particles of light, also exhibit wave-like behavior. Experiments like the double-slit experiment with light demonstrate interference patterns, a characteristic of waves.

Conversely, electrons, which were traditionally considered particles, also exhibit wave-like properties. The de Broglie hypothesis proposed that all matter has an associated wavelength, and this was experimentally confirmed by the observation of electron diffraction patterns, similar to those seen with light.

Therefore, both electrons and photons are not strictly either waves or particles, but rather possess both aspects. Which aspect is more prominent depends on the experiment being performed. For example, when light interacts with matter as in the photoelectric effect, it behaves like particles (photons). When light propagates through space, it behaves like a wave. Similarly, when electrons move through a conductor, their particle nature is evident, but when they pass through a narrow slit, their wave nature manifests as diffraction. This dual nature is a cornerstone of quantum mechanics and is essential for understanding the behavior of matter at the atomic and subatomic level.

How is the de Broglie wavelength related to the momentum of a particle?

The de Broglie wavelength ($\lambda$) is inversely proportional to the momentum ($p$) of a particle. This relationship is mathematically expressed by the de Broglie equation:

$\lambda = \frac{h}{p}$

where $h$ is Planck’s constant.

This equation indicates that:

• Higher momentum corresponds to a shorter de Broglie wavelength. A particle moving with a greater velocity or having a larger mass will have a larger momentum and thus a smaller associated wavelength.
• Lower momentum corresponds to a longer de Broglie wavelength. A particle moving slowly or having a smaller mass will have a smaller momentum and thus a larger associated wavelength.

This relationship highlights the wave nature of matter. The de Broglie wavelength provides a measure of the “waviness” of a particle. Particles with significant momentum have very small wavelengths, making their wave-like behavior less apparent in macroscopic systems. However, for very small particles like electrons moving at non-relativistic speeds, the de Broglie wavelength can be significant enough to produce observable wave phenomena. The de Broglie equation is a central concept in understanding the wave-particle duality of matter.

Explain why we don’t observe the wave nature of macroscopic objects.

The wave nature of objects is described by the de Broglie wavelength, given by $\lambda = \frac{h}{mv}$. Planck’s constant, $h$, is an extremely small number ($6.626 \times 10^{-34} , \text{J s}$). For macroscopic objects, their mass ($m$) is relatively large. Even for moderate velocities ($v$), the product $mv$ (the momentum) is a significant number compared to Planck’s constant.

As a result, the de Broglie wavelength ($\lambda$) for macroscopic objects is incredibly small, far too small to be detected or have any observable effects. For example, as calculated previously, a baseball has a de Broglie wavelength on the order of $10^{-34}$ meters, which is orders of magnitude smaller than the size of an atom’s nucleus.

The wave nature of an object becomes observable when its de Broglie wavelength is comparable to the size of the obstacles or apertures it interacts with. For macroscopic objects, the wavelengths are so tiny that they do not interact with everyday objects in a way that would manifest wave-like behavior like diffraction or interference. The “waviness” is still there, according to quantum mechanics, but its scale is far below our ability to observe it directly. Therefore, in our everyday experience, macroscopic objects behave according to classical mechanics, where their particle nature dominates.

Calculate the de Broglie wavelength of a neutron of kinetic energy 1 eV.

First, convert the kinetic energy from eV to Joules: $KE = 1 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} = 1.602 \times 10^{-19} , \text{J}$.

The kinetic energy is related to momentum by $KE = \frac{p^2}{2m}$, so $p = \sqrt{2mKE}$. The mass of a neutron $m_n = 1.675 \times 10^{-27} , \text{kg}$. $p = \sqrt{2 \times (1.675 \times 10^{-27} , \text{kg}) \times (1.602 \times 10^{-19} , \text{J})} = \sqrt{5.367 \times 10^{-46}} , \text{kg m/s}$ $p \approx 2.317 \times 10^{-23} , \text{kg m/s}$.

The de Broglie wavelength is $\lambda = \frac{h}{p}$. $\lambda = \frac{6.626 \times 10^{-34} , \text{J s}}{2.317 \times 10^{-23} , \text{kg m/s}} \approx 2.860 \times 10^{-11} , \text{m}$. $\lambda \approx 0.0286 , \text{nm}$.

Light of wavelength 200 nm is incident on a metal with a work function of 4.0 eV. What is the maximum velocity of the emitted photoelectrons? What is their de Broglie wavelength?

First, calculate the energy of the incident photon: $E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{200 \times 10^{-9} , \text{m}} = 9.939 \times 10^{-19} , \text{J}$.

Convert the photon energy to eV: $E (\text{eV}) = \frac{9.939 \times 10^{-19} , \text{J}}{1.602 \times 10^{-19} , \text{J/eV}} \approx 6.204 , \text{eV}$.

The maximum kinetic energy of the photoelectrons is $KE_{max} = E – \phi$. Given $\phi = 4.0 , \text{eV}$. $KE_{max} = 6.204 , \text{eV} – 4.0 , \text{eV} = 2.204 , \text{eV}$.

Convert the maximum kinetic energy to Joules: $KE_{max} = 2.204 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} \approx 3.531 \times 10^{-19} , \text{J}$.

To find the maximum velocity $v_{max}$, use $KE_{max} = \frac{1}{2}mv_{max}^2$: $v_{max}^2 = \frac{2 \times KE_{max}}{m} = \frac{2 \times 3.531 \times 10^{-19} , \text{J}}{9.109 \times 10^{-31} , \text{kg}} = 7.755 \times 10^{11} , \text{m}^2/\text{s}^2$. $v_{max} = \sqrt{7.755 \times 10^{11}} , \text{m/s} \approx 8.806 \times 10^5 , \text{m/s}$.

Now, calculate the de Broglie wavelength of the photoelectrons using $\lambda = \frac{h}{mv_{max}}$: $\lambda = \frac{6.626 \times 10^{-34} , \text{J s}}{(9.109 \times 10^{-31} , \text{kg})(8.806 \times 10^5 , \text{m/s})} = \frac{6.626 \times 10^{-34}}{80.21 \times 10^{-26}} , \text{m}$ $\lambda \approx 0.0826 \times 10^{-8} , \text{m} = 0.826 \times 10^{-9} , \text{m} = 0.826 , \text{nm}$.

A beam of electrons has a de Broglie wavelength of 0.05 nm. What is the momentum of each electron? What accelerating voltage was required to produce this beam?

The de Broglie wavelength is given by $\lambda = \frac{h}{p}$, so the momentum $p = \frac{h}{\lambda}$. Given $\lambda = 0.05 , \text{nm} = 0.05 \times 10^{-9} , \text{m} = 5 \times 10^{-11} , \text{m}$. $p = \frac{6.626 \times 10^{-34} , \text{J s}}{5 \times 10^{-11} , \text{m}} = 1.3252 \times 10^{-23} , \text{kg m/s}$.

The kinetic energy of the electron is $KE = \frac{p^2}{2m}$, where $m$ is the mass of the electron ($9.109 \times 10^{-31} , \text{kg}$). $KE = \frac{(1.3252 \times 10^{-23} , \text{kg m/s})^2}{2(9.109 \times 10^{-31} , \text{kg})} = \frac{1.75615 \times 10^{-46}}{1.8218 \times 10^{-30}} , \text{J}$ $KE \approx 9.6397 \times 10^{-17} , \text{J}$.

The kinetic energy gained by an electron accelerated through a potential difference $V$ is $KE = eV$. $V = \frac{KE}{e} = \frac{9.6397 \times 10^{-17} , \text{J}}{1.602 \times 10^{-19} , \text{C}} \approx 601.7 , \text{V}$.

If the energy of a photon is equal to the kinetic energy of an electron, what is the ratio of the wavelength of the photon to the de Broglie wavelength of the electron?

Let the energy of the photon be $E_{photon}$ and its wavelength be $\lambda_{photon}$. Then $E_{photon} = \frac{hc}{\lambda_{photon}}$. Let the kinetic energy of the electron be $KE_{electron}$ and its de Broglie wavelength be $\lambda_{deBroglie}$. Then $KE_{electron} = \frac{p^2}{2m} = \frac{(h/\lambda_{deBroglie})^2}{2m} = \frac{h^2}{2m\lambda_{deBroglie}^2}$.

Given $E_{photon} = KE_{electron}$: $\frac{hc}{\lambda_{photon}} = \frac{h^2}{2m\lambda_{deBroglie}^2}$.

We need to find the ratio $\frac{\lambda_{photon}}{\lambda_{deBroglie}}$. Rearranging the equation: $\frac{\lambda_{photon}}{\lambda_{deBroglie}^2} = \frac{2mc}{h}$ $\frac{\lambda_{photon}}{\lambda_{deBroglie}} = \frac{2mc\lambda_{deBroglie}}{h}$.

We also know that $KE_{electron} = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2KE_{electron}}{m}}$. And $\lambda_{deBroglie} = \frac{h}{mv} = \frac{h}{m\sqrt{\frac{2KE_{electron}}{m}}} = \frac{h}{\sqrt{2mKE_{electron}}}$. Since $KE_{electron} = E_{photon} = \frac{hc}{\lambda_{photon}}$, $\lambda_{deBroglie} = \frac{h}{\sqrt{2m\frac{hc}{\lambda_{photon}}}} = \frac{h}{\sqrt{\frac{2mhc}{\lambda_{photon}}}} = h \sqrt{\frac{\lambda_{photon}}{2mhc}} = \sqrt{\frac{h^2\lambda_{photon}}{2mhc}} = \sqrt{\frac{h\lambda_{photon}}{2mc}}$. Squaring both sides, $\lambda_{deBroglie}^2 = \frac{h\lambda_{photon}}{2mc}$. So, $\frac{\lambda_{photon}}{\lambda_{deBroglie}^2} = \frac{\lambda_{photon}}{\frac{h\lambda_{photon}}{2mc}} = \frac{2mc}{h}$.

From $\frac{hc}{\lambda_{photon}} = \frac{h^2}{2m\lambda_{deBroglie}^2}$, we get $\frac{\lambda_{photon}}{\lambda_{deBroglie}} = \frac{2mc\lambda_{deBroglie}}{h}$. We have $\lambda_{deBroglie} = \frac{h}{p} = \frac{h}{\sqrt{2mKE_{electron}}}$. Since $KE_{electron} = E_{photon}$, $\lambda_{deBroglie} = \frac{h}{\sqrt{2m\frac{hc}{\lambda_{photon}}}}$. $\frac{\lambda_{photon}}{\lambda_{deBroglie}} = \frac{\lambda_{photon}}{\frac{h}{\sqrt{2m\frac{hc}{\lambda_{photon}}}}} = \frac{\lambda_{photon}\sqrt{2m\frac{hc}{\lambda_{photon}}}}{h} = \frac{\sqrt{\lambda_{photon}^2 2m\frac{hc}{\lambda_{photon}}}}{h} = \frac{\sqrt{2mhc\lambda_{photon}}}{h} = \sqrt{\frac{2mc\lambda_{photon}}{h}}$. Using $E_{photon} = KE_{electron}$: $\frac{hc}{\lambda_{photon}} = \frac{1}{2}mv^2$, so $\lambda_{photon} = \frac{2hc}{mv^2}$. $\frac{\lambda_{photon}}{\lambda_{deBroglie}} = \frac{\lambda_{photon}}{h/mv} = \frac{mv\lambda_{photon}}{h} = \frac{mv}{h} \frac{2hc}{mv^2} = \frac{2hc}{hv} = \frac{2c}{v}$.

A proton and an electron have the same kinetic energy. What is the ratio of their de Broglie wavelengths?

The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$. For the proton, $\lambda_p = \frac{h}{\sqrt{2m_p KE}}$. For the electron, $\lambda_e = \frac{h}{\sqrt{2m_e KE}}$. The ratio of their de Broglie wavelengths is: $\frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2m_p KE}}}{\frac{h}{\sqrt{2m_e KE}}} = \sqrt{\frac{2m_e KE}{2m_p KE}} = \sqrt{\frac{m_e}{m_p}}$. The mass of an electron $m_e = 9.109 \times 10^{-31} , \text{kg}$. The mass of a proton $m_p = 1.672 \times 10^{-27} , \text{kg}$. $\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{9.109 \times 10^{-31}}{1.672 \times 10^{-27}}} = \sqrt{0.0005448} \approx 0.0233$.

A photon with a wavelength of 200nm strikes a metal with a work function of 3eV. What is the kinetic energy and de Broglie wavelength of the emitted electron?

First, calculate the energy of the incident photon: $E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{200 \times 10^{-9} , \text{m}} = 9.939 \times 10^{-19} , \text{J}$.

Convert the photon energy to eV: $E (\text{eV}) = \frac{9.939 \times 10^{-19} , \text{J}}{1.602 \times 10^{-19} , \text{J/eV}} \approx 6.20 , \text{eV}$.

The kinetic energy of the emitted electron is given by the photoelectric equation: $KE = E – \phi$. Given $\phi = 3 , \text{eV}$. $KE = 6.20 , \text{eV} – 3 , \text{eV} = 3.20 , \text{eV}$.

Convert the kinetic energy to Joules: $KE = 3.20 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} = 5.1264 \times 10^{-19} , \text{J}$.

Now, calculate the momentum of the electron using $KE = \frac{p^2}{2m}$: $p = \sqrt{2mKE} = \sqrt{2(9.109 \times 10^{-31} , \text{kg})(5.1264 \times 10^{-19} , \text{J})} = \sqrt{93.25 \times 10^{-50}} , \text{kg m/s}$ $p \approx 9.656 \times 10^{-25} , \text{kg m/s}$.

Finally, calculate the de Broglie wavelength of the electron: $\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} , \text{J s}}{9.656 \times 10^{-25} , \text{kg m/s}} \approx 6.86 \times 10^{-10} , \text{m} = 0.686 , \text{nm}$.

What is the ratio of the de Broglie wavelengths of a proton and an electron, if they have the same momentum?

The de Broglie wavelength is given by $\lambda = \frac{h}{p}$. For a proton, $\lambda_p = \frac{h}{p_p}$. For an electron, $\lambda_e = \frac{h}{p_e}$.

If the proton and electron have the same momentum, $p_p = p_e = p$. The ratio of their de Broglie wavelengths is: $\frac{\lambda_p}{\lambda_e} = \frac{h/p}{h/p} = 1$.

Explain the significance of the Davisson-Germer experiment in the context of de Broglie’s hypothesis.

The Davisson-Germer experiment, conducted in the 1920s, provided the first direct experimental evidence of the wave nature of electrons, thereby confirming de Broglie’s hypothesis.

De Broglie’s hypothesis, proposed in 1924, suggested that all matter exhibits wave-like properties, with a wavelength $\lambda = h/p$. This was a revolutionary idea as it extended the wave-particle duality, previously known for light, to matter.

The Davisson-Germer experiment involved firing a beam of electrons at a nickel target and observing the scattering pattern of the electrons. They observed that the scattered electrons showed a diffraction pattern, similar to what is observed when waves pass through a grating. Specifically, they found a peak in the intensity of scattered electrons at certain angles.

This diffraction pattern could only be explained by considering the wave nature of electrons. The wavelength of the electrons, calculated from the diffraction pattern, matched the wavelength predicted by de Broglie’s formula ($\lambda = h/p$), where $p$ was the momentum of the electrons determined by the accelerating voltage.

The significance of the Davisson-Germer experiment is that it:

1. Provided experimental confirmation of de Broglie’s hypothesis: It showed that electrons, previously considered particles, exhibit wave-like behavior.
2. Established the wave-particle duality for matter: It solidified the concept that matter, like light, has both wave and particle characteristics.
3. Opened the door for the development of wave mechanics: It played a crucial role in the development of quantum mechanics and our understanding of the behavior of matter at the atomic and subatomic level.

The experiment demonstrated that de Broglie’s theoretical concept was not just a mathematical abstraction but a physical reality, fundamentally changing our understanding of the nature of matter.

What happens to the de Broglie wavelength of an electron as its kinetic energy approaches the rest mass energy?

As the kinetic energy of an electron approaches its rest mass energy ($E_0 = mc^2$), its speed becomes a significant fraction of the speed of light, and we must use relativistic expressions for momentum and kinetic energy.

The relativistic kinetic energy is given by $KE = (\gamma – 1)mc^2$, where $\gamma = \frac{1}{\sqrt{1 – v^2/c^2}}$ is the Lorentz factor. As the kinetic energy approaches the rest mass energy, $\gamma – 1 \approx 1$, so $\gamma \approx 2$. This means the velocity of the electron is approaching the speed of light.

The relativistic momentum is given by $p = \gamma mv$. The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\gamma mv}$.

As the kinetic energy approaches the rest mass energy, the velocity $v$ approaches $c$, and the Lorentz factor $\gamma$ increases significantly. Since the momentum $p = \gamma mv$ increases as $\gamma$ increases, the de Broglie wavelength $\lambda = \frac{h}{p}$ will decrease.

In the extreme relativistic limit where the kinetic energy is much greater than the rest mass energy, the momentum becomes approximately $p \approx \frac{E}{c}$, where $E$ is the total energy. However, in the specific case where kinetic energy approaches the rest mass energy, the velocity is approaching but not equal to $c$, and $\gamma$ is finite but greater than 1. The momentum increases, and therefore, the de Broglie wavelength of the electron decreases, approaching a very small value.

Calculate the de Broglie wavelength of a 1 eV electron, a 1 eV photon, and a 1 eV neutron. What do you notice?

First, convert the kinetic energy/energy to Joules: $1 , \text{eV} = 1.602 \times 10^{-19} , \text{J}$.

Electron: $KE = \frac{p^2}{2m_e} \implies p = \sqrt{2m_e KE} = \sqrt{2(9.109 \times 10^{-31} , \text{kg})(1.602 \times 10^{-19} , \text{J})} \approx 5.40 \times 10^{-25} , \text{kg m/s}$. $\lambda_e = \frac{h}{p} = \frac{6.626 \times 10^{-34} , \text{J s}}{5.40 \times 10^{-25} , \text{kg m/s}} \approx 1.227 \times 10^{-9} , \text{m} = 1.227 , \text{nm}$.

Photon: $E = \frac{hc}{\lambda_{photon}} \implies \lambda_{photon} = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} , \text{J s})(3 \times 10^8 , \text{m/s})}{1.602 \times 10^{-19} , \text{J}} \approx 1.240 \times 10^{-6} , \text{m} = 1240 , \text{nm}$.

Neutron: $KE = \frac{p^2}{2m_n} \implies p = \sqrt{2m_n KE} = \sqrt{2(1.675 \times 10^{-27} , \text{kg})(1.602 \times 10^{-19} , \text{J})} \approx 7.32 \times 10^{-24} , \text{kg m/s}$. $\lambda_n = \frac{h}{p} = \frac{6.626 \times 10^{-34} , \text{J s}}{7.32 \times 10^{-24} , \text{kg m/s}} \approx 0.905 \times 10^{-10} , \text{m} = 0.0905 , \text{nm}$.

Notice: The de Broglie wavelengths are significantly different for the three particles, even though they have the same kinetic energy. This is because the de Broglie wavelength depends on the momentum ($\lambda = h/p$), and for the same kinetic energy, particles with different masses have different momenta ($KE = p^2/2m \implies p = \sqrt{2mKE}$).

• The photon has the longest wavelength because its momentum is the smallest ($E = pc \implies p = E/c$).
• The neutron, being much more massive than the electron, has a much larger momentum for the same kinetic energy, and therefore a much smaller de Broglie wavelength.
• The electron’s de Broglie wavelength is between that of the photon and the neutron.

A monochromatic light source emits photons with a wavelength of 400 nm at a rate of 10^15 photons per second. If all these photons are absorbed by a metal surface of area 1 cm², what is the radiation pressure exerted on the metal surface?

The momentum of a single photon is $p = \frac{h}{\lambda}$. Given $\lambda = 400 , \text{nm} = 400 \times 10^{-9} , \text{m}$. $p = \frac{6.626 \times 10^{-34} , \text{J s}}{400 \times 10^{-9} , \text{m}} = 1.6565 \times 10^{-27} , \text{kg m/s}$.

When a photon is absorbed by a surface, the change in momentum of the photon is $\Delta p = p – 0 = p$. The force exerted by a single photon is the rate of change of momentum, but here we consider the impulse delivered by each photon. The force exerted by $N$ photons per second is $F = N \frac{\Delta p}{\Delta t} = Np$, since $\Delta t = 1$ second. The number of photons per second $N = 10^{15}$. The total force $F = (10^{15}) \times (1.6565 \times 10^{-27} , \text{kg m/s}) = 1.6565 \times 10^{-12} , \text{N}$.

The area of the metal surface $A = 1 , \text{cm}^2 = (1 \times 10^{-2} , \text{m})^2 = 1 \times 10^{-4} , \text{m}^2$. The radiation pressure $P = \frac{F}{A} = \frac{1.6565 \times 10^{-12} , \text{N}}{1 \times 10^{-4} , \text{m}^2} = 1.6565 \times 10^{-8} , \text{Pa}$.

An electron microscope uses electrons with a kinetic energy of 20 keV. What is the resolving power of the microscope (i.e., the smallest resolvable distance)?

The resolving power of an electron microscope is approximately equal to the de Broglie wavelength of the electrons. The kinetic energy of the electrons is $KE = 20 , \text{keV} = 20 \times 10^3 , \text{eV}$. Convert the kinetic energy to Joules: $KE = 20 \times 10^3 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} = 3.204 \times 10^{-15} , \text{J}$.

The kinetic energy is related to momentum by $KE = \frac{p^2}{2m}$, so $p = \sqrt{2mKE}$. The mass of an electron $m_e = 9.109 \times 10^{-31} , \text{kg}$. $p = \sqrt{2(9.109 \times 10^{-31} , \text{kg})(3.204 \times 10^{-15} , \text{J})} = \sqrt{5.836 \times 10^{-45}} , \text{kg m/s}$ $p \approx 7.639 \times 10^{-23} , \text{kg m/s}$.

The de Broglie wavelength is $\lambda = \frac{h}{p}$. $\lambda = \frac{6.626 \times 10^{-34} , \text{J s}}{7.639 \times 10^{-23} , \text{kg m/s}} \approx 8.674 \times 10^{-12} , \text{m}$. $\lambda \approx 0.00867 , \text{nm}$.

The resolving power of the microscope is approximately equal to this de Broglie wavelength.

Calculate the uncertainty in the velocity of an electron if its position is known with an uncertainty of 1 nm, using the Heisenberg uncertainty principle.

The Heisenberg uncertainty principle states that $\Delta x \Delta p \ge \frac{\hbar}{2}$, where $\Delta x$ is the uncertainty in position and $\Delta p$ is the uncertainty in momentum. We are given $\Delta x = 1 , \text{nm} = 1 \times 10^{-9} , \text{m}$. The momentum of an electron is $p = mv$, so the uncertainty in momentum is $\Delta p = m \Delta v$, where $m$ is the mass of the electron ($9.109 \times 10^{-31} , \text{kg}$). Planck’s reduced constant $\hbar = \frac{h}{2\pi} = \frac{6.626 \times 10^{-34} , \text{J s}}{2\pi} \approx 1.054 \times 10^{-34} , \text{J s}$.

Substituting into the uncertainty principle: $(1 \times 10^{-9} , \text{m})(9.109 \times 10^{-31} , \text{kg} \times \Delta v) \ge \frac{1.054 \times 10^{-34} , \text{J s}}{2}$ $9.109 \times 10^{-40} , \text{kg m} \times \Delta v \ge 0.527 \times 10^{-34} , \text{J s}$ $\Delta v \ge \frac{0.527 \times 10^{-34} , \text{J s}}{9.109 \times 10^{-40} , \text{kg m}}$ $\Delta v \ge 0.0578 \times 10^{6} , \text{m/s}$ $\Delta v \ge 5.78 \times 10^{4} , \text{m/s}$.

A beam of electrons with energy of 100 eV is incident upon a crystal with an inter-atomic distance of 0.2 nm. At what angle would you expect to find the first diffraction maximum of the electrons?

For electron diffraction, we can use Bragg’s law: $n\lambda = 2d \sin \theta$, where $n$ is the order of the maximum, $\lambda$ is the de Broglie wavelength of the electrons, $d$ is the inter-atomic distance, and $\theta$ is the angle of incidence (and diffraction). For the first maximum, $n=1$.

First, calculate the kinetic energy of the electrons in Joules: $KE = 100 , \text{eV} \times 1.602 \times 10^{-19} , \text{J/eV} = 1.602 \times 10^{-17} , \text{J}$.

Next, calculate the momentum of the electrons using $KE = \frac{p^2}{2m}$: $p = \sqrt{2mKE} = \sqrt{2(9.109 \times 10^{-31} , \text{kg})(1.602 \times 10^{-17} , \text{J})} = \sqrt{2.918 \times 10^{-47}} , \text{kg m/s}$ $p \approx 5.402 \times 10^{-24} , \text{kg m/s}$.

Now, calculate the de Broglie wavelength of the electrons: $\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} , \text{J s}}{5.402 \times 10^{-24} , \text{kg m/s}} \approx 1.226 \times 10^{-10} , \text{m} = 0.1226 , \text{nm}$.

Given the inter-atomic distance $d = 0.2 , \text{nm}$. Using Bragg’s law for the first maximum ($n=1$): $1 \times \lambda = 2d \sin \theta$ $\sin \theta = \frac{\lambda}{2d} = \frac{0.1226 , \text{nm}}{2 \times 0.2 , \text{nm}} = \frac{0.1226}{0.4} = 0.3065$. $\theta = \arcsin(0.3065) \approx 17.87^\circ$.

Two light sources emit photons of the same energy. One source emits photons isotropically (equally in all directions), while the other emits photons in a collimated beam. Compare the force exerted by the two sources on a perfectly absorbing surface.

Let the number of photons emitted per second by both sources be $N$, and the energy of each photon be $E$. The momentum of each photon is $p = E/c$.

For the collimated beam, all $N$ photons are incident on the surface perpendicularly. When a photon is absorbed, the change in momentum is $\Delta p = p$. The total change in momentum per second (force) exerted by the collimated beam is $F_{collimated} = N \Delta p = N \frac{E}{c}$.

For the isotropic source, the photons are emitted in all directions. Only the component of the momentum perpendicular to the surface contributes to the force. If we consider a small area $dA$ of the surface, the number of photons hitting this area per second will be a fraction of the total number of photons emitted. The average component of momentum perpendicular to the surface is more complex to calculate precisely without integration over the hemisphere, but we can reason qualitatively. On average, the perpendicular component of momentum for isotropically emitted photons will be less than the full momentum of the photons in the collimated beam.

Consider a simplified case: for every photon heading directly towards the surface in the collimated beam, there will be photons emitted by the isotropic source in a range of directions. Only those photons directed towards the surface will contribute to the force, and their perpendicular component of momentum will be less than $E/c$.

Therefore, for the same number of photons with the same energy, the force exerted by the collimated beam will be greater than the force exerted by the isotropic source on a perfectly absorbing surface. The collimated beam directs all the photon momentum towards the surface, while the isotropic source spreads the momentum in all directions.

Explain how the Heisenberg uncertainty principle limits our ability to measure both the position and momentum of a particle.

The Heisenberg uncertainty principle is a fundamental principle in quantum mechanics that places a limit on the precision with which certain pairs of physical properties of a particle, known as conjugate variables, can be known simultaneously. The most well-known pair is position ($x$) and momentum ($p$), and the principle is mathematically expressed as:

$\Delta x \Delta p \ge \frac{\hbar}{2}$

where $\Delta x$ is the uncertainty in the measurement of position, $\Delta p$ is the uncertainty in the measurement of momentum, and $\hbar$ is the reduced Planck constant.

This inequality implies that the more accurately we know the position of a particle, the less accurately we can simultaneously know its momentum, and vice versa. It’s crucial to understand that this is not a limitation of our measurement equipment; it is an inherent property of quantum systems. No matter how sophisticated our measuring devices become, this fundamental limit will always exist.

The wave nature of particles provides a way to understand this principle. Consider an electron. According to de Broglie’s hypothesis, it has an associated wave.

• A pure sine wave, like $\sin(kx)$: This represents an electron with a perfectly well-defined momentum because the wave number $k$ is directly related to the momentum ($p = \hbar k$). However, a pure sine wave extends infinitely in space, meaning the position of the electron is completely uncertain – we have no idea where along the x-axis the electron is.

• Adding another sine wave, like $A\sin(kx+\phi)$: This begins the process of superposing waves. By adding waves with different phases and amplitudes, we can start to create interference patterns. When we add waves constructively in a certain region, the amplitude increases there, suggesting a higher probability of finding the electron in that region. This means we are gaining some information about the electron’s position. However, adding another wave introduces another possible momentum value (related to the wave number of that second wave), making the momentum less precisely defined.

• Creating a wave packet (by adding many waves): As we superpose more and more waves with different wave numbers (and thus different momenta), we can create a localized wave packet. A narrow wave packet signifies a high certainty in the electron’s position – we know it’s likely to be found within the confines of the packet. However, to construct such a localized packet, we need to combine waves with a range of different wave numbers (and therefore momenta). The more localized the packet becomes (better position knowledge), the wider the range of wave numbers needed, leading to a greater uncertainty in the electron’s momentum.

In essence, there’s a trade-off. A wave with a single, well-defined wavelength (and thus momentum) is spread out in space. To localize the particle in space, we need to combine waves of different wavelengths (and thus different momenta). The more localized we make the particle, the broader the range of momenta we need to include in the superposition, and thus the greater the uncertainty in the momentum. This inherent connection between the wave and particle nature of matter dictates the limitations described by the Heisenberg uncertainty principle.

A free electron is confined within a one dimensional region of 1 nm. Calculate the minimum kinetic energy of the electron.

The uncertainty in the position of the electron is $\Delta x = 1 , \text{nm} = 1 \times 10^{-9} , \text{m}$. According to the Heisenberg uncertainty principle, $\Delta x \Delta p \ge \frac{\hbar}{2}$. The minimum uncertainty in momentum is $\Delta p_{min} = \frac{\hbar}{2 \Delta x} = \frac{1.054 \times 10^{-34} , \text{J s}}{2 \times 1 \times 10^{-9} , \text{m}} = 0.527 \times 10^{-25} , \text{kg m/s}$.

The minimum momentum of the electron can be approximated by this uncertainty: $p_{min} \approx \Delta p_{min} = 0.527 \times 10^{-25} , \text{kg m/s}$. The minimum kinetic energy of the electron is $KE_{min} = \frac{p_{min}^2}{2m}$, where $m$ is the mass of the electron ($9.109 \times 10^{-31} , \text{kg}$). $KE_{min} = \frac{(0.527 \times 10^{-25} , \text{kg m/s})^2}{2 \times 9.109 \times 10^{-31} , \text{kg}} = \frac{0.2777 \times 10^{-50}}{18.218 \times 10^{-31}} , \text{J}$ $KE_{min} \approx 0.01524 \times 10^{-19} , \text{J} = 1.524 \times 10^{-21} , \text{J}$.

Convert the kinetic energy to electron volts: $KE_{min} (\text{eV}) = \frac{1.524 \times 10^{-21} , \text{J}}{1.602 \times 10^{-19} , \text{J/eV}} \approx 0.00951 , \text{eV} = 9.51 , \text{meV}$.

Derive the expression for the de Broglie wavelength of a relativistic particle.

The de Broglie wavelength is given by $\lambda = \frac{h}{p}$, where $p$ is the momentum of the particle. For a relativistic particle, the relationship between energy ($E$), momentum ($p$), and rest mass ($m_0$) is given by the relativistic energy-momentum relation: $E^2 = (pc)^2 + (m_0 c^2)^2$

The total energy of the particle is also given by $E = \gamma m_0 c^2$, where $\gamma = \frac{1}{\sqrt{1 – v^2/c^2}}$ is the Lorentz factor. The relativistic momentum is $p = \gamma m_0 v$.

From the energy-momentum relation, we can solve for $p$: $(pc)^2 = E^2 – (m_0 c^2)^2$ $p^2 c^2 = (\gamma m_0 c^2)^2 – (m_0 c^2)^2$ $p^2 c^2 = m_0^2 c^4 (\gamma^2 – 1)$

We know that $\gamma^2 – 1 = \frac{1}{1 – v^2/c^2} – 1 = \frac{1 – (1 – v^2/c^2)}{1 – v^2/c^2} = \frac{v^2/c^2}{1 – v^2/c^2} = \frac{v^2}{c^2(1 – v^2/c^2)} = \frac{v^2}{c^2} \gamma^2$. Substituting this back into the equation for $p^2 c^2$: $p^2 c^2 = m_0^2 c^4 \frac{v^2}{c^2} \gamma^2 = (m_0 \gamma v)^2 c^2$ $p^2 c^2 = p^2 c^2$, this doesn’t directly help us find $p$ in terms of $E$ and $m_0$.

Let’s go back to $p^2 c^2 = E^2 – (m_0 c^2)^2$: $p^2 = \frac{1}{c^2} [E^2 – m_0^2 c^4]$ $p = \frac{1}{c} \sqrt{E^2 – m_0^2 c^4}$

Substitute this into the de Broglie wavelength formula: $\lambda = \frac{h}{p} = \frac{h}{\frac{1}{c} \sqrt{E^2 – m_0^2 c^4}} = \frac{hc}{\sqrt{E^2 – m_0^2 c^4}}$

This is one form of the relativistic de Broglie wavelength. We can also express it in terms of kinetic energy ($KE = E – m_0 c^2$, so $E = KE + m_0 c^2$): $\lambda = \frac{hc}{\sqrt{(KE + m_0 c^2)^2 – m_0^2 c^4}}$ $\lambda = \frac{hc}{\sqrt{KE^2 + 2 KEm_0 c^2 + m_0^2 c^4 – m_0^2 c^4}}$ $\lambda = \frac{hc}{\sqrt{KE^2 + 2 KEm_0 c^2}}$ $\lambda = \frac{hc}{\sqrt{KE(KE + 2 m_0 c^2)}}$

Alternatively, we can use $p = \gamma m_0 v$: $\lambda = \frac{h}{\gamma m_0 v}$ We know that $E = \gamma m_0 c^2$, so $\gamma = \frac{E}{m_0 c^2}$. Also, $E^2 = p^2 c^2 + m_0^2 c^4$, so $p = \frac{\sqrt{E^2 – m_0^2 c^4}}{c}$. $\lambda = \frac{h}{\frac{\sqrt{E^2 – m_0^2 c^4}}{c}} = \frac{hc}{\sqrt{E^2 – m_0^2 c^4}}$. Final Answer:

$\boxed{\lambda = \frac{hc}{\sqrt{E^2 – m_0^2 c^4}}}$

Explain why a high accelerating voltage is needed in an electron microscope to achieve higher resolutions.

The resolving power of a microscope, which is the ability to distinguish between two closely spaced objects, is fundamentally limited by the wavelength of the illumination source. In an electron microscope, the “illumination” is a beam of electrons, and their effective wavelength is given by the de Broglie wavelength, $\lambda = h/p$, where $p$ is the momentum of the electron.

Electrons in an electron microscope are accelerated through a potential difference (accelerating voltage), gaining kinetic energy. The kinetic energy ($KE$) of an electron accelerated through a voltage $V$ is given by $KE = eV$.

For non-relativistic electrons, $KE = \frac{1}{2}mv^2$. Increasing the accelerating voltage increases the kinetic energy of the electrons, which in turn increases their velocity and momentum ($p = mv$). According to the de Broglie equation, an increase in momentum leads to a decrease in the de Broglie wavelength.

A shorter de Broglie wavelength means that the electrons can be used to resolve smaller features. Just as a shorter wavelength of light in an optical microscope allows for better resolution, a shorter de Broglie wavelength of electrons in an electron microscope allows for the imaging of finer details.

Therefore, a higher accelerating voltage imparts a higher kinetic energy and thus a greater momentum to the electrons, resulting in a shorter de Broglie wavelength. This shorter wavelength is crucial for achieving higher resolutions in an electron microscope, enabling the visualization of smaller structures. At very high accelerating voltages, relativistic effects become significant, but the fundamental principle remains the same: higher energy electrons have a shorter effective wavelength.

Describe how a photocell works and how the photoelectric effect is used in its operation.

A photocell, also known as a photoelectric cell, is a device that converts light energy into electrical energy. Its operation is based on the photoelectric effect.

Here’s how it works:

1. Components: A basic photocell consists of two electrodes within an evacuated glass or quartz tube:

   • Cathode: This electrode is made of a material with a low work function (e.g., alkali metals like sodium or potassium). It serves as the emitter of photoelectrons.
   • Anode: This electrode is typically a wire or mesh and is kept at a positive potential relative to the cathode. It acts as the collector of the emitted photoelectrons.

2. Photoelectric Effect: When light of a suitable frequency (above the threshold frequency of the cathode material) falls on the cathode, photons in the light transfer their energy to the electrons in the cathode material.

3. Electron Emission: If a photon has enough energy (greater than the work function of the cathode material), it can liberate an electron from the surface of the cathode. This emitted electron is called a photoelectron.

4. Current Generation: The positively charged anode attracts these photoelectrons. The flow of these electrons from the cathode to the anode constitutes an electric current in the external circuit connected to the photocell.

5. Current Dependence on Light:

   • Intensity: The number of photoelectrons emitted per unit time is directly proportional to the intensity of the incident light (provided the frequency is above the threshold frequency). A brighter light will result in a larger photocurrent.
   • Frequency: The kinetic energy of the emitted photoelectrons depends on the frequency of the incident light. Higher frequency light will result in photoelectrons with higher kinetic energy. However, the photocurrent is not directly dependent on the frequency (as long as the frequency is above the threshold).
   • Stopping Potential: A negative potential applied to the anode can stop the photocurrent. The minimum negative potential required to stop the photocurrent (stopping potential) is related to the maximum kinetic energy of the emitted photoelectrons.

In summary, the photocell utilizes the photoelectric effect to convert light into electricity. Incident light on the cathode causes the emission of photoelectrons, which are then attracted to the anode, creating a current. The magnitude of this current is dependent on the intensity of the incident light. Phototocells are used in various applications such as light meters, automatic door openers, solar cells (though solar cells are more complex semiconductor devices), and light-sensitive switches.

Explain the concept of a wave packet and its relation to the wave-particle duality.

A wave packet, also known as a wave pulse, is a localized wave disturbance formed by the superposition of multiple waves with slightly different frequencies, wave numbers, and amplitudes. Imagine adding several sine waves together; if you carefully choose their properties, they will interfere constructively in a small region of space and destructively elsewhere, creating a localized “packet” of wave energy.

The concept of a wave packet is crucial for understanding wave-particle duality because it provides a way to reconcile the seemingly contradictory wave-like and particle-like behaviors of quantum entities like electrons and photons.

Here’s how it relates:

• Particle-like Localization: A particle is characterized by its localization in space. A wave packet, by its very nature, is localized. The region where the constituent waves interfere constructively is the region where the “particle” is most likely to be found. The sharper the wave packet (more confined in space), the more precisely the particle’s position is known.

• Wave-like Properties within the Packet: The wave nature is still present within the wave packet. The constituent waves that form the packet have specific wavelengths and frequencies, which relate to the momentum and energy of the particle according to the de Broglie relations ($p = \hbar k$) and ($E = \hbar \omega$). The wave packet can exhibit wave-like behaviors such as diffraction and interference as it propagates.

• Uncertainty Principle: The more localized the wave packet is (smaller $\Delta x$), the wider the range of frequencies and wave numbers needed to create it (larger $\Delta k$ and $\Delta \omega$). This translates directly to the Heisenberg uncertainty principle. A narrower spatial extent implies a greater uncertainty in momentum (since momentum is related to wave number), and a shorter time duration of the packet implies a greater uncertainty in energy (since energy is related to frequency).

In essence, the wave packet is a mathematical construct that allows us to represent a quantum entity as something that is both localized (like a particle) and possesses wave-like properties (wavelength and frequency). The particle is not a tiny, hard sphere, nor is it a delocalized wave spread out in space. Instead, it’s a localized packet of wave energy, and the properties of this packet dictate the probability of finding the particle at a particular location and with a particular momentum. The wave packet is the embodiment of the wave-particle duality, a localized wave that carries the energy and momentum of the “particle”.

How can the Compton effect be used to demonstrate the particle nature of electromagnetic radiation?

The Compton effect, discovered by Arthur Compton in the 1920s, is the scattering of a photon by a charged particle, usually an electron, resulting in a decrease in energy (increase in wavelength) of the photon. This phenomenon provides compelling evidence for the particle nature of electromagnetic radiation.

Here’s why:

1. Classical Explanation Fails: Classically, when an electromagnetic wave interacts with a charged particle, the particle oscillates at the frequency of the incident wave and re-radiates electromagnetic waves at the same frequency. There should be no change in the wavelength (or frequency) of the scattered radiation. However, experiments showed that the scattered photons had a longer wavelength than the incident photons, a phenomenon that classical wave theory could not explain.

2. Particle-like Collision: The Compton effect can be accurately explained by treating the interaction as a collision between two particles: a photon and an electron. Just like billiard balls colliding, energy and momentum are conserved in this interaction.

3. Energy and Momentum Conservation:

   • Energy Conservation: The energy of the incident photon ($E = hf$) is partially transferred to the electron as kinetic energy ($KE_e$), and the remaining energy is carried by the scattered photon ($E’ = hf’$). Thus, $hf = hf’ + KE_e$. Since $f’ < f$, the scattered photon has lower energy.
   • Momentum Conservation: The momentum of the incident photon ($\vec{p} = \frac{h}{\lambda} \hat{n}$) is a vector. In the collision, momentum is conserved in both the direction of the incident photon and perpendicular to it. The incident photon’s momentum is shared between the scattered photon and the recoiling electron.

4. Compton Shift: By applying the laws of conservation of energy and momentum to the photon-electron collision, one can derive the Compton shift formula:

   $\Delta \lambda = \lambda’ – \lambda = \frac{h}{m_e c}(1 – \cos \theta)$

   where $\Delta \lambda$ is the change in wavelength, $h$ is Planck’s constant, $m_e$ is the mass of the electron, $c$ is the speed of light, and $\theta$ is the scattering angle of the photon. This formula precisely matches the experimental observations.

5. Photon as a Particle: The fact that the interaction can be described by the conservation laws of energy and momentum, which are fundamental to particle collisions, strongly supports the idea that photons, despite their wave-like properties in phenomena like interference, also possess particle-like properties, including momentum. The Compton effect demonstrates that photons behave like discrete packets of energy and momentum that can collide with other particles.

In conclusion, the Compton effect, with its observed shift in the wavelength of scattered photons, can only be consistently explained by treating the incident light as a stream of particles (photons) that collide with electrons, transferring energy and momentum in a manner consistent with the laws of conservation applicable to particle interactions. This provides strong evidence for the particle nature of electromagnetic radiation.

Derive an expression for the momentum of a photon in terms of its energy and the speed of light.

We can derive the expression for the momentum of a photon using the following relationships:

1. Energy of a photon: According to Planck’s quantum theory, the energy of a photon ($E$) is directly proportional to its frequency ($f$): $E = hf$ where $h$ is Planck’s constant.

2. Relationship between frequency, wavelength, and the speed of light: For any wave, the speed of propagation ($c$ for light in a vacuum) is related to its frequency ($f$) and wavelength ($\lambda$) by: $c = f\lambda$

3. Momentum of a photon (de Broglie relation for photons): Extending the concept of wave-particle duality, we can assign a momentum ($p$) to a photon based on its wavelength using the de Broglie relation: $p = \frac{h}{\lambda}$

Now, let’s combine these equations:

From the relationship between frequency, wavelength, and the speed of light, we can express the wavelength in terms of frequency and the speed of light: $\lambda = \frac{c}{f}$

Substitute this expression for $\lambda$ into the momentum equation: $p = \frac{h}{\frac{c}{f}} = \frac{hf}{c}$

Now, substitute the expression for the energy of a photon ($E = hf$) into this equation: $p = \frac{E}{c}$

Therefore, the momentum of a photon is given by the expression:

$\boxed{p = \frac{E}{c}}$

This equation shows that the momentum of a photon is directly proportional to its energy and inversely proportional to the speed of light. It’s a fundamental relationship in the physics of light and is essential for understanding phenomena like radiation pressure and the Compton effect.

Normalize the wavefunction $\psi(x) = A x(L-x)$ for a particle in a 1D box of length L (0 ≤ x ≤ L).

To normalize the wavefunction, we need to find the value of A such that the integral of the probability density over the domain is equal to 1: $$ \int_{0}^{L} |\psi(x)|^2 dx = 1 $$ Substitute the given wavefunction: $$ \int_{0}^{L} |A x(L-x)|^2 dx = 1 $$ $$ A^2 \int_{0}^{L} x^2 (L-x)^2 dx = 1 $$ $$ A^2 \int_{0}^{L} x^2 (L^2 – 2Lx + x^2) dx = 1 $$ $$ A^2 \int_{0}^{L} (L^2 x^2 – 2Lx^3 + x^4) dx = 1 $$ Now, integrate term by term: $$ A^2 \left[ L^2 \frac{x^3}{3} – 2L \frac{x^4}{4} + \frac{x^5}{5} \right]_{0}^{L} = 1 $$ $$ A^2 \left[ L^2 \frac{L^3}{3} – \frac{L}{2} L^4 + \frac{L^5}{5} \right] = 1 $$ $$ A^2 \left[ \frac{L^5}{3} – \frac{L^5}{2} + \frac{L^5}{5} \right] = 1 $$ Factor out $L^5$: $$ A^2 L^5 \left[ \frac{1}{3} – \frac{1}{2} + \frac{1}{5} \right] = 1 $$ Find a common denominator for the fractions (30): $$ A^2 L^5 \left[ \frac{10}{30} – \frac{15}{30} + \frac{6}{30} \right] = 1 $$ $$ A^2 L^5 \left[ \frac{10 – 15 + 6}{30} \right] = 1 $$ $$ A^2 L^5 \left[ \frac{1}{30} \right] = 1 $$ Solve for $A^2$: $$ A^2 = \frac{30}{L^5} $$ Take the square root to find A: $$ A = \sqrt{\frac{30}{L^5}} $$

A particle’s wavefunction is given by $\psi(x) = Ce^{-x/a}$ for x ≥ 0 and $\psi(x)=0$ for x less than 0. Find the normalization constant C.

To find the normalization constant C, we use the normalization condition: $$ \int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1 $$ Since $\psi(x) = 0$ for $x < 0$, the integral becomes: $$ \int_{0}^{\infty} |Ce^{-x/a}|^2 dx = 1 $$ $$ C^2 \int_{0}^{\infty} e^{-2x/a} dx = 1 $$ Now, evaluate the integral: $$ C^2 \left[ -\frac{a}{2} e^{-2x/a} \right]_{0}^{\infty} = 1 $$ $$ C^2 \left[ -\frac{a}{2} (e^{-\infty} – e^{0}) \right] = 1 $$ Since $e^{-\infty} = 0$ and $e^0 = 1$: $$ C^2 \left[ -\frac{a}{2} (0 – 1) \right] = 1 $$ $$ C^2 \left[ \frac{a}{2} \right] = 1 $$ Solve for $C^2$: $$ C^2 = \frac{2}{a} $$ Take the square root to find C: $$ C = \sqrt{\frac{2}{a}} $$ Note that we take the positive square root as the normalization constant is usually taken to be a positive real number.

The wavefunction for a particle is $\psi(x) = B e^{-|x|/a}$. Find the normalization constant B.

For a wavefunction to be normalized, the integral of the probability density over all space must be equal to 1: $\int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1$

Given $\psi(x) = B e^{-|x|/a}$, then $|\psi(x)|^2 = |B e^{-|x|/a}|^2 = B^2 |e^{-|x|/a}|^2 = B^2 e^{-2|x|/a}$. So, we have: $\int_{-\infty}^{\infty} B^2 e^{-2|x|/a} dx = 1$ $B^2 \int_{-\infty}^{\infty} e^{-2|x|/a} dx = 1$

We can split the integral into two parts due to the absolute value: $B^2 \left( \int_{-\infty}^{0} e^{2x/a} dx + \int_{0}^{\infty} e^{-2x/a} dx \right) = 1$

Evaluating the integrals: $\int_{-\infty}^{0} e^{2x/a} dx = \left[ \frac{a}{2} e^{2x/a} \right]{-\infty}^{0} = \frac{a}{2} (e^0 – \lim{x \to -\infty} e^{2x/a}) = \frac{a}{2} (1 – 0) = \frac{a}{2}$ $\int_{0}^{\infty} e^{-2x/a} dx = \left[ -\frac{a}{2} e^{-2x/a} \right]{0}^{\infty} = -\frac{a}{2} (\lim{x \to \infty} e^{-2x/a} – e^0) = -\frac{a}{2} (0 – 1) = \frac{a}{2}$

Substituting these back into the normalization condition: $B^2 \left( \frac{a}{2} + \frac{a}{2} \right) = 1$ $B^2 (a) = 1$ $B^2 = \frac{1}{a}$ $B = \frac{1}{\sqrt{a}}$

For the wavefunction $\psi(x) = A \sin(\pi x/L)$ for 0 ≤ x ≤ L, what is the probability of finding the particle between x = L/4 and x = 3L/4?

First, we need to normalize the wavefunction to find the value of A. The normalization condition is: $\int_{0}^{L} |\psi(x)|^2 dx = 1$ $\int_{0}^{L} |A \sin(\pi x/L)|^2 dx = 1$ $A^2 \int_{0}^{L} \sin^2(\pi x/L) dx = 1$

We use the identity $\sin^2(\theta) = \frac{1 – \cos(2\theta)}{2}$: $A^2 \int_{0}^{L} \frac{1 – \cos(2\pi x/L)}{2} dx = 1$ $\frac{A^2}{2} \left[ x – \frac{L}{2\pi} \sin(2\pi x/L) \right]_{0}^{L} = 1$ $\frac{A^2}{2} \left[ (L – \frac{L}{2\pi} \sin(2\pi)) – (0 – \frac{L}{2\pi} \sin(0)) \right] = 1$ $\frac{A^2}{2} (L – 0 – 0 + 0) = 1$ $\frac{A^2 L}{2} = 1$ $A^2 = \frac{2}{L}$ $A = \sqrt{\frac{2}{L}}$

Now we can calculate the probability of finding the particle between x = L/4 and x = 3L/4: $P(L/4 \le x \le 3L/4) = \int_{L/4}^{3L/4} |\psi(x)|^2 dx = \int_{L/4}^{3L/4} \frac{2}{L} \sin^2(\pi x/L) dx$ $P = \frac{2}{L} \int_{L/4}^{3L/4} \frac{1 – \cos(2\pi x/L)}{2} dx$ $P = \frac{1}{L} \left[ x – \frac{L}{2\pi} \sin(2\pi x/L) \right]_{L/4}^{3L/4}$ $P = \frac{1}{L} \left[ \left( \frac{3L}{4} – \frac{L}{2\pi} \sin\left(\frac{3\pi}{2}\right) \right) – \left( \frac{L}{4} – \frac{L}{2\pi} \sin\left(\frac{\pi}{2}\right) \right) \right]$ $P = \frac{1}{L} \left[ \frac{3L}{4} – \frac{L}{2\pi} (-1) – \frac{L}{4} + \frac{L}{2\pi} (1) \right]$ $P = \frac{1}{L} \left[ \frac{3L}{4} + \frac{L}{2\pi} – \frac{L}{4} + \frac{L}{2\pi} \right]$ $P = \frac{1}{L} \left[ \frac{2L}{4} + \frac{2L}{2\pi} \right]$ $P = \frac{1}{L} \left[ \frac{L}{2} + \frac{L}{\pi} \right]$ $P = \frac{1}{2} + \frac{1}{\pi}$ Final Answer: The final answer is: $\boxed{B = \frac{1}{\sqrt{a}}}$ and $\boxed{P = \frac{1}{2} + \frac{1}{\pi}}$

A particle in a 1D box has the wavefunction $\psi(x) = A\sin(\frac{3\pi x}{L})$. What is the probability density at x = L/2?

The probability density is given by $|\psi(x)|^2$. Given $\psi(x) = A\sin(\frac{3\pi x}{L})$. The probability density is $|\psi(x)|^2 = \left| A\sin\left(\frac{3\pi x}{L}\right) \right|^2 = A^2 \sin^2\left(\frac{3\pi x}{L}\right)$.

To find the probability density at $x = L/2$, we substitute $x = L/2$ into the expression for the probability density: $|\psi(L/2)|^2 = A^2 \sin^2\left(\frac{3\pi (L/2)}{L}\right) = A^2 \sin^2\left(\frac{3\pi}{2}\right)$.

We know that $\sin\left(\frac{3\pi}{2}\right) = -1$. So, $|\psi(L/2)|^2 = A^2 (-1)^2 = A^2 (1) = A^2$.

To have a physically meaningful probability density, the wavefunction should be normalized. For a particle in a 1D box of length L, the normalized wavefunction for the $n$-th state is $\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$. In this case, $n=3$, so $A = \sqrt{\frac{2}{L}}$.

Therefore, the probability density at x = L/2 is: $|\psi(L/2)|^2 = \left(\sqrt{\frac{2}{L}}\right)^2 = \frac{2}{L}$.

Given a normalized wavefunction, $\psi(x)$, what does the quantity $|\psi(x)|^2 dx$ represent?

For a normalized wavefunction, $\psi(x)$, the quantity $|\psi(x)|^2$ represents the probability density of finding the particle at a particular position $x$.

Therefore, the quantity $|\psi(x)|^2 dx$ represents the probability of finding the particle in the infinitesimal interval $dx$ around the position $x$.

Explain what is meant by the “Born interpretation” of the wavefunction.

The Born interpretation is a fundamental principle in quantum mechanics that provides the physical meaning of the wavefunction, $\psi(r, t)$, of a particle. It states that:

The probability of finding a particle in a small volume $dV$ around a point $r$ at time $t$ is given by the square of the absolute value of the wavefunction at that point and time, multiplied by the volume element $dV$:

$P(r \in dV) = |\psi(r, t)|^2 dV$

In one dimension, this becomes:

$P(x \in dx) = |\psi(x, t)|^2 dx$

Here’s a breakdown of the key aspects of the Born interpretation:

• Probability Density: $|\psi(r, t)|^2$ represents the probability density of finding the particle at a specific point in space and time. It’s a measure of the likelihood of the particle being present at that location. A higher value of $|\psi|^2$ indicates a greater probability of finding the particle there.
• Probability, not Position: The wavefunction itself does not directly represent the position of the particle. Instead, its square magnitude gives the probability distribution of the particle’s possible locations. Quantum mechanics is inherently probabilistic.
• Normalization: For the Born interpretation to be consistent with probability theory, the wavefunction must be normalized. This means that the integral of the probability density over all space must equal 1, reflecting the certainty that the particle exists somewhere in the universe: $\int_{\text{all space}} |\psi(r, t)|^2 dV = 1$
• Time Dependence: The wavefunction, and therefore the probability density, can change with time, reflecting the dynamic nature of quantum systems.
• Connection to Measurement: The Born interpretation provides the link between the abstract mathematical object (the wavefunction) and experimentally observable quantities (the probability of finding a particle in a certain region). It’s how we extract physical predictions from the quantum mechanical description of a system.

In essence, the Born interpretation tells us that the wavefunction is not a physical wave in the classical sense, but rather a probability amplitude. Its square modulus describes the probability distribution of finding the particle in different regions of space.

If the wavefunction of a particle is 0 at a given point, what does it mean physically?

If the wavefunction of a particle, $\psi(x)$, is 0 at a given point (let’s say $x = x_0$), it means that the probability density of finding the particle at that specific point is zero.

Since the probability density is given by $|\psi(x)|^2$, and if $\psi(x_0) = 0$, then $|\psi(x_0)|^2 = 0^2 = 0$.

This implies that there is no probability of finding the particle exactly at the position $x_0$. The particle will never be detected at that precise location.

It’s important to note that this does not mean the particle can’t pass through that point. The wavefunction describes the probability amplitude, and while the probability of being at that point is zero, the particle can still propagate through that region of space. The wavefunction can be zero at specific nodes or points while being non-zero in the surrounding regions. Think of a standing wave on a string; the displacement is zero at the nodes, but the wave exists and energy is transferred.

If the wavefunction of a particle is $\psi(x) = A e^{ikx}$, what does this tell you about the momentum of the particle?

The wavefunction $\psi(x) = A e^{ikx}$ represents a free particle moving in the positive x-direction with a definite momentum.

Here’s why:

• Plane Wave: The form $e^{ikx}$ represents a plane wave. In quantum mechanics, a free particle with a definite momentum is described by a plane wave.

• Momentum Eigenstate: This wavefunction is an eigenfunction of the momentum operator, $\hat{p} = -i\hbar \frac{\partial}{\partial x}$. Applying the momentum operator to the wavefunction: $\hat{p}\psi(x) = -i\hbar \frac{\partial}{\partial x} (A e^{ikx}) = -i\hbar (ik A e^{ikx}) = \hbar k (A e^{ikx}) = (\hbar k) \psi(x)$

  The result is the wavefunction multiplied by a constant, $\hbar k$. This means that the particle has a definite momentum with a value of $\hbar k$.

• De Broglie Relation: The wave number $k$ is related to the momentum $p$ by the de Broglie relation: $p = \hbar k$. Therefore, the wavefunction directly encodes the momentum of the particle.

• Direction of Motion: The positive sign in the exponent ($+ikx$) indicates that the wave is propagating in the positive x-direction. This corresponds to a positive momentum. If the wavefunction were $\psi(x) = A e^{-ikx}$, it would represent a particle moving in the negative x-direction with momentum $-\hbar k$.

In summary, the wavefunction $\psi(x) = A e^{ikx}$ tells you that the particle has a definite momentum in the positive x-direction, and the value of that momentum is $\hbar k$. This is a key example of how the wavefunction encapsulates information about the physical properties of a quantum system. It’s important to note that this wavefunction provides complete knowledge of the momentum but no information about the particle’s position, as the plane wave is delocalized throughout space.

Which of the following are acceptable wavefunctions (and why) for a particle free to move along the x axis:

(i) $\psi(x) = \cos(kx)$ (ii) $\psi(x) = e^{-x}$ (iii) $\psi(x) = x e^{-x^2}$

Let’s evaluate each function based on the requirements for an acceptable wavefunction:

(i) $\psi(x) = \cos(kx)$

• Finite: The cosine function oscillates between -1 and 1, so it is always finite.
• Single-valued: For every value of x, there is only one value of $\cos(kx)$.
• Continuous: The cosine function is continuous everywhere.
• Square-integrable: $\int_{-\infty}^{\infty} |\cos(kx)|^2 dx$ does not converge to a finite value. The average value of $\cos^2(kx)$ is 1/2, so the integral over an infinite range diverges.

Conclusion: While $\psi(x) = \cos(kx)$ is finite, single-valued, and continuous, it is not square-integrable. Therefore, it is not an acceptable wavefunction for a physically realistic state of a particle. However, it can represent a component of a momentum eigenstate for a free particle.

(ii) $\psi(x) = e^{-x}$

• Finite: The exponential function $e^{-x}$ is finite for all real values of x.
• Single-valued: For every value of x, there is only one value of $e^{-x}$.
• Continuous: The exponential function is continuous everywhere.
• Square-integrable: $\int_{-\infty}^{\infty} |e^{-x}|^2 dx = \int_{-\infty}^{\infty} e^{-2x} dx$. This integral diverges at the lower limit ($-\infty$).

Conclusion: While $\psi(x) = e^{-x}$ is finite, single-valued, and continuous, it is not square-integrable over the entire x-axis. Therefore, it is not an acceptable wavefunction for a particle free to move along the x-axis.

(iii) $\psi(x) = x e^{-x^2}$

• Finite: As $x \to \pm\infty$, $x e^{-x^2} \to 0$, so the function is always finite.
• Single-valued: For every value of x, there is only one value of $x e^{-x^2}$.
• Continuous: The function is a product of continuous functions, so it is continuous everywhere.
• Square-integrable: $\int_{-\infty}^{\infty} |x e^{-x^2}|^2 dx = \int_{-\infty}^{\infty} x^2 e^{-2x^2} dx$. This integral converges to a finite value.

Conclusion: $\psi(x) = x e^{-x^2}$ is finite, single-valued, continuous, and square-integrable. Therefore, it is an acceptable wavefunction for a particle free to move along the x-axis.

Explain why a wavefunction must be single-valued.

A wavefunction, $\psi(x, t)$, must be single-valued because its square magnitude, $|\psi(x, t)|^2$, represents the probability density of finding the particle at a specific position $x$ at time $t$.

If the wavefunction were multi-valued at a particular point in space, it would mean that there would be multiple values for the probability density at that single point. This would lead to the nonsensical situation where the particle has multiple different probabilities of being found at the same location at the same time.

Probability is a well-defined concept, and at any given point in space and time, the particle can only have one definite probability of being located there. A multi-valued wavefunction would violate this fundamental requirement of probability theory, making it impossible to interpret the wavefunction physically.

Therefore, for a wavefunction to provide a consistent and physically meaningful description of a particle’s probability distribution, it must be single-valued at every point in space and time.

What are the conditions for a wavefunction to be considered “well-behaved”?

For a wavefunction, $\psi(x, t)$, to be considered “well-behaved” and represent a physically realistic state of a quantum system, it must satisfy the following conditions:

1. Finite: The wavefunction must be finite everywhere. If the wavefunction were infinite at any point, the probability of finding the particle at that point would be infinite, which is physically impossible.
2. Single-valued: As explained previously, the wavefunction must have only one value at each point in space and time to ensure a unique probability density.
3. Continuous: The wavefunction must be continuous everywhere. A discontinuity in the wavefunction would imply an abrupt change in the probability of finding the particle, which is not physically realistic for finite potentials.
4. Square-integrable: The integral of the probability density over all space must be finite and can be normalized to unity: $\int_{-\infty}^{\infty} |\psi(x, t)|^2 dx < \infty$. This condition ensures that the total probability of finding the particle somewhere in space is finite. If the wavefunction is square-integrable, it can be normalized, meaning we can multiply it by a constant to ensure the total probability is exactly 1.
5. Its first derivative must be continuous (where the potential is finite): This condition arises from the requirement that the momentum of the particle be well-defined. The momentum operator involves the first derivative of the wavefunction. A discontinuity in the first derivative implies a discontinuity in the momentum, which is not physically realistic for finite potentials.

These conditions ensure that the wavefunction is mathematically well-defined and leads to physically meaningful probabilities and other observable quantities.

Why does the wavefunction (and its first derivative) need to be continuous?

The continuity of the wavefunction and its first derivative are crucial for the physical interpretation and mathematical consistency of quantum mechanics.

Continuity of the Wavefunction ($\psi(x)$):

• Single Probability: A discontinuity in the wavefunction would mean that there’s an abrupt jump in the probability amplitude, leading to a sudden, unphysical jump in the probability density of finding the particle. Probability densities should vary smoothly in space.
• Physical Accessibility: Imagine a potential barrier. If the wavefunction were discontinuous, it would imply that the particle could instantaneously appear or disappear at the discontinuity, which violates the principle of locality and conservation of probability. A continuous wavefunction ensures that the probability of finding the particle changes smoothly as it moves through space.

Continuity of the First Derivative of the Wavefunction ($\frac{d\psi}{dx}$):

• Well-defined Momentum: The momentum of a particle is related to the spatial derivative of the wavefunction through the momentum operator, $\hat{p} = -i\hbar \frac{d}{dx}$. A discontinuity in the first derivative would mean that the momentum is undefined or changes abruptly at that point, which is not physically realistic for finite potentials.
• Smooth Kinetic Energy: The kinetic energy of a particle is related to the second derivative of the wavefunction (through the Schrödinger equation). A discontinuity in the first derivative implies a discontinuity (or a Dirac delta function behavior) in the second derivative, which would correspond to an infinite kinetic energy at that point, again unphysical for finite potentials.
• Matching Solutions at Boundaries: When solving the Schrödinger equation for piecewise potentials, the continuity of the wavefunction and its first derivative are necessary conditions to match the solutions at the boundaries between different regions. This ensures a unique and physically consistent solution for the entire system.

In essence, the continuity of the wavefunction and its first derivative reflects the physical requirement that the probability of finding a particle and its momentum should change smoothly in space for finite potentials, avoiding unphysical scenarios like instantaneous jumps in probability or infinite kinetic energies.

What are the boundary conditions for a particle in a 1D box?

For a particle confined to a one-dimensional box of length $L$, with the potential being infinite outside the box (typically defined as $V(x) = 0$ for $0 < x < L$ and $V(x) = \infty$ otherwise), the boundary conditions on the wavefunction $\psi(x)$ are:

1. $\psi(0) = 0$: The wavefunction must be zero at the left edge of the box ($x = 0$).
2. $\psi(L) = 0$: The wavefunction must be zero at the right edge of the box ($x = L$).

Explanation:

• Infinite Potential: The infinite potential outside the box means that it would require infinite energy for the particle to exist outside the box. Therefore, the probability of finding the particle outside the box must be zero.
• Zero Probability Density: Since the probability density is $|\psi(x)|^2$, for the probability of finding the particle outside the box to be zero, the wavefunction itself must be zero outside the box.
• Continuity: Because the wavefunction must be continuous, if $\psi(x) = 0$ for $x < 0$ and $x > L$, then at the boundaries, we must have $\psi(0) = 0$ and $\psi(L) = 0$.

These boundary conditions are crucial for solving the time-independent Schrödinger equation for the particle in a 1D box. They lead to the quantization of energy levels, as only specific wavelengths (and thus energies) of the wavefunction will satisfy these conditions, resulting in standing wave solutions within the box.

A wavefunction is defined as $\psi(x) = Ax$ for $0\leq x \leq L/2$ and $\psi(x)=A(L-x)$ for $L/2\leq x\leq L$ and $\psi(x)=0$ elsewhere. Is this a valid wavefunction? Explain why or why not.

To determine if the given wavefunction is valid, we need to check the conditions for a well-behaved wavefunction:

1. Finiteness:

   • For $0 \leq x \leq L/2$, $|\psi(x)| = |Ax|$, which is finite.
   • For $L/2 \leq x \leq L$, $|\psi(x)| = |A(L-x)|$, which is finite.
   • Elsewhere, $\psi(x) = 0$, which is finite. Thus, the wavefunction is finite everywhere.

2. Single-valuedness: For each value of $x$, there is only one defined value of $\psi(x)$.

3. Continuity: We need to check for continuity at the boundaries:

   • At $x = 0$: $\lim_{x \to 0^+} \psi(x) = A(0) = 0$. The function is defined as 0 elsewhere for $x<0$. So, it’s continuous at $x=0$.
   • At $x = L/2$:
     • From the left: $\psi(L/2^-) = A(L/2)$.
     • From the right: $\psi(L/2^+) = A(L – L/2) = A(L/2)$. The wavefunction is continuous at $x = L/2$.
   • At $x = L$: $\lim_{x \to L^-} \psi(x) = A(L-L) = 0$. The function is defined as 0 elsewhere for $x>L$. So, it’s continuous at $x=L$. The wavefunction is continuous everywhere.

4. Square-integrability: We need to check if $\int_{-\infty}^{\infty} |\psi(x)|^2 dx$ is finite. $\int_{-\infty}^{\infty} |\psi(x)|^2 dx = \int_{0}^{L/2} (Ax)^2 dx + \int_{L/2}^{L} [A(L-x)]^2 dx$ $= A^2 \int_{0}^{L/2} x^2 dx + A^2 \int_{L/2}^{L} (L^2 – 2Lx + x^2) dx$ $= A^2 \left[ \frac{x^3}{3} \right]{0}^{L/2} + A^2 \left[ L^2x – Lx^2 + \frac{x^3}{3} \right]{L/2}^{L}$ $= A^2 \left( \frac{(L/2)^3}{3} \right) + A^2 \left[ (L^3 – L^3 + \frac{L^3}{3}) – (L^2\frac{L}{2} – L(\frac{L}{2})^2 + \frac{(L/2)^3}{3}) \right]$ $= A^2 \frac{L^3}{24} + A^2 \left[ \frac{L^3}{3} – (\frac{L^3}{2} – \frac{L^3}{4} + \frac{L^3}{24}) \right]$ $= A^2 \frac{L^3}{24} + A^2 \left[ \frac{L^3}{3} – \frac{12L^3 – 6L^3 + L^3}{24} \right]$ $= A^2 \frac{L^3}{24} + A^2 \left[ \frac{8L^3}{24} – \frac{7L^3}{24} \right] = A^2 \frac{L^3}{24} + A^2 \frac{L^3}{24} = A^2 \frac{L^3}{12}$ Since this integral is finite, the wavefunction is square-integrable.

5. Continuity of the first derivative: We need to check if the first derivative is continuous:

   • For $0 < x < L/2$: $\frac{d\psi}{dx} = A$
   • For $L/2 < x < L$: $\frac{d\psi}{dx} = -A$ At $x = L/2$:
   • From the left: $\lim_{x \to L/2^-} \frac{d\psi}{dx} = A$
   • From the right: $\lim_{x \to L/2^+} \frac{d\psi}{dx} = -A$ The first derivative is discontinuous at $x = L/2$.

Conclusion: While the wavefunction is finite, single-valued, continuous, and square-integrable, its first derivative is discontinuous at $x = L/2$. This discontinuity arises from the sharp corner in the wavefunction at this point. For potentials that are not infinitely large at a point, the first derivative of the wavefunction must be continuous. Therefore, this is not a valid wavefunction for a typical physical system where the potential is well-behaved. However, it could represent a physically realizable situation if the potential has a corresponding discontinuity in its derivative that compensates for this, which is not the case for a free particle or simple potentials.

For a particle confined in a 1D box with infinite walls, will the wavefunction have any nodes? What do they physically represent?

Yes, for a particle confined in a 1D box with infinite walls, the wavefunctions for the excited states ($n > 1$) will have nodes.

The normalized wavefunctions for a particle in a 1D box are given by: $\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$, for $n = 1, 2, 3, …$

A node is a point where the wavefunction $\psi_n(x)$ is equal to zero. For the $n$-th state, the nodes occur when: $\sin\left(\frac{n\pi x}{L}\right) = 0$ $\frac{n\pi x}{L} = m\pi$, where $m$ is an integer. $x = \frac{mL}{n}$

Since the particle is confined within the box $0 < x < L$, the possible values of $m$ are $1, 2, …, n-1$.

• For the ground state ($n=1$), there are no nodes within the box (the zeros are at the boundaries $x=0$ and $x=L$).
• For the first excited state ($n=2$), there is one node at $x = L/2$.
• For the second excited state ($n=3$), there are two nodes at $x = L/3$ and $x = 2L/3$.
• In general, the $n$-th state has $n-1$ nodes within the box.

Physical Representation of Nodes:

Physically, a node represents a point where the probability of finding the particle is zero. Since the probability density is given by $|\psi_n(x)|^2$, if $\psi_n(x) = 0$ at a node, then the probability density is also zero at that point. This means that for a particle in a particular energy state, there are specific locations within the box where the particle will never be found.

The existence of nodes is a direct consequence of the wave nature of the particle and the boundary conditions imposed by the infinite potential walls. The wavefunction must be zero at the walls, and to satisfy this for higher energy states, the wavefunction oscillates with increasing frequency, leading to the formation of nodes.

A potential energy function is given by $V(x) = \infty$ for x<0, $V(x)=0$ for 0<=x<=L, and $V(x)=V_0$ for x>L. What boundary conditions apply to the wavefunction in this potential?

We need to consider the behavior of the wavefunction at the points where the potential energy changes abruptly, which are $x=0$ and $x=L$. The wavefunction and its first derivative must be continuous everywhere where the potential is finite.

1. Boundary at x = 0:

   • For $x < 0$, $V(x) = \infty$. This means the wavefunction must be zero in this region: $\psi(x < 0) = 0$.
   • For $x \geq 0$, the wavefunction is non-zero.
   • Due to the continuity of the wavefunction at $x = 0$, we must have: $\lim_{x \to 0^-} \psi(x) = \lim_{x \to 0^+} \psi(x)$ $0 = \psi(0)$
   • Therefore, the boundary condition at $x = 0$ is: $\boxed{\psi(0) = 0}$

2. Boundary at x = L:

   • For $0 \leq x \leq L$, $V(x) = 0$. Let the wavefunction in this region be $\psi_{II}(x)$.
   • For $x > L$, $V(x) = V_0$. Let the wavefunction in this region be $\psi_{III}(x)$.
   • The wavefunction must be continuous at $x = L$: $\lim_{x \to L^-} \psi_{II}(x) = \lim_{x \to L^+} \psi_{III}(x)$ $\boxed{\psi_{II}(L) = \psi_{III}(L)}$
   • The first derivative of the wavefunction must also be continuous at $x = L$ (since the potential is finite): $\lim_{x \to L^-} \frac{d\psi_{II}(x)}{dx} = \lim_{x \to L^+} \frac{d\psi_{III}(x)}{dx}$ $\boxed{\frac{d\psi_{II}(x)}{dx}\Big|{x=L} = \frac{d\psi{III}(x)}{dx}\Big|_{x=L}}$

Summary of Boundary Conditions:

• $\boxed{\psi(0) = 0}$
• $\boxed{\psi_{II}(L) = \psi_{III}(L)}$
• $\boxed{\frac{d\psi_{II}(x)}{dx}\Big|{x=L} = \frac{d\psi{III}(x)}{dx}\Big|_{x=L}}$

These boundary conditions are essential for solving the time-independent Schrödinger equation for this potential and determining the allowed energy levels and corresponding wavefunctions of the particle.

Why do we say that the wavefunctions for a particle in a box are quantized?

We say that the wavefunctions for a particle in a box are quantized because only specific, discrete wavefunctions are allowed as solutions to the time-independent Schrödinger equation for this system, subject to the imposed boundary conditions.

Here’s a breakdown of why this quantization occurs:

1. Solving the Schrödinger Equation: The time-independent Schrödinger equation for a particle in a 1D box with infinite potential walls ($V(x) = 0$ for $0 < x < L$, and $V(x) = \infty$ otherwise) is: $-\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} = E\psi(x)$

   The general solution to this differential equation is of the form: $\psi(x) = A \sin(kx) + B \cos(kx)$, where $k = \sqrt{\frac{2mE}{\hbar^2}}$.

2. Applying Boundary Conditions: The confinement of the particle within the box imposes specific boundary conditions on the wavefunction:

   • $\psi(0) = 0$
   • $\psi(L) = 0$

   Applying the first boundary condition, $\psi(0) = 0$, to the general solution: $A \sin(0) + B \cos(0) = 0 \implies 0 + B(1) = 0 \implies B = 0$. This simplifies the wavefunction to $\psi(x) = A \sin(kx)$.

   Applying the second boundary condition, $\psi(L) = 0$: $A \sin(kL) = 0$.

   For a non-trivial solution (where $A \neq 0$), we must have: $\sin(kL) = 0$.

3. Quantization of Wave Number (k): The condition $\sin(kL) = 0$ is only satisfied when $kL$ is an integer multiple of $\pi$: $kL = n\pi$, where $n = 1, 2, 3, …$

   This means that the wave number $k$ can only take on specific, discrete values: $k_n = \frac{n\pi}{L}$

4. Quantization of Energy: Since $k = \sqrt{\frac{2mE}{\hbar^2}}$, the quantization of $k$ directly leads to the quantization of energy: $k_n^2 = \frac{2mE_n}{\hbar^2}$ $\left(\frac{n\pi}{L}\right)^2 = \frac{2mE_n}{\hbar^2}$ $E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$

   Only these specific, discrete energy values, $E_n$, are allowed for the particle in the box.

5. Quantization of Wavefunctions: Substituting the quantized values of $k_n$ back into the expression for the wavefunction, we get the allowed wavefunctions: $\psi_n(x) = A \sin\left(\frac{n\pi x}{L}\right)$

   After normalization, the constant $A$ is determined, giving the specific, normalized wavefunctions: $\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$

Therefore, the boundary conditions imposed by the confinement of the particle within the box restrict the possible solutions of the Schrödinger equation to a discrete set of wavefunctions, each corresponding to a specific quantized energy level. This is why we say the wavefunctions (and consequently the energy levels) for a particle in a box are quantized. Only these specific wave patterns can “fit” within the box while satisfying the requirement that the wavefunction must be zero at the impenetrable walls.

Consider the wavefunction $\psi(x) = A \sin^2(\pi x/L)$ for 0 ≤ x ≤ L. Normalize this wavefunction and determine the probability of finding the particle in the region x=L/4 and x = L/2.

First, we normalize the wavefunction: $\int_{0}^{L} |\psi(x)|^2 dx = 1$ $\int_{0}^{L} A^2 \sin^4(\pi x/L) dx = 1$ $A^2 \int_{0}^{L} \left(\frac{1 – \cos(2\pi x/L)}{2}\right)^2 dx = 1$ $\frac{A^2}{4} \int_{0}^{L} (1 – 2\cos(2\pi x/L) + \cos^2(2\pi x/L)) dx = 1$ $\frac{A^2}{4} \int_{0}^{L} \left(1 – 2\cos(2\pi x/L) + \frac{1 + \cos(4\pi x/L)}{2}\right) dx = 1$ $\frac{A^2}{4} \int_{0}^{L} \left(\frac{3}{2} – 2\cos(2\pi x/L) + \frac{1}{2}\cos(4\pi x/L)\right) dx = 1$ $\frac{A^2}{4} \left[ \frac{3}{2}x – \frac{2L}{2\pi}\sin(2\pi x/L) + \frac{L}{8\pi}\sin(4\pi x/L) \right]_{0}^{L} = 1$ $\frac{A^2}{4} \left[ \left(\frac{3}{2}L – 0 + 0\right) – (0 – 0 + 0) \right] = 1$ $\frac{A^2}{4} \frac{3}{2}L = 1$ $A^2 = \frac{8}{3L}$ $A = \sqrt{\frac{8}{3L}}$

Now, we determine the probability of finding the particle between x=L/4 and x=L/2: $P(L/4 \le x \le L/2) = \int_{L/4}^{L/2} |\psi(x)|^2 dx = \frac{8}{3L} \int_{L/4}^{L/2} \sin^4(\pi x/L) dx$ $P = \frac{8}{3L} \int_{L/4}^{L/2} \left(\frac{1 – \cos(2\pi x/L)}{2}\right)^2 dx$ $P = \frac{2}{3L} \int_{L/4}^{L/2} \left(1 – 2\cos(2\pi x/L) + \cos^2(2\pi x/L)\right) dx$ $P = \frac{2}{3L} \int_{L/4}^{L/2} \left(\frac{3}{2} – 2\cos(2\pi x/L) + \frac{1}{2}\cos(4\pi x/L)\right) dx$ $P = \frac{2}{3L} \left[ \frac{3}{2}x – \frac{L}{\pi}\sin(2\pi x/L) + \frac{L}{8\pi}\sin(4\pi x/L) \right]_{L/4}^{L/2}$ $P = \frac{2}{3L} \left[ \left(\frac{3}{4}L – \frac{L}{\pi}\sin(\pi) + \frac{L}{8\pi}\sin(2\pi)\right) – \left(\frac{3}{8}L – \frac{L}{\pi}\sin(\frac{\pi}{2}) + \frac{L}{8\pi}\sin(\pi)\right) \right]$ $P = \frac{2}{3L} \left[ \left(\frac{3}{4}L – 0 + 0\right) – \left(\frac{3}{8}L – \frac{L}{\pi} + 0\right) \right]$ $P = \frac{2}{3L} \left[ \frac{3}{4}L – \frac{3}{8}L + \frac{L}{\pi} \right] = \frac{2}{3L} \left[ \frac{6L – 3L}{8} + \frac{L}{\pi} \right] = \frac{2}{3L} \left[ \frac{3L}{8} + \frac{L}{\pi} \right]$ $P = \frac{1}{4} + \frac{2}{3\pi}$

A particle has the wavefunction $\psi(x) = A(1+x)e^{-ax}$ for x>0 and 0 otherwise. Determine the normalization constant A and calculate the average position $\langle x \rangle$ of the particle.

First, we determine the normalization constant A: $\int_{0}^{\infty} |\psi(x)|^2 dx = 1$ $\int_{0}^{\infty} A^2 (1+x)^2 e^{-2ax} dx = 1$ $A^2 \int_{0}^{\infty} (1+2x+x^2) e^{-2ax} dx = 1$ $A^2 \left[ \int_{0}^{\infty} e^{-2ax} dx + 2\int_{0}^{\infty} x e^{-2ax} dx + \int_{0}^{\infty} x^2 e^{-2ax} dx \right] = 1$ Using the standard integrals $\int_{0}^{\infty} x^n e^{-bx} dx = \frac{n!}{b^{n+1}}$: $A^2 \left[ \frac{0!}{(2a)^{1}} + 2\frac{1!}{(2a)^{2}} + \frac{2!}{(2a)^{3}} \right] = 1$ $A^2 \left[ \frac{1}{2a} + \frac{2}{4a^2} + \frac{2}{8a^3} \right] = 1$ $A^2 \left[ \frac{1}{2a} + \frac{1}{2a^2} + \frac{1}{4a^3} \right] = 1$ $A^2 \left[ \frac{2a^2 + 2a + 1}{4a^3} \right] = 1$ $A^2 = \frac{4a^3}{2a^2 + 2a + 1}$ $A = \sqrt{\frac{4a^3}{2a^2 + 2a + 1}} = 2a^{3/2} (2a^2 + 2a + 1)^{-1/2}$

Now, we calculate the average position $\langle x \rangle$: $\langle x \rangle = \int_{0}^{\infty} x |\psi(x)|^2 dx = A^2 \int_{0}^{\infty} x (1+x)^2 e^{-2ax} dx$ $\langle x \rangle = A^2 \int_{0}^{\infty} (x+2x^2+x^3) e^{-2ax} dx$ $\langle x \rangle = A^2 \left[ \int_{0}^{\infty} x e^{-2ax} dx + 2\int_{0}^{\infty} x^2 e^{-2ax} dx + \int_{0}^{\infty} x^3 e^{-2ax} dx \right]$ $\langle x \rangle = A^2 \left[ \frac{1!}{(2a)^{2}} + 2\frac{2!}{(2a)^{3}} + \frac{3!}{(2a)^{4}} \right]$ $\langle x \rangle = A^2 \left[ \frac{1}{4a^2} + \frac{4}{8a^3} + \frac{6}{16a^4} \right] = A^2 \left[ \frac{1}{4a^2} + \frac{1}{2a^3} + \frac{3}{8a^4} \right]$ $\langle x \rangle = A^2 \left[ \frac{2a^2 + 4a + 3}{8a^4} \right] = \frac{4a^3}{2a^2 + 2a + 1} \frac{2a^2 + 4a + 3}{8a^4} = \frac{2a^2 + 4a + 3}{2a(2a^2 + 2a + 1)}$

The wavefunction of a particle is given as $\psi(x) = C \exp(-x^2/(2a^2))$. Calculate the normalization constant C. Then calculate the probability of finding the particle within the region -a<x<a.

First, calculate the normalization constant C: $\int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1$ $\int_{-\infty}^{\infty} C^2 \exp\left(-\frac{x^2}{a^2}\right) dx = 1$ $C^2 \int_{-\infty}^{\infty} \exp\left(-\left(\frac{x}{a}\right)^2\right) dx = 1$ Let $u = x/a$, then $du = dx/a$, so $dx = a du$. $C^2 \int_{-\infty}^{\infty} e^{-u^2} a du = 1$ $C^2 a \int_{-\infty}^{\infty} e^{-u^2} du = 1$ We know that $\int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{\pi}$. $C^2 a \sqrt{\pi} = 1$ $C^2 = \frac{1}{a\sqrt{\pi}}$ $C = \frac{1}{\sqrt{a\sqrt{\pi}}} = \left(a\pi^{1/2}\right)^{-1/2} = a^{-1/2} \pi^{-1/4}$

Now, calculate the probability of finding the particle within the region -a<x<a: $P(-a < x < a) = \int_{-a}^{a} |\psi(x)|^2 dx = \int_{-a}^{a} \frac{1}{a\sqrt{\pi}} \exp\left(-\frac{x^2}{a^2}\right) dx$ Let $u = x/a$, then $du = dx/a$. When $x=-a$, $u=-1$. When $x=a$, $u=1$. $P = \frac{1}{\sqrt{\pi}} \int_{-1}^{1} e^{-u^2} du$ We know that the error function is defined as $\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{-t^2} dt$. $\int_{-1}^{1} e^{-u^2} du = \int_{-1}^{0} e^{-u^2} du + \int_{0}^{1} e^{-u^2} du$ Since $e^{-u^2}$ is an even function, $\int_{-1}^{0} e^{-u^2} du = \int_{0}^{1} e^{-u^2} du$. So, $\int_{-1}^{1} e^{-u^2} du = 2 \int_{0}^{1} e^{-u^2} du$. $P = \frac{1}{\sqrt{\pi}} \cdot 2 \int_{0}^{1} e^{-u^2} du = \text{erf}(1)$. The value of $\text{erf}(1) \approx 0.8427$.

A particle in a 1D box has the following normalized wavefunction: $\psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right)$. Calculate the probability density at position x = L/4 for the cases n=1, n=2, and n=3.

The probability density is given by $|\psi(x)|^2$. In this case, since $\psi(x)$ is real, $|\psi(x)|^2 = \psi(x)^2$. $\psi(x)^2 = \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right)\right)^2 = \frac{2}{L} \sin^2\left(\frac{n \pi x}{L}\right)$

For x = L/4: $\psi(L/4)^2 = \frac{2}{L} \sin^2\left(\frac{n \pi (L/4)}{L}\right) = \frac{2}{L} \sin^2\left(\frac{n \pi}{4}\right)$

Case n=1: $\psi_1(L/4)^2 = \frac{2}{L} \sin^2\left(\frac{1 \cdot \pi}{4}\right) = \frac{2}{L} \sin^2\left(\frac{\pi}{4}\right) = \frac{2}{L} \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{2}{L} \left(\frac{1}{2}\right) = \frac{1}{L}$

Case n=2: $\psi_2(L/4)^2 = \frac{2}{L} \sin^2\left(\frac{2 \cdot \pi}{4}\right) = \frac{2}{L} \sin^2\left(\frac{\pi}{2}\right) = \frac{2}{L} (1)^2 = \frac{2}{L}$

Case n=3: $\psi_3(L/4)^2 = \frac{2}{L} \sin^2\left(\frac{3 \cdot \pi}{4}\right) = \frac{2}{L} \sin^2\left(\frac{3\pi}{4}\right) = \frac{2}{L} \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{2}{L} \left(\frac{1}{2}\right) = \frac{1}{L}$

For a particle in a box of length L, what is the average value of the square of the momentum ($<p^2>$) when the particle is in the nth energy level? (Use the correct wave functions.)

The average value of an operator $\hat{A}$ for a state described by the normalized wavefunction $\psi(x)$ is given by: $\langle A \rangle = \int_{0}^{L} \psi^*(x) \hat{A} \psi(x) dx$

In this case, the operator is the square of the momentum operator, $\hat{p}^2$. The momentum operator in one dimension is $\hat{p} = -i\hbar \frac{d}{dx}$, so $\hat{p}^2 = \left(-i\hbar \frac{d}{dx}\right)^2 = -\hbar^2 \frac{d^2}{dx^2}$.

The normalized wavefunction for a particle in a 1D box is $\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right)$.

$\langle p^2 \rangle = \int_{0}^{L} \psi_n^*(x) \left(-\hbar^2 \frac{d^2}{dx^2}\right) \psi_n(x) dx$ $\langle p^2 \rangle = -\hbar^2 \int_{0}^{L} \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right) \frac{d^2}{dx^2} \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right)\right) dx$ $\langle p^2 \rangle = -\hbar^2 \frac{2}{L} \int_{0}^{L} \sin\left(\frac{n \pi x}{L}\right) \frac{d^2}{dx^2} \left(\sin\left(\frac{n \pi x}{L}\right)\right) dx$

First, calculate the second derivative: $\frac{d}{dx} \left(\sin\left(\frac{n \pi x}{L}\right)\right) = \frac{n \pi}{L} \cos\left(\frac{n \pi x}{L}\right)$ $\frac{d^2}{dx^2} \left(\sin\left(\frac{n \pi x}{L}\right)\right) = \frac{n \pi}{L} \left(-\frac{n \pi}{L}\right) \sin\left(\frac{n \pi x}{L}\right) = -\left(\frac{n \pi}{L}\right)^2 \sin\left(\frac{n \pi x}{L}\right)$

Substitute this back into the integral: $\langle p^2 \rangle = -\hbar^2 \frac{2}{L} \int_{0}^{L} \sin\left(\frac{n \pi x}{L}\right) \left(-\left(\frac{n \pi}{L}\right)^2 \sin\left(\frac{n \pi x}{L}\right)\right) dx$ $\langle p^2 \rangle = \hbar^2 \frac{2}{L} \left(\frac{n \pi}{L}\right)^2 \int_{0}^{L} \sin^2\left(\frac{n \pi x}{L}\right) dx$ We know that $\int_{0}^{L} \sin^2\left(\frac{n \pi x}{L}\right) dx = \frac{L}{2}$. $\langle p^2 \rangle = \hbar^2 \frac{2}{L} \frac{n^2 \pi^2}{L^2} \frac{L}{2} = \frac{\hbar^2 n^2 \pi^2}{L^2}$

Alternatively, we know that the energy of the particle in a box is $E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$. Also, for a free particle, $E = \frac{p^2}{2m}$, so $p^2 = 2mE$. Therefore, $\langle p^2 \rangle = 2m E_n = 2m \frac{n^2 \pi^2 \hbar^2}{2mL^2} = \frac{n^2 \pi^2 \hbar^2}{L^2}$.

For a normalized wavefunction $\psi(x)$, what does the expectation value <x²> represent? How do you calculate it?

For a normalized wavefunction $\psi(x)$, the expectation value of the square of the position operator, denoted as $<x^2>$, represents the average value of the square of the particle’s position. In simpler terms, it tells us about the spread or dispersion of the particle’s probability distribution around the origin (x=0).

Here’s a more detailed explanation:

• Probability Density: The quantity $|\psi(x)|^2$ gives the probability density of finding the particle at position $x$.
• Weighting by $x^2$: When calculating $<x^2>$, we multiply the probability density at each point $x$ by $x^2$. This weighting emphasizes positions that are further from the origin, either in the positive or negative direction.
• A Measure of Spread: A larger value of $<x^2>$ indicates that the probability distribution is more spread out, meaning there’s a higher likelihood of finding the particle further away from the origin. Conversely, a smaller value suggests the particle is more localized around the origin.

Calculation:

For a one-dimensional system with a normalized wavefunction $\psi(x)$, the expectation value $<x^2>$ is calculated using the following integral:

$<x^2> = \int_{-\infty}^{\infty} \psi^*(x) \hat{x}^2 \psi(x) dx$

Where:

• $\psi^*(x)$ is the complex conjugate of the wavefunction.
• $\hat{x}^2$ is the operator for the square of the position, which simply means multiplying by $x^2$. So, $\hat{x}^2 \psi(x) = x^2 \psi(x)$.

Therefore, the formula becomes:

$<x^2> = \int_{-\infty}^{\infty} |\psi(x)|^2 x^2 dx$

For a three-dimensional system, the calculation extends to:

$<x^2> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |\psi(x, y, z)|^2 x^2 dx , dy , dz$

In essence, we are averaging the squared position of the particle, weighted by the probability of finding it at each position. The square root of $<x^2>$ is related to the standard deviation of the position, providing a more direct measure of the spatial uncertainty in the particle’s location.

For a free particle, explain why the wavefunction cannot be normalized to unity over all space.

For a free particle, the wavefunction that represents a state of definite momentum is a plane wave, typically expressed as:

$\psi(x) = A e^{ikx}$

where $A$ is the amplitude and $k$ is the wave number related to the momentum by $p = \hbar k$.

The probability density for this wavefunction is:

$|\psi(x)|^2 = |A e^{ikx}|^2 = |A|^2 |e^{ikx}|^2 = |A|^2 (\cos^2(kx) + \sin^2(kx)) = |A|^2$

The probability density is constant everywhere in space.

To normalize the wavefunction, we require that the total probability of finding the particle somewhere in space is unity:

$\int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1$

Substituting the probability density for the free particle:

$\int_{-\infty}^{\infty} |A|^2 dx = 1$ $|A|^2 \int_{-\infty}^{\infty} dx = 1$

The integral $\int_{-\infty}^{\infty} dx$ diverges (it equals infinity). Therefore, the only way for the normalization condition to be satisfied would be if $|A|^2 = 0$, which means $A = 0$, resulting in a trivial wavefunction $\psi(x) = 0$, indicating no particle at all.

The reason for this non-normalizability is inherent to the nature of a free particle with a definite momentum:

• Definite Momentum Implies Uncertainty in Position: According to the Heisenberg uncertainty principle, $\Delta x \Delta p \ge \hbar/2$. A free particle with a precisely defined momentum ($\Delta p = 0$) must have a completely uncertain position ($\Delta x = \infty$).
• Plane Wave Represents Complete Delocalization: The plane wave $e^{ikx}$ extends infinitely in space. This reflects the fact that if we know the momentum of the free particle exactly, we have no information about its location. The probability of finding it in any specific finite region is non-zero, but summed over all infinite space, it remains infinite unless the amplitude is zero.
• Physical Interpretation: A truly free particle with a precisely defined momentum is an idealized concept. In reality, particles are always subject to some form of confinement or interaction, even if very weak. These subtle effects would allow for wave packets that are normalizable.

While a perfect plane wave is not normalizable, it is still a useful concept in quantum mechanics. It can be used as a basis state for constructing more realistic, localized wave packets (which are normalizable) through superposition. Alternatively, one can work with a “box normalization” approach, where the particle is assumed to be confined within a very large box, and then the limit of the box size going to infinity is considered.

Explain the physical meaning of $\langle \psi | \hat{A} | \psi \rangle$ where $\hat{A}$ is an operator.

The expression $\langle \psi | \hat{A} | \psi \rangle$, also written as $\langle \hat{A} \rangle$ or $\langle A \rangle_\psi$, represents the expectation value or average value of the physical observable corresponding to the operator $\hat{A}$, for a system described by the quantum state $|\psi \rangle$.

Here’s a breakdown of its physical meaning:

1. Operator and Observable: In quantum mechanics, each physically measurable quantity (like position, momentum, energy, etc.) is associated with a linear operator. The operator $\hat{A}$ represents the physical observable we are interested in.

2. Quantum State: The ket $|\psi \rangle$ (and its conjugate bra $\langle \psi |$) describes the quantum state of the system. This state contains all the information we can know about the system. We assume the wavefunction is normalized, i.e., $\langle \psi | \psi \rangle = 1$.

3. Expectation Value as a Statistical Average: Due to the probabilistic nature of quantum mechanics, we cannot predict with certainty the outcome of a single measurement of the observable $A$ on a system in state $|\psi \rangle$. Instead, the expectation value $\langle \hat{A} \rangle$ represents the average of the results of a large number of identical measurements performed on identically prepared systems, all in the same state $|\psi \rangle$.

4. Calculation: In the position representation, where the wavefunction is $\psi(x)$, the expectation value is calculated as: $\langle \hat{A} \rangle = \int_{-\infty}^{\infty} \psi^(x) \hat{A} \psi(x) dx$ The operator $\hat{A}$ acts on the wavefunction $\psi(x)$, and the result is then multiplied by the complex conjugate of the wavefunction $\psi^(x)$, and integrated over all space.

5. Information about the Distribution: The expectation value provides a single, representative number for the observable, but it doesn’t tell the whole story. It’s like the mean of a probability distribution. While it gives the central tendency, it doesn’t describe the spread or shape of the distribution of measurement outcomes.

In summary, $\langle \psi | \hat{A} | \psi \rangle$ is the theoretically predicted average value you would obtain if you measured the physical quantity associated with the operator $\hat{A}$ many times on a system prepared in the state $|\psi \rangle$. It’s a fundamental concept linking the mathematical formalism of quantum mechanics to experimentally measurable quantities.

Given a wavefunction $\psi(x)$, what does $\int_{-\infty}^{\infty} \psi^*(x) \psi(x) dx$ represent and why is it important?

Given a wavefunction $\psi(x)$, the integral $\int_{-\infty}^{\infty} \psi^*(x) \psi(x) dx$ represents the total probability of finding the particle somewhere in space.

Here’s a breakdown:

1. Probability Density: The product $\psi^*(x) \psi(x) = |\psi(x)|^2$ represents the probability density of finding the particle at a specific position $x$. It’s the probability per unit length (in 1D) of finding the particle at that point.

2. Integrating over All Space: The integral $\int_{-\infty}^{\infty} |\psi(x)|^2 dx$ sums up the probabilities of finding the particle in all infinitesimal intervals $dx$ across the entire range of possible positions (from $-\infty$ to $\infty$).

3. Total Probability: Since the particle must exist somewhere in space, the sum of the probabilities of finding it in all possible locations must equal 1. This leads to the normalization condition: $\int_{-\infty}^{\infty} \psi^*(x) \psi(x) dx = \int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1$

Why is it important?

The integral $\int_{-\infty}^{\infty} \psi^*(x) \psi(x) dx$ and the normalization condition are crucial for several reasons:

1. Probabilistic Interpretation: The Born interpretation of the wavefunction relies on the idea that $|\psi(x)|^2$ represents a probability density. For this interpretation to be consistent, the total probability must be unity.

2. Physical Realism: A wavefunction that cannot be normalized to unity (i.e., the integral diverges) does not represent a physically realizable state of a particle. It would imply an infinite probability of finding the particle somewhere, which is nonsensical.

3. Meaningful Probabilities: Only normalized wavefunctions allow us to calculate meaningful probabilities of finding the particle within a specific region. The probability of finding the particle between $a$ and $b$ is given by $\int_{a}^{b} |\psi(x)|^2 dx$. If the wavefunction is not normalized, these probabilities would not be correctly scaled.

4. Expectation Values: The calculation of expectation values of physical observables involves integrals that rely on the wavefunction being normalized. The formula $\langle \hat{A} \rangle = \int_{-\infty}^{\infty} \psi^*(x) \hat{A} \psi(x) dx$ assumes that $\psi(x)$ is normalized.

5. Mathematical Consistency: Normalization ensures the mathematical consistency of quantum mechanics. It allows us to treat the wavefunction as representing a probability distribution.

In essence, $\int_{-\infty}^{\infty} \psi^*(x) \psi(x) dx$ being equal to 1 is a fundamental requirement for a wavefunction to have a physically meaningful probabilistic interpretation. It ensures that the total probability of finding the particle somewhere is 1, which is a cornerstone of quantum mechanics.

Write down the time-dependent Schrodinger equation for a single particle in one dimension.

The time-dependent Schrödinger equation for a single particle in one dimension is:

$i\hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} + V(x,t) \Psi(x,t)$

where:

• $i$ is the imaginary unit
• $\hbar$ is the reduced Planck constant
• $\Psi(x,t)$ is the time-dependent wavefunction
• $m$ is the mass of the particle
• $V(x,t)$ is the potential energy function
• $t$ is time

What is the Hamiltonian operator and how is it related to the total energy of a system?

The Hamiltonian operator, denoted by $\hat{H}$, is a central operator in quantum mechanics that corresponds to the total energy of a system. It is the quantum mechanical analogue of the classical Hamiltonian function.

For a single particle in one dimension, the Hamiltonian operator is given by:

$\hat{H} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x,t)$

where:

• $-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$ is the kinetic energy operator.
• $V(x,t)$ is the potential energy operator (which is just multiplication by the potential energy function).

Relation to Total Energy:

The time-dependent Schrödinger equation can be written concisely using the Hamiltonian operator:

$i\hbar \frac{\partial \Psi(x,t)}{\partial t} = \hat{H} \Psi(x,t)$

This equation shows that the Hamiltonian operator acting on the wavefunction describes how the wavefunction evolves in time, which is directly related to the energy of the system.

Furthermore, in the time-independent Schrödinger equation, which describes stationary states (states with definite energy), the Hamiltonian operator acting on the wavefunction yields the energy eigenvalue multiplied by the wavefunction:

$\hat{H} \psi(x) = E \psi(x)$

Here, $E$ represents the specific, constant value of the total energy for that stationary state. This eigenvalue equation explicitly demonstrates that the Hamiltonian operator is the quantum mechanical representation of the total energy.

How does the time-dependent Schrodinger equation differ from the time-independent Schrodinger equation?

The time-dependent Schrödinger equation and the time-independent Schrödinger equation are two related but distinct equations in quantum mechanics that describe different aspects of quantum systems.

Time-Dependent Schrödinger Equation:

• Equation: $i\hbar \frac{\partial \Psi(x,t)}{\partial t} = \hat{H} \Psi(x,t)$
• Purpose: Describes how the quantum state of a system changes with time. It governs the temporal evolution of the wavefunction $\Psi(x,t)$.
• Wavefunction: The wavefunction $\Psi(x,t)$ is a function of both position ($x$) and time ($t$).
• Applicability: Applicable to all quantum systems, whether or not the potential energy is time-dependent. It is the more general form of the Schrödinger equation.
• Solutions: Solutions to the time-dependent Schrödinger equation describe the full dynamics of the quantum system, including how probabilities and expectation values evolve over time.

Time-Independent Schrödinger Equation:

• Equation: $\hat{H} \psi(x) = E \psi(x)$
• Purpose: Determines the allowed energy levels and the corresponding stationary states (time-independent states) of a system where the potential energy is time-independent ($V(x)$ only).
• Wavefunction: The wavefunction $\psi(x)$ is a function of position ($x$) only. The time dependence is a simple oscillatory factor: $\Psi(x,t) = \psi(x) e^{-iEt/\hbar}$.
• Applicability: Only applicable to systems where the potential energy does not explicitly depend on time.
• Solutions: Solutions to the time-independent Schrödinger equation are the energy eigenstates, representing states with a definite, constant energy $E$.

Key Differences Summarized:

In essence, the time-independent Schrödinger equation is a special case of the time-dependent Schrödinger equation, applicable when the system’s potential energy is constant in time. Solving the time-independent equation helps us find the fundamental energy states of a system, while the time-dependent equation describes how the system transitions between these states or evolves under the influence of time-varying potentials.

Explain what is meant by an “eigenfunction” of an operator, and what is an “eigenvalue”?

In quantum mechanics, an eigenfunction of an operator $\hat{A}$ is a specific function (let’s call it $f$) that, when the operator acts upon it, results in the same function multiplied by a constant. Mathematically, this is expressed as:

$\hat{A} f = \lambda f$

Here:

• $\hat{A}$ is the operator.
• $f$ is the eigenfunction of the operator $\hat{A}$.
• $\lambda$ is a constant called the eigenvalue associated with that eigenfunction.

Physical Meaning:

• Eigenfunction as a State with a Definite Value: When a system is in a state described by an eigenfunction of an operator corresponding to a physical observable, measuring that observable will always yield the associated eigenvalue. In other words, the eigenfunction represents a state where the physical quantity represented by the operator has a definite, precise value.

• Eigenvalue as the Measured Value: The eigenvalue $\lambda$ is the specific value of the physical observable that is obtained when the measurement is performed on the system in the eigenstate.

Analogy:

Think of an operator as a machine that transforms functions. An eigenfunction is a special input to this machine that doesn’t change its basic form; it only gets scaled by a constant (the eigenvalue).

Example:

Consider the momentum operator $\hat{p}_x = -i\hbar \frac{d}{dx}$. The function $f(x) = e^{ikx}$ is an eigenfunction of the momentum operator because:

$\hat{p}_x e^{ikx} = -i\hbar \frac{d}{dx} (e^{ikx}) = -i\hbar (ik e^{ikx}) = \hbar k e^{ikx}$

Here, $e^{ikx}$ is the eigenfunction, and $\hbar k$ is the eigenvalue (which corresponds to the momentum of the particle).

What is the momentum operator in quantum mechanics in one dimension?

The momentum operator in quantum mechanics for a single particle moving in one dimension (along the x-axis) is given by:

$\hat{p}_x = -i\hbar \frac{\partial}{\partial x}$

where:

• $\hat{p}_x$ is the momentum operator in the x-direction.
• $i$ is the imaginary unit.
• $\hbar$ is the reduced Planck constant (h/2π).
• $\frac{\partial}{\partial x}$ is the partial derivative with respect to position $x$.

This operator, when acting on the wavefunction of a particle, is used to determine the momentum properties of that particle.

How are operators used to obtain physical observables in quantum mechanics?

Operators play a fundamental role in obtaining physical observables in quantum mechanics through the concept of expectation values.

Here’s how it works:

1. Observable and Operator Correspondence: For every physical observable (a quantity that can be measured), such as position, momentum, energy, angular momentum, etc., there is a corresponding linear Hermitian operator in quantum mechanics.

2. Expectation Value: The expectation value of a physical observable represented by the operator $\hat{A}$, for a system in a state described by the normalized wavefunction $\psi$, is given by:

   $\langle A \rangle = \langle \psi | \hat{A} | \psi \rangle = \int_{-\infty}^{\infty} \psi^*(x) \hat{A} \psi(x) dx$ (in one dimension, position representation)

   The expectation value represents the average value of the observable that would be obtained from a large number of measurements on identically prepared systems.

3. Eigenvalues as Possible Measurement Outcomes: When a measurement of a physical observable is performed, the possible outcomes are the eigenvalues of the corresponding operator. If the system is in an eigenstate of the operator, the measurement will yield the corresponding eigenvalue with certainty. If the system is in a superposition of eigenstates, the outcome of a single measurement is probabilistic, but the expectation value gives the average of these possible outcomes.

4. Applying the Operator: To calculate the expectation value, the operator $\hat{A}$ acts on the wavefunction $\psi(x)$. The result is then multiplied by the complex conjugate of the wavefunction, $\psi^*(x)$, and integrated over all space.

In summary, operators are the mathematical entities in quantum mechanics that represent physical observables. The expectation value of an operator, calculated using the wavefunction of the system, provides the average value of the corresponding observable. The eigenvalues of the operator represent the possible discrete values that can be obtained when measuring that observable.

If $\hat{A}$ is an operator with eigenvalue $a$ and $\psi$ is the corresponding eigenfunction, what is the result of applying $\hat{A}^2$ to $\psi$?

Given that $\hat{A}$ is an operator with eigenvalue $a$ and $\psi$ is the corresponding eigenfunction, by definition:

$\hat{A} \psi = a \psi$

Now, let’s apply the operator $\hat{A}^2$ to $\psi$. Note that $\hat{A}^2$ means applying the operator $\hat{A}$ twice: $\hat{A}^2 = \hat{A} \hat{A}$.

So, applying $\hat{A}^2$ to $\psi$:

$\hat{A}^2 \psi = \hat{A} (\hat{A} \psi)$

Since we know that $\hat{A} \psi = a \psi$, we can substitute this into the equation:

$\hat{A}^2 \psi = \hat{A} (a \psi)$

Since $a$ is a scalar (the eigenvalue), it can be moved outside the operator:

$\hat{A}^2 \psi = a (\hat{A} \psi)$

Again, we know that $\hat{A} \psi = a \psi$, so we substitute this in:

$\hat{A}^2 \psi = a (a \psi)$

$\hat{A}^2 \psi = a^2 \psi$

Therefore, the result of applying $\hat{A}^2$ to $\psi$ is the eigenfunction $\psi$ multiplied by the square of the eigenvalue, $a^2$. This shows that $\psi$ is also an eigenfunction of the operator $\hat{A}^2$, with the corresponding eigenvalue $a^2$.

What is the expectation value of the energy for a state described by an energy eigenfunction?

Let $\hat{H}$ be the Hamiltonian operator (the operator corresponding to energy), and let $\phi_E$ be an energy eigenfunction with corresponding eigenvalue $E$. This means:

$\hat{H} \phi_E = E \phi_E$

The expectation value of the energy, denoted as $\langle E \rangle$, for a state described by the normalized wavefunction $\phi_E$ is given by:

$\langle E \rangle = \langle \phi_E | \hat{H} | \phi_E \rangle$

Substituting $\hat{H} \phi_E = E \phi_E$:

$\langle E \rangle = \langle \phi_E | (E \phi_E) \rangle$

Since $E$ is a scalar (the energy eigenvalue), it can be moved outside the bra-ket:

$\langle E \rangle = E \langle \phi_E | \phi_E \rangle$

Since the eigenfunction is assumed to be normalized, we have $\langle \phi_E | \phi_E \rangle = 1$.

Therefore, the expectation value of the energy is:

$\langle E \rangle = E \cdot 1 = E$

The expectation value of the energy for a state described by an energy eigenfunction is simply the energy eigenvalue associated with that eigenfunction. This makes intuitive sense: if a system is in a state with a definite energy, then the average value of the energy measured will be that definite energy value.