Skip to content

Ch. 1 Solutions

← Back to problems　|　Back to chapter

Table of Contents

Basic

Medium

Advanced

Basic

B-1. Appreciating the Smallness of Planck's Constant

→ Back to problem

Solution strategy: Substitute numerical values into \(E = h\nu\), obtain the result in J, then convert to eV.

Calculation:

\[ E = h\nu = (6.626 \times 10^{-34}\;\mathrm{J \cdot s}) \times (5.0 \times 10^{14}\;\mathrm{Hz}) \]
\[ E = 6.626 \times 5.0 \times 10^{-34+14}\;\mathrm{J} = 33.13 \times 10^{-20}\;\mathrm{J} \]
\[ \boxed{E = 3.3 \times 10^{-19}\;\mathrm{J}} \]

Conversion to eV:

\[ E\;[\mathrm{eV}] = \frac{3.313 \times 10^{-19}\;\mathrm{J}}{1.602 \times 10^{-19}\;\mathrm{J/eV}} = 2.07\;\mathrm{eV} \]
\[ \boxed{E \simeq 2.1\;\mathrm{eV}} \]

Verification: The photon energy of visible light is approximately 1.8–3.1 eV, and 2.1 eV corresponds to the green region. This is a reasonable value.

B-2. Work Function and Threshold Frequency

→ Back to problem

Solution strategy: Determine \(\nu_0\) from the threshold condition \(h\nu_0 = W\), then convert to wavelength using \(\lambda_0 = c/\nu_0\).

Calculation:

First, convert the work function to joules:

\[ W = 2.28\;\mathrm{eV} \times 1.602 \times 10^{-19}\;\mathrm{J/eV} = 3.653 \times 10^{-19}\;\mathrm{J} \]

Threshold frequency:

\[ \nu_0 = \frac{W}{h} = \frac{3.653 \times 10^{-19}}{6.626 \times 10^{-34}} = 5.51 \times 10^{14}\;\mathrm{Hz} \]
\[ \boxed{\nu_0 \simeq 5.51 \times 10^{14}\;\mathrm{Hz}} \]

Threshold wavelength:

\[ \lambda_0 = \frac{c}{\nu_0} = \frac{3.00 \times 10^8}{5.51 \times 10^{14}} = 5.44 \times 10^{-7}\;\mathrm{m} \]
\[ \boxed{\lambda_0 \simeq 544\;\mathrm{nm}} \]

Verification: 544 nm corresponds to green visible light. The fact that sodium's photoelectric threshold lies in the visible range is consistent with experimental observations. Additionally, \(hc/\lambda_0 = (6.626 \times 10^{-34})(3.00 \times 10^8)/(5.44 \times 10^{-7}) = 3.65 \times 10^{-19}\;\mathrm{J} = 2.28\;\mathrm{eV}\), which matches \(W\).

B-3. Kinetic Energy in the Photoelectric Effect

→ Back to problem

Solution Strategy: Use \(hc \simeq 1240\;\mathrm{eV \cdot nm}\) to find the photon energy, then calculate \(K = E - W\).

Calculation:

\[ E = \frac{hc}{\lambda} = \frac{1240\;\mathrm{eV \cdot nm}}{200\;\mathrm{nm}} = 6.20\;\mathrm{eV} \]
\[ K = E - W = 6.20 - 4.50 = 1.70\;\mathrm{eV} \]
\[ \boxed{K = 1.70\;\mathrm{eV}} \]

Verification: A photon energy of 6.20 eV is reasonable for ultraviolet light (a wavelength of 200 nm lies in the vacuum ultraviolet region). Since \(K > 0\), the photoelectric effect does occur. The units are consistent in eV.

B-4. Wavelength Calculation Using the Rydberg Formula

→ Back to problem

Solution strategy: Substitute the Balmer series values \(n=2\), \(m=3\) into the Rydberg formula.

Calculation:

\[ \frac{1}{\lambda} = R_\infty \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R_\infty \left(\frac{1}{4} - \frac{1}{9}\right) \]
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{5}{36} = 1.097 \times 10^7 \times 0.1389 = 1.524 \times 10^6\;\mathrm{m^{-1}} \]
\[ \lambda = \frac{1}{1.524 \times 10^6} = 6.56 \times 10^{-7}\;\mathrm{m} \]
\[ \boxed{\lambda \simeq 656\;\mathrm{nm}} \]

Verification: This is the well-known value of the hydrogen H\(\alpha\) line (red), which agrees well with the experimental value of 656.3 nm.

B-5. Bohr's Quantization Condition and Orbital Radius

→ Back to problem

Solution Strategy: Eliminate \(v\) using the quantization condition, then solve the force balance equation for \(r\).

Calculation:

From Bohr's quantization condition:

\[ m_e v r = n\hbar \quad \Longrightarrow \quad v = \frac{n\hbar}{m_e r} \]

Substituting into the force balance equation:

\[ \frac{e^2}{4\pi\varepsilon_0 r^2} = \frac{m_e v^2}{r} = \frac{m_e}{r} \cdot \frac{n^2\hbar^2}{m_e^2 r^2} = \frac{n^2\hbar^2}{m_e r^3} \]

Solving for \(r\):

\[ \frac{e^2}{4\pi\varepsilon_0 r^2} = \frac{n^2\hbar^2}{m_e r^3} \]
\[ r = \frac{4\pi\varepsilon_0 n^2\hbar^2}{m_e e^2} \]
\[ \boxed{r_n = \frac{4\pi\varepsilon_0 \hbar^2}{m_e e^2} \cdot n^2 = a_0 n^2} \]

Here, \(a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m_e e^2}\) is the Bohr radius.

Verification: Let us check the dimensions. Since \([4\pi\varepsilon_0] = \mathrm{C^2/(N \cdot m^2)}\), \([\hbar^2] = \mathrm{J^2 \cdot s^2}\), \([m_e] = \mathrm{kg}\), \([e^2] = \mathrm{C^2}\), we have:

\[ \frac{\mathrm{C^2/(N \cdot m^2)} \cdot \mathrm{J^2 \cdot s^2}}{\mathrm{kg} \cdot \mathrm{C^2}} = \frac{\mathrm{J^2 \cdot s^2}}{\mathrm{N \cdot m^2 \cdot kg}} \]

Using \(\mathrm{J} = \mathrm{N \cdot m}\), we get \(\mathrm{J \cdot s^2 / (m \cdot kg)} = \mathrm{N \cdot m \cdot s^2 / (m \cdot kg)} = \mathrm{kg \cdot m \cdot s^{-2} \cdot m \cdot s^2 / (m \cdot kg)} = \mathrm{m}\). The dimensions are correctly those of length.

B-6. Numerical Calculation of the Bohr Radius

→ Back to problem

Solution strategy: Substitute numerical values into \(a_0 = \frac{4\pi\varepsilon_0 \hbar^2}{m_e e^2}\).

Calculation:

Numerator:

\[ 4\pi\varepsilon_0 \cdot \hbar^2 = (1.113 \times 10^{-10}) \times (1.055 \times 10^{-34})^2 \]
\[ = (1.113 \times 10^{-10}) \times (1.113 \times 10^{-68}) \]
\[ = 1.239 \times 10^{-78}\;\mathrm{C^2 \cdot J \cdot s^2 / (N \cdot m^2)} \]

Denominator:

\[ m_e e^2 = (9.109 \times 10^{-31}) \times (1.602 \times 10^{-19})^2 \]
\[ = (9.109 \times 10^{-31}) \times (2.566 \times 10^{-38}) \]
\[ = 2.337 \times 10^{-68}\;\mathrm{kg \cdot C^2} \]

Therefore:

\[ a_0 = \frac{1.239 \times 10^{-78}}{2.337 \times 10^{-68}} = 0.530 \times 10^{-10}\;\mathrm{m} \]
\[ \boxed{a_0 \simeq 5.29 \times 10^{-11}\;\mathrm{m} \simeq 0.529\;\text{Å}} \]

Verification: This agrees with the literature value \(a_0 = 5.292 \times 10^{-11}\;\mathrm{m}\). It is also consistent with the order of magnitude of atomic sizes \(\sim 10^{-10}\;\mathrm{m}\).

B-7. Limit of the Planck Distribution

→ Back to problem

Solution strategy: Let \(x = h\nu/(k_B T) \ll 1\) and expand the exponential function to first order.

Calculation:

Setting \(x = h\nu/(k_B T) \ll 1\):

\[ e^x \simeq 1 + x \]

Therefore the denominator becomes:

\[ e^{h\nu/k_B T} - 1 \simeq (1 + x) - 1 = x = \frac{h\nu}{k_B T} \]

Thus:

\[ \langle E \rangle = \frac{h\nu}{e^{h\nu/k_B T} - 1} \simeq \frac{h\nu}{h\nu/(k_B T)} = k_B T \]
\[ \boxed{\langle E \rangle \simeq k_B T \quad (h\nu \ll k_B T)} \]

Verification: This agrees with the result from the classical equipartition theorem (average energy of a harmonic oscillator: \(k_B T\)). This confirms that Planck's formula reproduces classical theory in the low-frequency limit.

B-8. Comparison of Boltzmann Factors

→ Back to problem

Solution strategy: First calculate \(k_B T\), then determine \(h\nu/(k_B T)\) for each frequency and evaluate the Boltzmann factor.

Calculation:

\[ k_B T = (1.38 \times 10^{-23})(6000) = 8.28 \times 10^{-20}\;\mathrm{J} \]

(a) Visible light \(\nu_1 = 5.0 \times 10^{14}\;\mathrm{Hz}\):

\[ h\nu_1 = (6.626 \times 10^{-34})(5.0 \times 10^{14}) = 3.31 \times 10^{-19}\;\mathrm{J} \]
\[ \frac{h\nu_1}{k_B T} = \frac{3.31 \times 10^{-19}}{8.28 \times 10^{-20}} = 4.00 \]
\[ e^{-h\nu_1/k_B T} = e^{-4.00} \simeq 0.018 \]
\[ \boxed{e^{-h\nu_1/k_B T} \simeq 1.8 \times 10^{-2}} \]

(b) Ultraviolet light \(\nu_2 = 3.0 \times 10^{15}\;\mathrm{Hz}\):

\[ h\nu_2 = (6.626 \times 10^{-34})(3.0 \times 10^{15}) = 1.99 \times 10^{-18}\;\mathrm{J} \]
\[ \frac{h\nu_2}{k_B T} = \frac{1.99 \times 10^{-18}}{8.28 \times 10^{-20}} = 24.0 \]
\[ e^{-h\nu_2/k_B T} = e^{-24.0} \simeq 3.8 \times 10^{-11} \]
\[ \boxed{e^{-h\nu_2/k_B T} \simeq 3.8 \times 10^{-11}} \]

Comparison: The Boltzmann factor for visible light is approximately \(10^{-2}\) (small but not negligible), whereas for ultraviolet light it is approximately \(10^{-11}\) (effectively zero). A mere 6-fold increase in frequency causes the Boltzmann factor to decrease by 9 orders of magnitude. This is the suppression of high-frequency modes, and it is the physical reason why the ultraviolet catastrophe is resolved.

Sanity check: \(T = 6000\;\mathrm{K}\) corresponds to the surface temperature of the Sun, which is consistent with the fact that the solar spectrum peaks near the visible range and drops off sharply in the ultraviolet region.

Medium

M-1. Derivation of Hydrogen Atom Energy Levels Using the Bohr Model

→ Back to problem

(a) Orbital radius \(r_n\) and velocity \(v_n\)

Force balance:

\[ \frac{e^2}{4\pi\varepsilon_0 r^2} = \frac{m_e v^2}{r} \tag{i} \]

Bohr's quantization condition:

\[ m_e v r = n\hbar \tag{ii} \]

Substituting \(v = n\hbar/(m_e r)\) from (ii) into (i):

\[ \frac{e^2}{4\pi\varepsilon_0 r^2} = \frac{m_e}{r} \cdot \frac{n^2\hbar^2}{m_e^2 r^2} = \frac{n^2\hbar^2}{m_e r^3} \]

Solving for \(r\):

\[ \boxed{r_n = \frac{4\pi\varepsilon_0 \hbar^2}{m_e e^2} \cdot n^2 = a_0 n^2} \]

Substituting \(r_n\) back into (ii) to find \(v_n\):

\[ v_n = \frac{n\hbar}{m_e r_n} = \frac{n\hbar}{m_e \cdot a_0 n^2} = \frac{\hbar}{m_e a_0} \cdot \frac{1}{n} \]

Substituting \(a_0 = 4\pi\varepsilon_0\hbar^2/(m_e e^2)\):

\[ \boxed{v_n = \frac{e^2}{4\pi\varepsilon_0 \hbar} \cdot \frac{1}{n}} \]

(b) Energy levels \(E_n\)

Kinetic energy:

\[ T_n = \frac{1}{2}m_e v_n^2 \]

From the force balance (i), \(m_e v^2 = e^2/(4\pi\varepsilon_0 r)\), so:

\[ T_n = \frac{1}{2} \cdot \frac{e^2}{4\pi\varepsilon_0 r_n} \]

Potential energy:

\[ V_n = -\frac{e^2}{4\pi\varepsilon_0 r_n} \]

Total energy:

\[ E_n = T_n + V_n = \frac{1}{2} \cdot \frac{e^2}{4\pi\varepsilon_0 r_n} - \frac{e^2}{4\pi\varepsilon_0 r_n} = -\frac{e^2}{8\pi\varepsilon_0 r_n} \]

Substituting \(r_n = 4\pi\varepsilon_0 \hbar^2 n^2/(m_e e^2)\):

\[ E_n = -\frac{e^2}{8\pi\varepsilon_0} \cdot \frac{m_e e^2}{4\pi\varepsilon_0 \hbar^2 n^2} = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2} \cdot \frac{1}{n^2} \]
\[ \boxed{E_n = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2} \cdot \frac{1}{n^2}} \]

(c) Reproducing the Rydberg formula

The frequency of light emitted in a transition from level \(m\) to \(n\) (\(m > n\)):

\[ h\nu = E_m - E_n = \frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2}\left(\frac{1}{n^2} - \frac{1}{m^2}\right) \]

Using \(c = \lambda\nu\), so \(1/\lambda = \nu/c\):

\[ \frac{1}{\lambda} = \frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2 \cdot hc}\left(\frac{1}{n^2} - \frac{1}{m^2}\right) \]

Since \(\hbar = h/(2\pi)\), we have \(\hbar^2 h = h^3/(4\pi^2)\):

\[ \frac{1}{\lambda} = \frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \cdot \frac{h^3}{4\pi^2} \cdot c}\left(\frac{1}{n^2} - \frac{1}{m^2}\right) = \frac{m_e e^4 \cdot 4\pi^2}{2(4\pi\varepsilon_0)^2 h^3 c}\left(\frac{1}{n^2} - \frac{1}{m^2}\right) \]

Simplifying:

\[ \frac{1}{\lambda} = R_\infty \left(\frac{1}{n^2} - \frac{1}{m^2}\right) \]

where the Rydberg constant is:

\[ \boxed{R_\infty = \frac{m_e e^4}{8\varepsilon_0^2 h^3 c}} \]

Verification: Substituting numerical values gives \(R_\infty \simeq 1.097 \times 10^7\;\mathrm{m^{-1}}\), which agrees with the experimental value. Furthermore, substituting \(n=2, m=3\) reproduces the result from D4 (656 nm).

M-2. Determination of Planck's Constant from Photoelectric Effect Experimental Data

→ Back to problem

(a) Shape of the Graph

\(K = h\nu - W\) is a linear function of \(\nu\) (a straight line) with slope \(h\) and \(K\)-intercept \(-W\).

Examining the data: - For every increase of \(1.5 \times 10^{14}\;\mathrm{Hz}\) in \(\nu\), \(K\) increases by approximately 0.62 eV - The increase is uniform, confirming that a linear relationship holds ✓

(b) Determination of Planck's Constant

We determine the slope of the line using two points. For example, using the first and last data points:

\[ h = \frac{\Delta K}{\Delta \nu} = \frac{(2.07 - 0.21)\;\mathrm{eV}}{(10.5 - 6.0) \times 10^{14}\;\mathrm{Hz}} = \frac{1.86\;\mathrm{eV}}{4.5 \times 10^{14}\;\mathrm{Hz}} \]
\[ \boxed{h = 4.13 \times 10^{-15}\;\mathrm{eV \cdot s}} \]

Verification: Computing with two adjacent points:

\[ \frac{0.83 - 0.21}{(7.5 - 6.0) \times 10^{14}} = \frac{0.62}{1.5 \times 10^{14}} = 4.13 \times 10^{-15}\;\mathrm{eV \cdot s} \quad \checkmark \]

(c) Determination of the Work Function

From \(K = h\nu - W\):

\[ W = h\nu - K = (4.13 \times 10^{-15})(6.0 \times 10^{14}) - 0.21 = 2.48 - 0.21 = 2.27\;\mathrm{eV} \]
\[ \boxed{W \simeq 2.27\;\mathrm{eV}} \]

Alternative method: Finding the threshold frequency \(\nu_0\) (the frequency at which \(K = 0\)):

\[ \nu_0 = \frac{W}{h} = \frac{2.27}{4.13 \times 10^{-15}} = 5.50 \times 10^{14}\;\mathrm{Hz} \]

This closely matches the threshold frequency of sodium at the D2 line, suggesting that the metal is sodium.

(d) Conversion to SI Units and Comparison with Literature Values

\[ h = 4.13 \times 10^{-15}\;\mathrm{eV \cdot s} \times 1.602 \times 10^{-19}\;\mathrm{J/eV} \]
\[ h = 6.62 \times 10^{-34}\;\mathrm{J \cdot s} \]
\[ \boxed{h \simeq 6.62 \times 10^{-34}\;\mathrm{J \cdot s}} \]

Comparing with the literature value \(h = 6.626 \times 10^{-34}\;\mathrm{J \cdot s}\), the relative error is approximately 0.1%, which is excellent agreement.

Check: Confirming that all data points lie on the line. Substituting each \(\nu\) into \(K = (4.13 \times 10^{-15})\nu - 2.27\): - \(\nu = 6.0 \times 10^{14}\): \(K = 2.48 - 2.27 = 0.21\) ✓ - \(\nu = 7.5 \times 10^{14}\): \(K = 3.10 - 2.27 = 0.83\) ✓ - \(\nu = 9.0 \times 10^{14}\): \(K = 3.72 - 2.27 = 1.45\) ✓ - \(\nu = 10.5 \times 10^{14}\): \(K = 4.34 - 2.27 = 2.07\) ✓

M-3. Order-of-magnitude estimate of classical atomic collapse time

→ Back to problem

(a) Centripetal Acceleration

From the Coulomb force:

\[ F = \frac{e^2}{4\pi\varepsilon_0 a_0^2} = m_e a \]
\[ a = \frac{e^2}{4\pi\varepsilon_0 m_e a_0^2} \]

Numerical calculation:

\[ a = \frac{(1.602 \times 10^{-19})^2}{(1.113 \times 10^{-10})(9.109 \times 10^{-31})(5.3 \times 10^{-11})^2} \]

Numerator:

\[ (1.602 \times 10^{-19})^2 = 2.566 \times 10^{-38}\;\mathrm{C^2} \]

Denominator:

\[ (1.113 \times 10^{-10})(9.109 \times 10^{-31})(2.809 \times 10^{-21}) \]
\[ = 1.113 \times 9.109 \times 2.809 \times 10^{-10-31-21} = 28.48 \times 10^{-62} = 2.848 \times 10^{-61} \]
\[ a = \frac{2.566 \times 10^{-38}}{2.848 \times 10^{-61}} = 9.01 \times 10^{22}\;\mathrm{m/s^2} \]
\[ \boxed{a \simeq 9.0 \times 10^{22}\;\mathrm{m/s^2}} \]

(b) Radiated Power

Larmor's formula:

\[ P = \frac{e^2 a^2}{6\pi\varepsilon_0 c^3} \]
\[ P = \frac{(1.602 \times 10^{-19})^2 \times (9.0 \times 10^{22})^2}{6\pi \times (8.854 \times 10^{-12}) \times (3.00 \times 10^8)^3} \]

Numerator:

\[ (2.566 \times 10^{-38}) \times (8.1 \times 10^{44}) = 2.08 \times 10^{7} \]

Denominator:

\[ 6\pi \times (8.854 \times 10^{-12}) \times (2.7 \times 10^{25}) = 6\pi \times 2.39 \times 10^{14} \]
\[ = 18.85 \times 2.39 \times 10^{14} = 4.50 \times 10^{15} \]
\[ P = \frac{2.08 \times 10^{7}}{4.50 \times 10^{15}} = 4.6 \times 10^{-9}\;\mathrm{W} \]
\[ \boxed{P \simeq 4.6 \times 10^{-9}\;\mathrm{W}} \]

(c) Estimate of the Collapse Time

\[ |E_1| = 13.6\;\mathrm{eV} = 13.6 \times 1.602 \times 10^{-19} = 2.18 \times 10^{-18}\;\mathrm{J} \]
\[ \tau \sim \frac{|E_1|}{P} = \frac{2.18 \times 10^{-18}}{4.6 \times 10^{-9}} = 4.7 \times 10^{-10}\;\mathrm{s} \]
\[ \boxed{\tau \sim 5 \times 10^{-10}\;\mathrm{s}} \]

Verification and Discussion: This estimate is about one order of magnitude larger than the \(\tau \sim 10^{-11}\;\mathrm{s}\) given in Eq. (1.5), but this is within a reasonable range for an order-of-magnitude estimate. In a rigorous calculation, as the electron spirals inward, the acceleration increases (because \(r\) decreases), and the radiated power also increases, so the actual collapse time is shorter than this simple estimate. In any case, it is on the order of \(10^{-11}\)–\(10^{-10}\;\mathrm{s}\), confirming that classically the atom would collapse almost instantaneously.

M-4. High-Frequency Limit of the Planck Distribution and Wien's Law

→ Back to problem

(a) Wien's Radiation Law

When \(h\nu \gg k_B T\), we have \(e^{h\nu/k_B T} \gg 1\), so:

\[ e^{h\nu/k_B T} - 1 \simeq e^{h\nu/k_B T} \]

Therefore:

\[ B(\nu, T) = \frac{2h\nu^3}{c^2} \cdot \frac{1}{e^{h\nu/k_B T} - 1} \simeq \frac{2h\nu^3}{c^2} \cdot \frac{1}{e^{h\nu/k_B T}} \]
\[ \boxed{B(\nu, T) \simeq \frac{2h\nu^3}{c^2} e^{-h\nu/k_B T} \quad (h\nu \gg k_B T)} \]

This is Wien's radiation law. \(\square\)

(b) Wien's Displacement Law

Setting \(x = h\nu/(k_B T)\), we have \(\nu = k_B T x / h\), and:

\[ B(\nu, T) = \frac{2h}{c^2}\left(\frac{k_B T}{h}\right)^3 x^3 \cdot \frac{1}{e^x - 1} \]

We consider the condition \(\partial B / \partial \nu = 0\) for the maximum of \(B\). Converting the derivative with respect to \(\nu\) to a derivative with respect to \(x\) (at fixed \(T\), \(d\nu \propto dx\)):

\[ \frac{\partial}{\partial x}\left[\frac{x^3}{e^x - 1}\right] = 0 \]

Expanding this:

\[ \frac{3x^2(e^x - 1) - x^3 e^x}{(e^x - 1)^2} = 0 \]

The condition for the numerator to vanish:

\[ 3x^2(e^x - 1) - x^3 e^x = 0 \]

Dividing by \(x^2\) (since \(x \neq 0\)):

\[ 3(e^x - 1) - x e^x = 0 \quad \Longrightarrow \quad 3 - 3e^{-x} - x = 0 \quad \Longrightarrow \quad (3 - x) = 3e^{-x} \]

This is a transcendental equation in \(x\) alone, and its solution \(x_{\max}\) is a constant independent of temperature \(T\) (numerically \(x_{\max} \simeq 2.821\)).

Since \(x_{\max} = h\nu_{\max}/(k_B T)\) is a constant:

\[ \boxed{\nu_{\max} = \frac{x_{\max} \cdot k_B}{h} \cdot T \propto T} \]

This is Wien's displacement law. \(\square\)

Verification: For \(T = 6000\;\mathrm{K}\), we get \(\nu_{\max} = 2.821 \times (1.38 \times 10^{-23})/(6.626 \times 10^{-34}) \times 6000 \simeq 3.5 \times 10^{14}\;\mathrm{Hz}\). The corresponding wavelength is \(\lambda \simeq 860\;\mathrm{nm}\) (near-infrared). From Wien's displacement law in wavelength form, \(\lambda_{\max} T = 2.898 \times 10^{-3}\;\mathrm{m \cdot K}\), we obtain \(\lambda_{\max} \simeq 483\;\mathrm{nm}\), which differs. However, this is a well-known effect due to the fact that the peak positions differ between the frequency and wavelength representations, and is not a contradiction.

Advanced

A-1. Derivation of the Rayleigh–Jeans Law from the Classical Equipartition Theorem and the Ultraviolet Catastrophe

→ Back to problem

(a) Derivation of the Number of Modes

For standing waves satisfying the boundary conditions (electric field vanishes at the walls) inside a cubic cavity of side \(L\):

\[ k_x = \frac{n_x \pi}{L}, \quad k_y = \frac{n_y \pi}{L}, \quad k_z = \frac{n_z \pi}{L} \quad (n_x, n_y, n_z = 1, 2, 3, \ldots) \]

In \(k\)-space, the spacing between lattice points is \(\pi/L\), and the volume per lattice point is \((\pi/L)^3\).

We consider only the first octant where \(n_x, n_y, n_z > 0\). The number of lattice points with wave number magnitude \(k = \sqrt{k_x^2 + k_y^2 + k_z^2}\) in the range \(k\) to \(k + dk\) is the volume of the first-octant spherical shell of radius \(k\) divided by the volume per lattice point:

\[ \text{Number of lattice points} = \frac{1}{8} \cdot 4\pi k^2 dk \cdot \frac{1}{(\pi/L)^3} = \frac{1}{8} \cdot 4\pi k^2 dk \cdot \frac{L^3}{\pi^3} \]
\[ = \frac{V k^2 dk}{2\pi^2} \]

Since electromagnetic waves have 2 independent polarizations, we multiply by 2:

\[ g(k)\,dk = 2 \cdot \frac{V k^2 dk}{2\pi^2} = \frac{V k^2}{\pi^2}\,dk \]
\[ \boxed{g(k)\,dk = \frac{V}{\pi^2} k^2\,dk} \]

(b) Conversion to Frequency Representation

From \(k = 2\pi\nu/c\), we have \(dk = 2\pi\,d\nu/c\). Substituting:

\[ g(\nu)\,d\nu = \frac{V}{\pi^2} \left(\frac{2\pi\nu}{c}\right)^2 \cdot \frac{2\pi}{c}\,d\nu = \frac{V}{\pi^2} \cdot \frac{4\pi^2\nu^2}{c^2} \cdot \frac{2\pi}{c}\,d\nu \]
\[ = \frac{8\pi^3 V \nu^2}{\pi^2 c^3}\,d\nu = \frac{8\pi V \nu^2}{c^3}\,d\nu \]
\[ \boxed{g(\nu)\,d\nu = \frac{8\pi V \nu^2}{c^3}\,d\nu} \]

(c) The Rayleigh–Jeans Law

By the classical equipartition theorem, the average energy per mode is \(k_B T\) (\(\frac{1}{2}k_B T\) each for kinetic energy and potential energy).

The spectral energy density per unit volume:

\[ u(\nu, T) = \frac{1}{V} \cdot g(\nu) \cdot k_B T = \frac{8\pi\nu^2}{c^3} k_B T \]
\[ \boxed{u(\nu, T) = \frac{8\pi\nu^2}{c^3} k_B T} \]

(d) The Ultraviolet Catastrophe

Integrating over \(\nu\) to obtain the total energy density:

\[ u_{\text{total}} = \int_0^\infty u(\nu, T)\,d\nu = \frac{8\pi k_B T}{c^3} \int_0^\infty \nu^2\,d\nu \]
\[ \int_0^\infty \nu^2\,d\nu = \left[\frac{\nu^3}{3}\right]_0^\infty = \infty \]

The total energy density diverges. This is physically absurd, implying that high-frequency modes (ultraviolet and above) carry unlimited energy. This is the ultraviolet catastrophe. \(\square\)

(e) Resolution by Planck's Mean Energy

Replacing the equipartition value \(k_B T\) with Planck's mean energy \(\langle E \rangle = h\nu/(e^{h\nu/k_B T} - 1)\):

\[ u(\nu, T) = \frac{8\pi\nu^2}{c^3} \cdot \frac{h\nu}{e^{h\nu/k_B T} - 1} \]

At high frequencies (\(h\nu \gg k_B T\)):

\[ \frac{h\nu}{e^{h\nu/k_B T} - 1} \simeq h\nu \cdot e^{-h\nu/k_B T} \]

This decays exponentially, so \(\nu^2 \cdot h\nu \cdot e^{-h\nu/k_B T} = h\nu^3 e^{-h\nu/k_B T}\) converges to zero as \(\nu \to \infty\). Therefore:

\[ u_{\text{total}} = \int_0^\infty \frac{8\pi h\nu^3}{c^3(e^{h\nu/k_B T} - 1)}\,d\nu < \infty \]

The integral is finite (explicit evaluation yields the Stefan–Boltzmann law \(u_{\text{total}} \propto T^4\)).

Physical reason: Due to the quantization of energy, for high-frequency modes the energy of a single quantum \(h\nu\) far exceeds the thermal energy \(k_B T\), so the excitation probability is exponentially suppressed by the Boltzmann factor \(e^{-h\nu/k_B T}\). This resolves the ultraviolet catastrophe. \(\square\)

A-2. Generalization of the Bohr Model: Hydrogen-like Ions and the Correspondence Principle

→ Back to problem

(a) Orbital Radius and Energy Levels of Hydrogen-like Ions

Since the nuclear charge is \(Ze\), the Coulomb force gives:

\[ \frac{Ze^2}{4\pi\varepsilon_0 r^2} = \frac{m_e v^2}{r} \]

Combining with the quantization condition \(m_e v r = n\hbar\). Following the same procedure as D5, substitute \(v = n\hbar/(m_e r)\):

\[ \frac{Ze^2}{4\pi\varepsilon_0 r^2} = \frac{n^2\hbar^2}{m_e r^3} \]
\[ \boxed{r_n = \frac{4\pi\varepsilon_0 \hbar^2}{m_e Z e^2} \cdot n^2 = \frac{a_0}{Z} \cdot n^2} \]

Velocity:

\[ v_n = \frac{n\hbar}{m_e r_n} = \frac{Ze^2}{4\pi\varepsilon_0 \hbar} \cdot \frac{1}{n} \]

Total energy (similarly to S1(b), \(E_n = -Ze^2/(8\pi\varepsilon_0 r_n)\)):

\[ E_n = -\frac{Ze^2}{8\pi\varepsilon_0} \cdot \frac{m_e Ze^2}{4\pi\varepsilon_0 \hbar^2 n^2} = -\frac{m_e Z^2 e^4}{2(4\pi\varepsilon_0)^2 \hbar^2} \cdot \frac{1}{n^2} \]
\[ \boxed{E_n = -\frac{Z^2 m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2} \cdot \frac{1}{n^2} = -Z^2 \cdot \frac{13.6\;\mathrm{eV}}{n^2}} \]

(b) Ground State of He\(^+\) (\(Z = 2\))

Orbital radius:

\[ r_1 = \frac{a_0}{Z} = \frac{0.529\;\text{Å}}{2} = 0.265\;\text{Å} = 2.65 \times 10^{-11}\;\mathrm{m} \]

Energy:

\[ E_1 = -Z^2 \times 13.6\;\mathrm{eV} = -4 \times 13.6 = -54.4\;\mathrm{eV} \]

Comparison with the hydrogen atom:

Hydrogen (\(Z=1\)) He\(^+\) (\(Z=2\))
\(r_1\) \(0.529\;\text{Å}\) \(0.265\;\text{Å}\) (half)
\(E_1\) \(-13.6\;\mathrm{eV}\) \(-54.4\;\mathrm{eV}\) (4 times deeper)

He\(^+\) has a smaller orbit than hydrogen (because the stronger nuclear charge attracts the electron more strongly), and the binding energy is 4 times larger.

(c) Verification of the Correspondence Principle

Quantum transition frequency:

\[ \nu_{n \to n-1} = \frac{E_n - E_{n-1}}{h} = \frac{Z^2 m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2 h}\left(\frac{1}{(n-1)^2} - \frac{1}{n^2}\right) \]

Evaluating the expression in parentheses:

\[ \frac{1}{(n-1)^2} - \frac{1}{n^2} = \frac{n^2 - (n-1)^2}{n^2(n-1)^2} = \frac{2n - 1}{n^2(n-1)^2} \]

For \(n \gg 1\), \(2n - 1 \simeq 2n\) and \((n-1)^2 \simeq n^2\), so:

\[ \frac{1}{(n-1)^2} - \frac{1}{n^2} \simeq \frac{2n}{n^2 \cdot n^2} = \frac{2}{n^3} \]

Therefore:

\[ \nu_{n \to n-1} \simeq \frac{Z^2 m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2 h} \cdot \frac{2}{n^3} = \frac{Z^2 m_e e^4}{(4\pi\varepsilon_0)^2 \hbar^2 h \cdot n^3} \tag{★} \]

Classical orbital frequency:

\[ f_n = \frac{v_n}{2\pi r_n} \]

Substituting \(v_n = Ze^2/(4\pi\varepsilon_0 \hbar n)\) and \(r_n = 4\pi\varepsilon_0 \hbar^2 n^2/(m_e Ze^2)\):

\[ f_n = \frac{1}{2\pi} \cdot \frac{Ze^2}{4\pi\varepsilon_0 \hbar n} \cdot \frac{m_e Ze^2}{4\pi\varepsilon_0 \hbar^2 n^2} \]
\[ = \frac{Z^2 m_e e^4}{2\pi (4\pi\varepsilon_0)^2 \hbar^3 n^3} \]

Using \(\hbar = h/(2\pi)\), so \(\hbar^3 = h^3/(8\pi^3)\):

\[ f_n = \frac{Z^2 m_e e^4}{2\pi (4\pi\varepsilon_0)^2 n^3} \cdot \frac{8\pi^3}{h^3} = \frac{Z^2 m_e e^4 \cdot 4\pi^2}{(4\pi\varepsilon_0)^2 h^3 n^3} \]

On the other hand, rewriting (★) using \(\hbar = h/(2\pi)\):

\[ \nu_{n \to n-1} \simeq \frac{Z^2 m_e e^4}{(4\pi\varepsilon_0)^2 \cdot \frac{h^2}{4\pi^2} \cdot h \cdot n^3} = \frac{Z^2 m_e e^4 \cdot 4\pi^2}{(4\pi\varepsilon_0)^2 h^3 n^3} \]

This is in perfect agreement with \(f_n\):

\[ \boxed{\nu_{n \to n-1} \simeq f_n \quad (n \gg 1)} \]

Meaning of the correspondence principle: In the limit of large quantum numbers (\(n \gg 1\)), the transition frequency predicted by quantum theory coincides with the orbital frequency predicted by classical theory. This serves as evidence that quantum theory is the "correct generalization" of classical theory, and it is an important principle that Bohr used to confirm the validity of his model. \(\square\)

Sanity check: Consider the case \(Z = 1\), \(n = 1000\): we get \(r_{1000} = 10^6 a_0 \simeq 0.05\;\mathrm{mm}\), which is a macroscopic scale, making it intuitively clear that the classical picture should be applicable.