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Ch. 1 Solutions

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Basic

Medium

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Basic

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

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Problem:

Let a typical frequency of visible light be \(\nu = 5.0 \times 10^{14}\;\mathrm{Hz}\). Calculate the energy of a single photon \(E = h\nu\) in SI units (J), and then convert it to electron volts (eV). Use \(h = 6.626 \times 10^{-34}\;\mathrm{J \cdot s}\) and \(1\;\mathrm{eV} = 1.602 \times 10^{-19}\;\mathrm{J}\).

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

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Problem:

The work function of sodium (Na) is \(W = 2.28\;\mathrm{eV}\). Find the threshold frequency \(\nu_0\) for the photoelectric effect to occur. Also, find the corresponding threshold wavelength \(\lambda_0\) in units of nm. Use \(c = 3.00 \times 10^8\;\mathrm{m/s}\).

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

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Problem:

Ultraviolet light with wavelength \(\lambda = 200\;\mathrm{nm}\) is incident on a metal with work function \(W = 4.50\;\mathrm{eV}\). Find the maximum kinetic energy \(K\) of the emitted electrons in units of eV.

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

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Problem:

Using the Rydberg formula (1.6), find the wavelength \(\lambda\) in nm of the spectral line corresponding to \(m = 3\) in the Balmer series (\(n = 2\)) of the hydrogen atom. Use \(R_\infty = 1.097 \times 10^7\;\mathrm{m^{-1}}\).

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

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Problem:

In the Bohr model of the hydrogen atom, the force balance for the electron's circular motion (Coulomb force = centripetal force) is given by

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

By combining this with Bohr's quantization condition \(m_e v r = n\hbar\), express the \(n\)-th orbital radius \(r_n\) in terms of \(n\), \(\hbar\), \(m_e\), \(e\), and \(\varepsilon_0\).

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

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Problem:

Setting \(n = 1\) in the result of D5 gives the value called the Bohr radius \(a_0\). Using the following constants, calculate the value of \(a_0\) to 3 significant figures.

  • \(\hbar = 1.055 \times 10^{-34}\;\mathrm{J \cdot s}\)
  • \(m_e = 9.109 \times 10^{-31}\;\mathrm{kg}\)
  • \(e = 1.602 \times 10^{-19}\;\mathrm{C}\)
  • \(4\pi\varepsilon_0 = 1.113 \times 10^{-10}\;\mathrm{C^2/(N \cdot m^2)}\)

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

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Problem:

Regarding Planck's formula for the average energy

\[ \langle E \rangle = \frac{h\nu}{e^{h\nu / k_B T} - 1} \]

show that in the limit \(h\nu \ll k_B T\), we obtain \(\langle E \rangle \simeq k_B T\), using the approximation \(e^x \simeq 1 + x\) (\(x \ll 1\)).

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

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Problem:

At temperature \(T = 6000\;\mathrm{K}\) (approximately the temperature of the Sun's surface), calculate the Boltzmann factor \(e^{-h\nu / k_B T}\) for each of the following two frequencies.

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

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

Compare the results and confirm how high-frequency modes are suppressed.

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

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Problem:

In the Bohr model of the hydrogen atom, derive the energy levels \(E_n\) following the steps below.

(a) By simultaneously solving the force balance for the electron's circular motion (Coulomb force = centripetal force) and the Bohr quantization condition \(L = n\hbar\), find the radius \(r_n\) and velocity \(v_n\) of the \(n\)-th orbit.

(b) Express the total energy of the electron (kinetic energy + Coulomb potential energy) in terms of \(r_n\), and derive

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

(c) Using the frequency condition \(h\nu = E_m - E_n\) and \(c = \lambda\nu\), reproduce the Rydberg formula (1.6) and express the Rydberg constant \(R_\infty\) in terms of fundamental constants.

(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

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Problem:

In a photoelectric effect experiment, a metal is irradiated with light of various frequencies \(\nu\), and the maximum kinetic energy \(K\) of the ejected electrons is measured. Suppose the following data are obtained.

\(\nu\;[10^{14}\;\mathrm{Hz}]\) \(K\;[\mathrm{eV}]\)
6.0 0.21
7.5 0.83
9.0 1.45
10.5 2.07

(a) Sketch the general shape of the graph of \(K\) as a function of \(\nu\), and confirm that the linear relationship \(K = h\nu - W\) holds.

(b) Determine the value of Planck's constant \(h\) in units of eV·s from the data.

(c) Determine the work function \(W\) of this metal in units of eV.

(d) Convert the obtained value of \(h\) to SI units (J·s) and compare it with the literature value \(h = 6.626 \times 10^{-34}\;\mathrm{J \cdot s}\).

(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

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Problem:

In classical electromagnetism, the energy radiated per unit time by a charge \(e\) moving with acceleration \(a\) (Larmor's formula) is given by

\[ P = \frac{e^2 a^2}{6\pi\varepsilon_0 c^3} \]

Assume that the electron in a hydrogen atom is in a circular orbit at the Bohr radius \(a_0 \simeq 5.3 \times 10^{-11}\;\mathrm{m}\), and answer the following.

(a) Express the centripetal acceleration \(a\) of the electron in terms of \(a_0\), \(m_e\), \(e\), and \(\varepsilon_0\) from the force balance, and calculate its numerical value.

(b) Use Larmor's formula to calculate the radiated power \(P\).

(c) Using the total energy of the electron \(E_1 = -13.6\;\mathrm{eV}\), estimate the classical collapse time \(\tau\) from the order of magnitude of \(|E_1|/P\), and confirm that it is consistent with \(\tau \sim 10^{-11}\;\mathrm{s}\) in Eq. (1.5).

(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

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Problem:

The spectral radiance (per unit frequency) of Planck's blackbody radiation is given by

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

(a) Show that in the limit \(h\nu \gg k_B T\), \(B(\nu, T) \simeq \frac{2h\nu^3}{c^2} e^{-h\nu/k_B T}\) (Wien's radiation law).

(b) Show that the frequency \(\nu_{\max}\) at which \(B(\nu, T)\) is maximized is proportional to the temperature \(T\) (Wien's displacement law: \(\nu_{\max} \propto T\)) by rewriting \(\partial B / \partial \nu = 0\) in terms of \(x = h\nu/(k_B T)\). (It is not necessary to solve the transcendental equation; it suffices to show that \(x\) is a constant.)

(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

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Problem:

Consider the number of standing wave modes of the electromagnetic field inside a cubic cavity of side length \(L\).

(a) Show that the number of modes (including the 2 degrees of freedom of polarization) in the wavenumber range from \(k\) to \(k + dk\) is

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

where \(V = L^3\) is the volume of the cavity. (Use the boundary conditions \(k_x = n_x \pi / L\), \(k_y = n_y \pi / L\), \(k_z = n_z \pi / L\) (\(n_x, n_y, n_z = 1, 2, 3, \ldots\)) for each direction.)

(b) Using the relation \(k = 2\pi\nu/c\), rewrite the number of modes in the frequency range from \(\nu\) to \(\nu + d\nu\) as

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

(c) By assigning an average energy \(k_B T\) to each mode according to the classical equipartition theorem, show that the spectral energy density per unit volume \(u(\nu, T)\) is

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

(the Rayleigh–Jeans law).

(d) Show that integrating \(u(\nu, T)\) from \(\nu = 0\) to \(\infty\) diverges, and explain that this is the ultraviolet catastrophe.

(e) Qualitatively explain why replacing the equipartition theorem with Planck's mean energy \(\langle E \rangle = h\nu/(e^{h\nu/k_B T} - 1)\) yields a finite total energy density.

(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

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Problem:

Apply the Bohr model to a hydrogen-like ion (atomic number \(Z\), one electron).

(a) Taking into account that the nuclear charge is \(Ze\), derive the \(n\)-th orbital radius \(r_n\) and energy level \(E_n\) in a form that includes \(Z\).

(b) Calculate the energy and orbital radius of the ground state (\(n = 1\)) of \(\mathrm{He^+}\) (\(Z = 2\)), and compare them with those of the hydrogen atom.

(c) Bohr's correspondence principle: Show that when the quantum number \(n\) is sufficiently large, the frequency \(\nu_{n \to n-1}\) of light emitted in a transition between adjacent levels coincides with the classical orbital frequency \(f_n\) of the electron in the \(n\)-th orbit. That is, confirm that in the limit \(n \gg 1\),

\[ \nu_{n \to n-1} \simeq f_n \]

holds.

(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.