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Ch. 2 Solutions

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Basic

B-1. In Compton scattering, suppose the wavelength of the incident X-ray is nm. For a scattering angle (back

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Problem:

In Compton scattering, suppose the wavelength of the incident X-ray is \(\lambda = 0.0711\) nm. Find the wavelength \(\lambda'\) of the scattered X-ray when the scattering angle is \(\theta = 180°\) (backscattering). You may use the Compton wavelength \(h/(m_e c) = 2.43 \times 10^{-12}\) m.

Compton Scattering: Scattered Wavelength at Backscattering (\(\theta = 180°\))

Solution Strategy:

Substitute \(\theta = 180°\) into the Compton scattering formula \(\lambda' - \lambda = \dfrac{h}{m_e c}(1 - \cos\theta)\).

Calculation:

\[\cos 180° = -1\]
\[\lambda' - \lambda = \frac{h}{m_e c}(1 - (-1)) = 2 \times \frac{h}{m_e c} = 2 \times 2.43 \times 10^{-12}\ \mathrm{m} = 4.86 \times 10^{-12}\ \mathrm{m}\]
\[\lambda' = \lambda + 4.86 \times 10^{-12}\ \mathrm{m} = 0.0711 \times 10^{-9}\ \mathrm{m} + 4.86 \times 10^{-12}\ \mathrm{m}\]
\[= 71.1 \times 10^{-12}\ \mathrm{m} + 4.86 \times 10^{-12}\ \mathrm{m} = 75.96 \times 10^{-12}\ \mathrm{m}\]

Final Answer:

\[\boxed{\lambda' \approx 0.0760\ \mathrm{nm}}\]

Verification:

The wavelength shift \(\Delta\lambda = 4.86 \times 10^{-12}\) m is twice the Compton wavelength, which corresponds to the maximum wavelength shift for backscattering. This represents approximately a 6.8% change relative to the original wavelength of \(71.1\) pm, which is a readily measurable value in X-ray experiments. This is reasonable.

B-2. In the Compton scattering formula (2.1), find the wavelength shift for scattering angle

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Problem:

In the Compton scattering formula (2.1), find the wavelength shift \(\Delta\lambda = \lambda' - \lambda\) for a scattering angle of \(\theta = 60°\).

Compton Scattering: Wavelength Shift at \(\theta = 60°\)

Calculation:

\[\cos 60° = 0.5\]
\[\Delta\lambda = \frac{h}{m_e c}(1 - \cos 60°) = 2.43 \times 10^{-12} \times (1 - 0.5) = 2.43 \times 10^{-12} \times 0.5\]

Final Answer:

\[\boxed{\Delta\lambda = 1.22 \times 10^{-12}\ \mathrm{m} = 1.22\ \mathrm{pm}}\]

Verification:

\(\theta = 60°\) is intermediate between \(\theta = 90°\) (\(\Delta\lambda = \lambda_C\)) and \(\theta = 0°\) (\(\Delta\lambda = 0\)). \(\Delta\lambda = \lambda_C / 2\) is half the Compton wavelength, which is reasonable.

B-3. Find the de Broglie wavelength of a proton with mass kg moving at velocity m/s

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Problem:

Find the de Broglie wavelength \(\lambda = h/(mv)\) of a proton with mass \(m = 1.67 \times 10^{-27}\) kg moving at velocity \(v = 3.0 \times 10^{4}\) m/s.

de Broglie Wavelength of a Proton

Calculation:

\[\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 3.0 \times 10^4}\]

Denominator:

\[mv = 1.67 \times 10^{-27} \times 3.0 \times 10^4 = 5.01 \times 10^{-23}\ \mathrm{kg \cdot m/s}\]
\[\lambda = \frac{6.626 \times 10^{-34}}{5.01 \times 10^{-23}} = 1.32 \times 10^{-11}\ \mathrm{m}\]

Final Answer:

\[\boxed{\lambda \approx 1.32 \times 10^{-11}\ \mathrm{m} = 0.132\ \mathrm{\AA}}\]

Verification:

Dimensional check: \([h]/([m][v]) = \mathrm{J \cdot s}/(\mathrm{kg \cdot m/s}) = \mathrm{kg \cdot m^2/s}/(\mathrm{kg \cdot m/s}) = \mathrm{m}\). ✓ The value is shorter than interatomic spacings (a few Å), and since the proton is heavier than the electron, a shorter wavelength is reasonable.

B-4. de Broglie wavelength of an electron accelerated through voltage V, using the simplified formula

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Problem:

Find the de Broglie wavelength of an electron accelerated through an accelerating voltage \(V_{\mathrm{acc}} = 200\) V, using the simplified formula

\[ \lambda \approx \frac{1.226}{\sqrt{V_{\mathrm{acc}}}}\ \mathrm{nm}\]

Also, convert the result to Å (angstrom) units.

de Broglie wavelength of an electron at an accelerating voltage of 200 V

Calculation:

\[\sqrt{200} = \sqrt{4 \times 50} = 2\sqrt{50} = 2 \times 7.071 = 14.14\]
\[\lambda \approx \frac{1.226}{14.14}\ \mathrm{nm} = 0.0867\ \mathrm{nm}\]

Conversion to Å units:

\[\lambda = 0.0867\ \mathrm{nm} \times 10\ \mathrm{\AA/nm} = 0.867\ \mathrm{\AA}\]

Final Answer:

\[\boxed{\lambda \approx 0.0867\ \mathrm{nm} = 0.867\ \mathrm{\AA}}\]

Verification:

For \(V_{\mathrm{acc}} = 150\) V, \(\lambda \approx 1.00\) Å (from the text). Since \(V_{\mathrm{acc}} = 200\) V is greater than 150 V, the wavelength should be shorter. \(1.226/\sqrt{150} = 1.226/12.25 = 0.100\) nm = 1.00 Å. At 200 V, 0.867 Å < 1.00 Å. ✓

B-5. When an electron (mass kg) is accelerated through an accelerating voltage V, find the electron's velocity using Eq. (2.9)

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Problem:

When an electron (mass \(m_e = 9.109 \times 10^{-31}\) kg) is accelerated through an accelerating voltage \(V_{\mathrm{acc}} = 54\) V, find the electron's velocity \(v\) using Eq. (2.9)

\[ \frac{1}{2}m_e v^2 = eV_{\mathrm{acc}}\]

where \(e = 1.602 \times 10^{-19}\) C.

Velocity of an electron accelerated through an accelerating voltage of 54 V

Calculation:

\[v = \sqrt{\frac{2eV_{\mathrm{acc}}}{m_e}}\]

Numerator:

\[2eV_{\mathrm{acc}} = 2 \times 1.602 \times 10^{-19} \times 54 = 1.730 \times 10^{-17}\ \mathrm{J}\]
\[\frac{2eV_{\mathrm{acc}}}{m_e} = \frac{1.730 \times 10^{-17}}{9.109 \times 10^{-31}} = 1.899 \times 10^{13}\ \mathrm{m^2/s^2}\]
\[v = \sqrt{1.899 \times 10^{13}} = 4.36 \times 10^6\ \mathrm{m/s}\]

Final Answer:

\[\boxed{v \approx 4.36 \times 10^6\ \mathrm{m/s}}\]

Verification:

\(v/c = 4.36 \times 10^6 / 3.0 \times 10^8 \approx 0.015\). The non-relativistic approximation is valid. In the text, for \(V_{\mathrm{acc}} = 100\) V, \(v \approx 5.93 \times 10^6\) m/s. Since \(v \propto \sqrt{V_{\mathrm{acc}}}\), we have \(v(54\mathrm{V})/v(100\mathrm{V}) = \sqrt{54/100} = 0.735\). \(5.93 \times 10^6 \times 0.735 = 4.36 \times 10^6\). ✓

B-6. Finding the wavelength of a diffracted wave given crystal plane spacing in Å, angle of incidence, and order in the Bragg condition

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Problem:

In the Bragg condition \(2d\sin\theta = n\lambda\), find the wavelength \(\lambda\) of the diffracted wave in Å when the crystal plane spacing is \(d = 0.91\) Å, the angle of incidence is \(\theta = 65°\), and the order is \(n = 1\).

Finding the wavelength from the Bragg condition

Calculation:

\[\lambda = \frac{2d\sin\theta}{n} = \frac{2 \times 0.91 \times \sin 65°}{1}\]
\[\sin 65° \approx 0.906\]
\[\lambda = 2 \times 0.91 \times 0.906 = 1.649\ \mathrm{\AA}\]

Final Answer:

\[\boxed{\lambda \approx 1.65\ \mathrm{\AA}}\]

Verification:

This value is on the same order as the electron wavelength obtained in the Davisson-Germer experiment (\(\sim 1.65\) Å) and falls within the observable range for crystal diffraction. The condition \(\lambda < 2d = 1.82\) Å also satisfies the requirement of the Bragg condition. ✓

B-7. Find the de Broglie wavelength of a neutron (mass in kg) with kinetic energy in eV, given the conversion in J

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Problem:

Find the de Broglie wavelength of a neutron (mass \(m_n = 1.675 \times 10^{-27}\) kg) with kinetic energy \(K = 1.0\) eV. Assume \(1\ \mathrm{eV} = 1.602 \times 10^{-19}\) J.

de Broglie wavelength of a neutron with kinetic energy 1.0 eV

Calculation:

Converting kinetic energy to J:

\[K = 1.0\ \mathrm{eV} = 1.0 \times 1.602 \times 10^{-19}\ \mathrm{J} = 1.602 \times 10^{-19}\ \mathrm{J}\]

Momentum:

\[p = \sqrt{2m_n K} = \sqrt{2 \times 1.675 \times 10^{-27} \times 1.602 \times 10^{-19}}\]
\[= \sqrt{5.367 \times 10^{-46}} = \sqrt{53.67 \times 10^{-47}} = 7.326 \times 10^{-23.5}\]

Calculating more carefully:

\[2m_n K = 2 \times 1.675 \times 10^{-27} \times 1.602 \times 10^{-19} = 5.367 \times 10^{-46}\ \mathrm{kg^2 \cdot m^2/s^2}\]
\[p = \sqrt{5.367 \times 10^{-46}} = \sqrt{5.367} \times 10^{-23} = 2.317 \times 10^{-23}\ \mathrm{kg \cdot m/s}\]

de Broglie wavelength:

\[\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{2.317 \times 10^{-23}} = 2.86 \times 10^{-11}\ \mathrm{m}\]

Final Answer:

\[\boxed{\lambda \approx 2.86 \times 10^{-11}\ \mathrm{m} = 0.286\ \mathrm{\AA}}\]

Verification:

A 1 eV neutron has higher energy than a "thermal neutron" (\(\sim 0.025\) eV), so its wavelength should be shorter. The wavelength of a thermal neutron is \(\sim 1.8\) Å. Since \(\lambda \propto 1/\sqrt{K}\), we have \(\lambda(1\mathrm{eV})/\lambda(0.025\mathrm{eV}) = \sqrt{0.025/1} = 0.158\). \(1.8 \times 0.158 \approx 0.28\) Å. ✓

B-8. Using de Broglie's relations expressed in terms of angular frequency and wavenumber, E = ℏω and p = ℏk

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Problem:

Using de Broglie's relations expressed in terms of angular frequency \(\omega\) and wavenumber \(k\), namely \(E = \hbar\omega\) and \(p = \hbar k\), find the wavenumber \(k\) in SI units (m\(^{-1}\)) corresponding to an electron with wavelength \(\lambda = 2.0\) Å.

Wavenumber \(k\) corresponding to an electron with wavelength 2.0 Å

Calculation:

\[\lambda = 2.0\ \mathrm{\AA} = 2.0 \times 10^{-10}\ \mathrm{m}\]
\[k = \frac{2\pi}{\lambda} = \frac{2\pi}{2.0 \times 10^{-10}} = \frac{6.283}{2.0 \times 10^{-10}} = 3.14 \times 10^{10}\ \mathrm{m^{-1}}\]

Final Answer:

\[\boxed{k \approx 3.14 \times 10^{10}\ \mathrm{m^{-1}}}\]

Verification:

\(p = \hbar k = 1.055 \times 10^{-34} \times 3.14 \times 10^{10} = 3.31 \times 10^{-24}\) kg·m/s. \(\lambda = h/p = 6.626 \times 10^{-34} / 3.31 \times 10^{-24} = 2.0 \times 10^{-10}\) m = 2.0 Å. ✓

Medium

M-1. Derivation of the Relationship Between Accelerating Voltage and de Broglie Wavelength

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Problem:

Derivation of the Relationship Between Accelerating Voltage and de Broglie Wavelength

When an electron (mass \(m_e\), charge \(e\)) is accelerated from rest through a voltage \(V_{\mathrm{acc}}\), follow the steps below to derive that the de Broglie wavelength is expressed as

\[ \lambda = \frac{h}{\sqrt{2m_e eV_{\mathrm{acc}}}}\]

(a) From energy conservation, express the electron's momentum \(p\) in terms of \(m_e\), \(e\), and \(V_{\mathrm{acc}}\).

(b) Substitute into the de Broglie relation \(\lambda = h/p\) to obtain the above expression.

(c) Substitute numerical values of the constants to verify that \(\lambda \approx 1.226/\sqrt{V_{\mathrm{acc}}}\) nm is obtained.

Derivation of the Relationship Between Accelerating Voltage and de Broglie Wavelength

(a) Expressing momentum \(p\) in terms of \(m_e\), \(e\), \(V_{\mathrm{acc}}\):

From energy conservation, the kinetic energy of an electron accelerated from rest through an accelerating voltage \(V_{\mathrm{acc}}\) is:

\[\frac{1}{2}m_e v^2 = eV_{\mathrm{acc}}\]

Using the momentum \(p = m_e v\), we have \(\frac{1}{2}m_e v^2 = \frac{p^2}{2m_e}\), so:

\[\frac{p^2}{2m_e} = eV_{\mathrm{acc}}\]
\[\boxed{p = \sqrt{2m_e eV_{\mathrm{acc}}}}\]

(b) Derivation of the de Broglie wavelength:

Substituting the result from (a) into the de Broglie relation \(\lambda = h/p\):

\[\boxed{\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_e eV_{\mathrm{acc}}}}}\]

(c) Verification of the convenient formula by numerical substitution:

Computing the constant factor:

\[\frac{h}{\sqrt{2m_e e}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19}}}\]

The quantity inside the square root:

\[2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} = 2.919 \times 10^{-49}\]
\[\sqrt{2.919 \times 10^{-49}} = \sqrt{2.919} \times 10^{-24.5} = 1.709 \times 10^{-24.5}\]

Since \(10^{-24.5} = 10^{-25} \times \sqrt{10} = 3.162 \times 10^{-25}\):

\[\sqrt{2.919 \times 10^{-49}} = 1.709 \times 3.162 \times 10^{-25} = 5.403 \times 10^{-25}\]
\[\frac{h}{\sqrt{2m_e e}} = \frac{6.626 \times 10^{-34}}{5.403 \times 10^{-25}} = 1.226 \times 10^{-9}\ \mathrm{m \cdot V^{1/2}}\]

Therefore:

\[\lambda = \frac{1.226 \times 10^{-9}}{\sqrt{V_{\mathrm{acc}}}}\ \mathrm{m} = \frac{1.226}{\sqrt{V_{\mathrm{acc}}}}\ \mathrm{nm}\]
\[\boxed{\lambda \approx \frac{1.226}{\sqrt{V_{\mathrm{acc}}}}\ \mathrm{nm}}\]

Verification:

For \(V_{\mathrm{acc}} = 150\) V, \(\lambda = 1.226/\sqrt{150} = 1.226/12.25 = 0.1001\) nm = 1.00 Å. This agrees with the example in the text. ✓

M-2. Reproduction Calculation of the Davisson-Germer Experiment

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Problem:

Reproduction Calculation of the Davisson-Germer Experiment

In the Davisson-Germer experiment, electrons accelerated through a voltage of \(V_{\mathrm{acc}} = 54\) V were directed at a nickel single crystal (surface atomic spacing \(d = 2.15\) Å), and a strong diffraction peak was observed at a scattering angle of \(\phi = 50°\).

(a) Using Eq. (2.12), find the de Broglie wavelength \(\lambda_{\mathrm{dB}}\) of electrons accelerated through 54 V.

(b) Substituting \(n = 1\) and \(\phi = 50°\) into the diffraction condition \(d\sin\phi = n\lambda\), find the wavelength \(\lambda_{\mathrm{exp}}\) obtained from the experiment.

(c) Compare \(\lambda_{\mathrm{dB}}\) and \(\lambda_{\mathrm{exp}}\), and discuss the validity of the de Broglie hypothesis.

Reproducing the Davisson-Germer Experiment Calculation

(a) Calculation of the de Broglie Wavelength:

Using Equation (2.12):

\[\lambda_{\mathrm{dB}} = \frac{1.226}{\sqrt{54}}\ \mathrm{nm} = \frac{1.226}{7.348}\ \mathrm{nm} = 0.1668\ \mathrm{nm} = 1.668\ \mathrm{\AA}\]
\[\boxed{\lambda_{\mathrm{dB}} \approx 1.67\ \mathrm{\AA}}\]

(b) Wavelength Obtained from the Diffraction Condition:

Substituting \(d = 2.15\) Å, \(\phi = 50°\), \(n = 1\) into the diffraction condition \(d\sin\phi = n\lambda\):

\[\lambda_{\mathrm{exp}} = d\sin\phi = 2.15 \times \sin 50° = 2.15 \times 0.766 = 1.647\ \mathrm{\AA}\]
\[\boxed{\lambda_{\mathrm{exp}} \approx 1.65\ \mathrm{\AA}}\]

(c) Comparison and Discussion:

The difference between the two values:

\[\frac{|\lambda_{\mathrm{dB}} - \lambda_{\mathrm{exp}}|}{\lambda_{\mathrm{dB}}} \times 100\% = \frac{|1.67 - 1.65|}{1.67} \times 100\% \approx 1.2\%\]

The wavelength predicted by the de Broglie hypothesis \(\lambda_{\mathrm{dB}} \approx 1.67\) Å and the wavelength measured from the diffraction experiment \(\lambda_{\mathrm{exp}} \approx 1.65\) Å agree to within approximately 1%. This level of discrepancy falls within the range of experimental measurement precision (angle reading errors, uncertainty in lattice spacing, etc.), and we can conclude that de Broglie's matter wave hypothesis \(\lambda = h/p\) has been experimentally confirmed.

Verification:

A possible source of the slight discrepancy is the refraction effect due to the inner potential when electrons enter the crystal interior (correction to the accelerating voltage). In more precise analyses, including this correction yields even better agreement.

M-3. Structural Analysis of the Compton Scattering Formula

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Problem:

Structural Analysis of the Compton Scattering Formula

Regarding the Compton scattering formula

\[ \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)\]

answer the following questions.

(a) Determine the range of possible values of \(\Delta\lambda = \lambda' - \lambda\) as a function of \(\theta\) (for \(0 \le \theta \le 180°\)).

(b) Calculate the Compton wavelength \(h/(m_p c)\) for the case where the scattering target is a proton (mass \(m_p = 1.673 \times 10^{-27}\) kg) instead of an electron, and compare it with the electron case.

(c) Using the result from (b), explain why "the wavelength shift in Compton scattering is larger for lighter particles."

Structural Analysis of the Compton Scattering Formula

(a) Range of Possible Values of \(\Delta\lambda\):

\[\Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta)\]

In the range \(0 \le \theta \le 180°\), \(\cos\theta\) varies from \(+1\) to \(-1\), so \((1 - \cos\theta)\) varies from \(0\) to \(2\).

\[\boxed{0 \le \Delta\lambda \le \frac{2h}{m_e c} = 4.86 \times 10^{-12}\ \mathrm{m}}\]
  • \(\theta = 0°\) (forward scattering): \(\Delta\lambda = 0\) (no wavelength change)
  • \(\theta = 90°\): \(\Delta\lambda = h/(m_e c) = 2.43 \times 10^{-12}\) m
  • \(\theta = 180°\) (backscattering): \(\Delta\lambda = 2h/(m_e c) = 4.86 \times 10^{-12}\) m (maximum)

(b) Compton Wavelength of the Proton:

\[\frac{h}{m_p c} = \frac{6.626 \times 10^{-34}}{1.673 \times 10^{-27} \times 2.998 \times 10^8}\]
\[= \frac{6.626 \times 10^{-34}}{5.015 \times 10^{-19}} = 1.321 \times 10^{-15}\ \mathrm{m}\]
\[\boxed{\frac{h}{m_p c} \approx 1.32 \times 10^{-15}\ \mathrm{m} = 1.32\ \mathrm{fm}}\]

Ratio to the electron Compton wavelength:

\[\frac{h/(m_e c)}{h/(m_p c)} = \frac{m_p}{m_e} = \frac{1.673 \times 10^{-27}}{9.109 \times 10^{-31}} \approx 1836\]

The proton Compton wavelength is approximately \(1/1836\) times that of the electron, about 1840 times smaller (\(1.32\) fm) compared to the electron case (\(2.43\) pm).

(c) Explanation of "Lighter Particles Produce Larger Wavelength Shifts":

The wavelength shift in Compton scattering is:

\[\Delta\lambda = \frac{h}{mc}(1 - \cos\theta)\]

Here, \(h/(mc)\) is inversely proportional to the mass \(m\) of the target particle. Therefore:

  • Scattering from a light particle (electron): The Compton wavelength is large, so the wavelength shift is large
  • Scattering from a heavy particle (proton): The Compton wavelength is small, so the wavelength shift is small

Physically, a lighter particle receives momentum from the photon and recoils more significantly, so the energy lost by the photon (= increase in wavelength) is larger. Conversely, for a very heavy particle (\(m \to \infty\)), \(\Delta\lambda \to 0\), recovering classical Thomson scattering (no wavelength change).

Verification:

In the limit \(m \to \infty\), \(\Delta\lambda \to 0\) is consistent with the classical intuition that a ball bouncing off a heavy wall rebounds at nearly the same speed (no energy change). ✓

M-4. Analysis of Electron Diffraction Using the Bragg Condition

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Problem:

Analysis of Electron Diffraction Using the Bragg Condition

Consider a situation modeled after G. P. Thomson's experiment. Electrons accelerated through a voltage of \(V_{\mathrm{acc}} = 10{,}000\) V are incident on an aluminum crystal with interplanar spacing \(d = 2.34\) Å.

(a) Find the de Broglie wavelength of the electrons.

(b) Using the Bragg condition \(2d\sin\theta = n\lambda\), find the incident angle \(\theta\) at which \(n = 1\) diffraction occurs.

(c) When the accelerating voltage is doubled (\(20{,}000\) V), how does the diffraction angle \(\theta\) for \(n = 1\) change? Provide a qualitative explanation and then calculate the specific value.

Analysis of Electron Diffraction Modeled After G. P. Thomson's Experiment

(a) de Broglie Wavelength of the Electron:

\[\lambda = \frac{1.226}{\sqrt{V_{\mathrm{acc}}}}\ \mathrm{nm} = \frac{1.226}{\sqrt{10000}}\ \mathrm{nm} = \frac{1.226}{100}\ \mathrm{nm} = 0.01226\ \mathrm{nm}\]
\[\boxed{\lambda = 0.1226\ \mathrm{\AA}}\]

(b) Diffraction Angle \(\theta\) for \(n = 1\):

From the Bragg condition \(2d\sin\theta = n\lambda\):

\[\sin\theta = \frac{n\lambda}{2d} = \frac{1 \times 0.1226}{2 \times 2.34} = \frac{0.1226}{4.68} = 0.02620\]
\[\theta = \arcsin(0.02620) \approx 1.50°\]

Since \(\sin\theta \approx 0.0262\) is sufficiently small, the small-angle approximation \(\sin\theta \approx \theta\) (in radians) is also applicable: \(\theta \approx 0.0262\) rad \(= 1.50°\).

\[\boxed{\theta \approx 1.50°}\]

(c) When the Accelerating Voltage Is Increased to 20,000 V:

Qualitative explanation: Since \(\lambda \propto 1/\sqrt{V_{\mathrm{acc}}}\), doubling the accelerating voltage reduces the wavelength by a factor of \(1/\sqrt{2}\). From the Bragg condition \(\sin\theta = \lambda/(2d)\), \(\sin\theta\) is also reduced by a factor of \(1/\sqrt{2}\), so the diffraction angle decreases.

Specific calculation:

\[\lambda' = \frac{1.226}{\sqrt{20000}}\ \mathrm{nm} = \frac{1.226}{141.4}\ \mathrm{nm} = 0.008671\ \mathrm{nm} = 0.08671\ \mathrm{\AA}\]
\[\sin\theta' = \frac{0.08671}{2 \times 2.34} = \frac{0.08671}{4.68} = 0.01853\]
\[\theta' = \arcsin(0.01853) \approx 1.06°\]
\[\boxed{\theta' \approx 1.06°}\]

Verification: \(\theta'/\theta = 1.06°/1.50° = 0.707 = 1/\sqrt{2}\) (within the range where the small-angle approximation holds, \(\theta \propto \lambda \propto 1/\sqrt{V_{\mathrm{acc}}}\)). ✓

Consistency Check:

The fact that higher-energy electrons have shorter wavelengths and smaller diffraction angles is physically reasonable. In G. P. Thomson's experiment, accelerating voltages of tens of thousands of volts were used, and diffraction rings were observed at small angles, which is consistent with these calculated results.

Advanced

A-1. Derivation of the Compton Scattering Formula

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Problem:

Derivation of the Compton Scattering Formula

Consider the case where an X-ray photon of wavelength \(\lambda\) collides with an electron at rest (mass \(m_e\)). Let \(\lambda'\) be the wavelength of the scattered photon, \(\theta\) be the scattering angle, \(p_e\) be the magnitude of the recoil electron's momentum, and \(\varphi\) be the recoil angle (the angle between the incident direction and the electron's direction of motion).

Following the steps below, derive the Compton scattering formula

\[ \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)\]

The photon energy is given by \(E = pc\) (the relativistic relation for a massless particle), and the electron energy is given relativistically by \(E_e = \sqrt{(p_e c)^2 + (m_e c^2)^2}\).

(a) Write down the energy conservation equation. Express it using \(h c/\lambda\) for the incident photon energy, \(h c/\lambda'\) for the scattered photon energy, and \(m_e c^2\) for the rest energy of the recoil electron.

(b) Write down the momentum conservation equations, separating them into the component along the incident direction (\(x\) direction) and the perpendicular component (\(y\) direction).

(c) Eliminate the recoil angle \(\varphi\) from the two equations in (b), and express \(p_e^2\) in terms of \(\lambda\), \(\lambda'\), and \(\theta\).

(d) From the energy conservation in (a), express \(p_e^2 c^2\) in terms of \(\lambda\) and \(\lambda'\), and by equating it with the result from (c), derive the Compton scattering formula.

Derivation of the Compton Scattering Formula

(a) Conservation of Energy:

The energy of the incident photon is \(hc/\lambda\), and the energy of the scattered photon is \(hc/\lambda'\). Since the electron is initially at rest, its initial energy is only the rest energy \(m_e c^2\). The energy of the electron after scattering is \(E_e = \sqrt{(p_e c)^2 + (m_e c^2)^2}\).

Conservation of energy:

\[\boxed{\frac{hc}{\lambda} + m_e c^2 = \frac{hc}{\lambda'} + \sqrt{(p_e c)^2 + (m_e c^2)^2}} \tag{A1.1}\]

(b) Conservation of Momentum:

Let the incident direction be the \(x\)-axis and the perpendicular direction be the \(y\)-axis. The momentum of the incident photon is \(h/\lambda\) (in the \(x\) direction). The scattered photon travels at angle \(\theta\), and the recoil electron travels at angle \(\varphi\).

\(x\) component:

\[\frac{h}{\lambda} = \frac{h}{\lambda'}\cos\theta + p_e\cos\varphi \tag{A1.2}\]

\(y\) component:

\[0 = \frac{h}{\lambda'}\sin\theta - p_e\sin\varphi \tag{A1.3}\]

(c) Eliminating \(\varphi\) and Deriving \(p_e^2\):

Rearranging equation (A1.2):

\[p_e\cos\varphi = \frac{h}{\lambda} - \frac{h}{\lambda'}\cos\theta \tag{A1.4}\]

Rearranging equation (A1.3):

\[p_e\sin\varphi = \frac{h}{\lambda'}\sin\theta \tag{A1.5}\]

Squaring (A1.4) and (A1.5) and adding them (using \(\cos^2\varphi + \sin^2\varphi = 1\)):

\[p_e^2 = \left(\frac{h}{\lambda} - \frac{h}{\lambda'}\cos\theta\right)^2 + \left(\frac{h}{\lambda'}\sin\theta\right)^2\]

Expanding:

\[p_e^2 = \frac{h^2}{\lambda^2} - \frac{2h^2\cos\theta}{\lambda\lambda'} + \frac{h^2\cos^2\theta}{\lambda'^2} + \frac{h^2\sin^2\theta}{\lambda'^2}\]
\[= \frac{h^2}{\lambda^2} - \frac{2h^2\cos\theta}{\lambda\lambda'} + \frac{h^2(\cos^2\theta + \sin^2\theta)}{\lambda'^2}\]
\[\boxed{p_e^2 = \frac{h^2}{\lambda^2} + \frac{h^2}{\lambda'^2} - \frac{2h^2\cos\theta}{\lambda\lambda'}} \tag{A1.6}\]

(d) Obtaining \(p_e^2 c^2\) from Energy Conservation and Deriving the Compton Formula:

Rearranging equation (A1.1) to isolate the recoil electron energy on the left side:

\[\sqrt{(p_e c)^2 + (m_e c^2)^2} = \frac{hc}{\lambda} - \frac{hc}{\lambda'} + m_e c^2\]

Squaring both sides:

\[(p_e c)^2 + (m_e c^2)^2 = \left(\frac{hc}{\lambda} - \frac{hc}{\lambda'} + m_e c^2\right)^2\]

Expanding the right side. Let \(A = hc/\lambda - hc/\lambda'\) and \(B = m_e c^2\):

\[(A + B)^2 = A^2 + 2AB + B^2\]
\[p_e^2 c^2 + m_e^2 c^4 = \left(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\right)^2 + 2\left(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\right)m_e c^2 + m_e^2 c^4\]

The \(m_e^2 c^4\) terms cancel on both sides:

\[p_e^2 c^2 = \left(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\right)^2 + 2m_e c^2\left(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\right)\]

Expanding \(A^2\) on the left:

\[\left(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\right)^2 = h^2c^2\left(\frac{1}{\lambda^2} - \frac{2}{\lambda\lambda'} + \frac{1}{\lambda'^2}\right)\]

Therefore:

\[p_e^2 c^2 = h^2c^2\left(\frac{1}{\lambda^2} + \frac{1}{\lambda'^2} - \frac{2}{\lambda\lambda'}\right) + 2m_e c^3 h\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right) \tag{A1.7}\]

On the other hand, multiplying (A1.6) by \(c^2\):

\[p_e^2 c^2 = h^2c^2\left(\frac{1}{\lambda^2} + \frac{1}{\lambda'^2} - \frac{2\cos\theta}{\lambda\lambda'}\right) \tag{A1.8}\]

Equating (A1.7) and (A1.8):

\[h^2c^2\left(\frac{1}{\lambda^2} + \frac{1}{\lambda'^2} - \frac{2}{\lambda\lambda'}\right) + 2m_e c^3 h\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right) = h^2c^2\left(\frac{1}{\lambda^2} + \frac{1}{\lambda'^2} - \frac{2\cos\theta}{\lambda\lambda'}\right)\]

Canceling \(h^2c^2\left(\frac{1}{\lambda^2} + \frac{1}{\lambda'^2}\right)\) from both sides:

\[-\frac{2h^2c^2}{\lambda\lambda'} + 2m_e c^3 h\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right) = -\frac{2h^2c^2\cos\theta}{\lambda\lambda'}\]

Rearranging:

\[2m_e c^3 h\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right) = -\frac{2h^2c^2\cos\theta}{\lambda\lambda'} + \frac{2h^2c^2}{\lambda\lambda'}\]
\[2m_e c^3 h\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right) = \frac{2h^2c^2}{\lambda\lambda'}(1 - \cos\theta)\]

Dividing both sides by \(2hc^2\):

\[m_e c\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right) = \frac{h}{\lambda\lambda'}(1 - \cos\theta)\]

Transforming the left side:

\[\frac{1}{\lambda} - \frac{1}{\lambda'} = \frac{\lambda' - \lambda}{\lambda\lambda'}\]

Substituting:

\[m_e c \cdot \frac{\lambda' - \lambda}{\lambda\lambda'} = \frac{h}{\lambda\lambda'}(1 - \cos\theta)\]

Multiplying both sides by \(\lambda\lambda'\):

\[m_e c(\lambda' - \lambda) = h(1 - \cos\theta)\]
\[\boxed{\lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)}\]

Verification:

  • Case \(\theta = 0\): \(\lambda' - \lambda = 0\). No wavelength change for forward scattering. This is physically reasonable since the photon merely grazes the electron. ✓
  • Dimensional analysis: \([h/(m_e c)] = \mathrm{J \cdot s}/(\mathrm{kg} \cdot \mathrm{m/s}) = \mathrm{kg \cdot m^2/s}/(\mathrm{kg \cdot m/s}) = \mathrm{m}\). This has dimensions of wavelength. ✓
  • Case \(\theta = 180°\): \(\Delta\lambda = 2h/(m_e c)\). Maximum wavelength change. This is reasonable since backscattering transfers the maximum energy to the electron. ✓

A-2. Relativistic Electron de Broglie Wavelength and Application to Electron Microscopy

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Problem:

Relativistic Electron de Broglie Wavelength and Application to Electron Microscopy

Equation (2.11) in this chapter is based on the non-relativistic approximation (\(v \ll c\)). When the accelerating voltage is very high (e.g., \(V_{\mathrm{acc}} = 200\) kV as used in electron microscopes), the kinetic energy of the electron is no longer negligible compared to the rest energy \(m_e c^2 \approx 511\) keV, and relativistic corrections become necessary.

(a) In the relativistic energy-momentum relation

\[ E^2 = (pc)^2 + (m_e c^2)^2\]

use the fact that the total energy is \(E = m_e c^2 + eV_{\mathrm{acc}}\) (rest energy + kinetic energy) to express the momentum \(p\) in terms of \(m_e\), \(c\), \(e\), and \(V_{\mathrm{acc}}\).

(b) Show that the relativistic de Broglie wavelength can be written as $$

\lambda = \frac{h}{\sqrt{2m_e eV_{\mathrm{acc}}\left(1 + \dfrac{eV_{\mathrm{acc}}}{2m_e c^2}\right)}}$$

(c) For \(V_{\mathrm{acc}} = 200\) kV, calculate the wavelength numerically using both the non-relativistic formula (2.11) and the relativistic formula from (b), and evaluate the difference between them as a percentage.

(d) Assume that the resolution of an electron microscope is approximately equal to the wavelength used. Discuss whether an electron microscope with \(V_{\mathrm{acc}} = 200\) kV can resolve atomic-level structures (\(\sim 1\) Å).

Relativistic Electron de Broglie Wavelength and Application to Electron Microscopy

(a) Derivation of Relativistic Momentum:

The relativistic energy-momentum relation:

\[E^2 = (pc)^2 + (m_e c^2)^2\]

When an electron is accelerated through an accelerating voltage \(V_{\mathrm{acc}}\), the total energy is:

\[E = m_e c^2 + eV_{\mathrm{acc}}\]

Substituting and solving for \(p\):

\[(m_e c^2 + eV_{\mathrm{acc}})^2 = (pc)^2 + (m_e c^2)^2\]
\[(pc)^2 = (m_e c^2 + eV_{\mathrm{acc}})^2 - (m_e c^2)^2\]

Expanding the right-hand side:

\[(m_e c^2)^2 + 2m_e c^2 \cdot eV_{\mathrm{acc}} + (eV_{\mathrm{acc}})^2 - (m_e c^2)^2 = 2m_e c^2 \cdot eV_{\mathrm{acc}} + (eV_{\mathrm{acc}})^2\]
\[p^2 c^2 = 2m_e c^2 \cdot eV_{\mathrm{acc}} + (eV_{\mathrm{acc}})^2\]
\[\boxed{p = \frac{1}{c}\sqrt{2m_e c^2 \cdot eV_{\mathrm{acc}} + (eV_{\mathrm{acc}})^2}}\]

(b) Relativistic de Broglie Wavelength:

\[\lambda = \frac{h}{p} = \frac{hc}{\sqrt{2m_e c^2 \cdot eV_{\mathrm{acc}} + (eV_{\mathrm{acc}})^2}}\]

Factoring out \(2m_e eV_{\mathrm{acc}}\) from the denominator:

\[p^2 = \frac{1}{c^2}\left[2m_e c^2 \cdot eV_{\mathrm{acc}}\left(1 + \frac{eV_{\mathrm{acc}}}{2m_e c^2}\right)\right] = 2m_e eV_{\mathrm{acc}}\left(1 + \frac{eV_{\mathrm{acc}}}{2m_e c^2}\right)\]
\[p = \sqrt{2m_e eV_{\mathrm{acc}}\left(1 + \frac{eV_{\mathrm{acc}}}{2m_e c^2}\right)}\]

Therefore:

\[\boxed{\lambda = \frac{h}{\sqrt{2m_e eV_{\mathrm{acc}}\left(1 + \dfrac{eV_{\mathrm{acc}}}{2m_e c^2}\right)}}}\]

In the non-relativistic limit (\(eV_{\mathrm{acc}} \ll m_e c^2\)), the correction term inside the parentheses becomes negligible, reducing to \(\lambda = h/\sqrt{2m_e eV_{\mathrm{acc}}}\). ✓

(c) Numerical Calculation for \(V_{\mathrm{acc}} = 200\) kV:

Non-relativistic wavelength:

\[\lambda_{\mathrm{NR}} = \frac{1.226}{\sqrt{200{,}000}}\ \mathrm{nm} = \frac{1.226}{447.2}\ \mathrm{nm} = 2.742 \times 10^{-3}\ \mathrm{nm} = 0.02742\ \mathrm{\AA}\]

Relativistic correction factor:

\[eV_{\mathrm{acc}} = 200\ \mathrm{keV},\quad m_e c^2 = 511\ \mathrm{keV}\]
\[\frac{eV_{\mathrm{acc}}}{2m_e c^2} = \frac{200}{2 \times 511} = \frac{200}{1022} = 0.1957\]
\[1 + \frac{eV_{\mathrm{acc}}}{2m_e c^2} = 1.1957\]

Relativistic wavelength:

\[\lambda_{\mathrm{rel}} = \frac{\lambda_{\mathrm{NR}}}{\sqrt{1 + \dfrac{eV_{\mathrm{acc}}}{2m_e c^2}}} = \frac{0.02742}{\sqrt{1.1957}}\ \mathrm{\AA} = \frac{0.02742}{1.0935}\ \mathrm{\AA} = 0.02508\ \mathrm{\AA}\]
\[\boxed{\lambda_{\mathrm{NR}} \approx 0.0274\ \mathrm{\AA},\quad \lambda_{\mathrm{rel}} \approx 0.0251\ \mathrm{\AA}}\]

Percentage difference:

\[\frac{\lambda_{\mathrm{NR}} - \lambda_{\mathrm{rel}}}{\lambda_{\mathrm{rel}}} \times 100\% = \frac{0.0274 - 0.0251}{0.0251} \times 100\% \approx 9.2\%\]

The non-relativistic formula overestimates the correct relativistic value by approximately 9%. At an accelerating voltage of 200 kV, the relativistic correction cannot be neglected.

(d) Observability of Atomic-Scale Structure:

The de Broglie wavelength of electrons used in an electron microscope with \(V_{\mathrm{acc}} = 200\) kV is:

\[\lambda_{\mathrm{rel}} \approx 0.0251\ \mathrm{\AA} = 2.51 \times 10^{-12}\ \mathrm{m}\]

Since the size of atomic-scale structures is \(\sim 1\) Å \(= 10^{-10}\) m, the electron wavelength is approximately 40 times shorter than this.

Since the diffraction-limited resolution is on the order of the wavelength used, from the wavelength perspective, atomic-scale structures can be sufficiently resolved. In actual electron microscopes, the resolution is worse than the theoretical limit due to lens aberrations and other factors, but recent aberration-correction techniques have achieved atomic resolution (\(\sim 1\) Å or better).

\[\boxed{\text{Wavelength } 0.025\ \mathrm{\AA} \ll 1\ \mathrm{\AA}\text{, so atomic-scale observation is fully possible in principle.}}\]

Verification:

  • Check of non-relativistic limit: \(eV_{\mathrm{acc}}/(2m_e c^2) = 0.196\) is not negligible, consistent with the need for relativistic correction.
  • Limit as \(V_{\mathrm{acc}} \to 0\): The correction factor \(\to 1\), reducing to the non-relativistic formula. ✓
  • Order of magnitude of the wavelength: A wavelength of \(\sim 0.025\) Å for a 200 kV electron microscope agrees with textbook values for electron microscopy. ✓