IChO 1991 · Theory problem 4

Hydrogen atom energy levels, Lyman series and the photoelectric effect

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The energy of stable states of the hydrogen atom is given by: \(E_n=-2.18\times10^{-18}/n^2\ [\mathrm{J}]\)

where \(n\) denotes the principal quantum number.

4.1 Calculate the energy differences between \(n=2\) (first excited state) and \(n=1\) (ground state) and between \(n=7\) and \(n=1\).

4.2 In what spectral range is the Lyman series lying?

4.3 Can a single photon, emitted in the first and/or sixth line of the Lyman series, ionize:

a) another hydrogen atom in its ground state?

b) a copper atom in the \(\ce{Cu}\) crystal?

The electron work function of \(\ce{Cu}\) is \(\Phi_{\ce{Cu}}=7.44\times10^{-19}\ \mathrm{J}\).

4.4 Calculate the de Broglie wavelength of the electrons emitted from a copper crystal when irradiated by photons from the first line and the sixth line of the Lyman series.

\(h=6.6256\times10^{-34}\ \mathrm{J\,s};\qquad m_e=9.1091\times10^{-31}\ \mathrm{kg};\qquad c=2.99792\times10^8\ \mathrm{m\,s^{-1}}\)

Official English text of the problem.

Answers

  • 4.1 \(E_2-E_1=\boxed{1.64\times10^{-18}\ \mathrm{J}}\); \(E_7-E_1=\boxed{2.14\times10^{-18}\ \mathrm{J}}\).
  • 4.2 The ultraviolet, specifically vacuum ultraviolet: approximately \(91.1\)–\(121\ \mathrm{nm}\), with \(91.1\ \mathrm{nm}\) the series limit.
  • 4.3(a) Neither photon can ionize a ground-state hydrogen atom.
  • 4.3(b) Both photons can eject an electron from the copper crystal.
  • 4.4 First line: \(\boxed{\lambda_e=5.20\times10^{-10}\ \mathrm{m}=0.520\ \mathrm{nm}}\). Sixth line: \(\boxed{\lambda_e=4.16\times10^{-10}\ \mathrm{m}=0.416\ \mathrm{nm}}\). These correspond to the maximum-energy photoelectrons.

Solution

4.1 — Energy differences

For an excited level \(n\), its energy above the ground state is

\[ \Delta E_{n,1}=E_n-E_1 =2.18\times10^{-18}\left(1-\frac{1}{n^2}\right)\ \mathrm{J}. \]

Thus,

\[ \Delta E_{2,1} =2.18\times10^{-18}\left(1-\frac14\right) =1.635\times10^{-18}\ \mathrm{J} \approx\boxed{1.64\times10^{-18}\ \mathrm{J}}, \]

and

\[ \Delta E_{7,1} =2.18\times10^{-18}\left(1-\frac1{49}\right) =2.135510\ldots\times10^{-18}\ \mathrm{J} \approx\boxed{2.14\times10^{-18}\ \mathrm{J}}. \]

These positive energy differences equal the photon energies emitted in the downward transitions. The atom’s energy changes during emission have the opposite sign.

4.2 — Spectral range of the Lyman series

The Lyman series consists of transitions from \(n=2,3,\ldots\) to \(n=1\). Its first line is therefore \(2\to1\), and its sixth line is \(7\to1\).

Using \(E_\gamma=hc/\lambda_\gamma\), the longest wavelength is that of the first line:

\[ \lambda_{\gamma,1} =\frac{(6.6256\times10^{-34})(2.99792\times10^8)} {1.635\times10^{-18}} =1.21486\ldots\times10^{-7}\ \mathrm{m} \approx121\ \mathrm{nm}. \]

As \(n\to\infty\), the photon energy approaches \(2.18\times10^{-18}\ \mathrm{J}\), giving the short-wavelength limit:

\[ \lambda_{\mathrm{limit}} =\frac{hc}{2.18\times10^{-18}} =9.11148\ldots\times10^{-8}\ \mathrm{m} \approx91.1\ \mathrm{nm}. \]

Thus, the entire series lies in the ultraviolet, specifically the vacuum-ultraviolet region.

4.3 — Can the photons cause ionization?

(a) Ground-state hydrogen

Ionization requires raising the electron from \(E_1=-2.18\times10^{-18}\ \mathrm{J}\) to the continuum at zero energy. The required energy is therefore

\[ E_{\mathrm{ion}}=0-E_1=2.18\times10^{-18}\ \mathrm{J}. \]

Both photon energies are smaller:

\[ E_{\gamma,1}=1.635\times10^{-18}\ \mathrm{J}<E_{\mathrm{ion}}, \]

\[ E_{\gamma,6}=2.135510\ldots\times10^{-18}\ \mathrm{J}<E_{\mathrm{ion}}. \]

Consequently, neither photon can ionize a ground-state hydrogen atom. They can excite it to the corresponding bound level.

(b) Copper crystal

For electron emission from the copper crystal, the relevant threshold is the work function:

\[ \Phi_{\ce{Cu}}=7.44\times10^{-19}\ \mathrm{J}. \]

Since

\[ E_{\gamma,1}>\Phi_{\ce{Cu}} \qquad\text{and}\qquad E_{\gamma,6}>\Phi_{\ce{Cu}}, \]

both photons can eject an electron by the photoelectric effect. The supplied work function describes electron removal from the solid.

4.4 — De Broglie wavelengths of the emitted electrons

The photoelectric equation gives the maximum kinetic energy:

\[ K_{\max}=E_\gamma-\Phi_{\ce{Cu}}. \]

For a nonrelativistic electron,

\[ K=\frac{p^2}{2m_e}, \qquad \lambda_e=\frac{h}{p} =\frac{h}{\sqrt{2m_eK}}. \]

For irradiation by the first Lyman line,

\[ K_{\max,1} =1.635\times10^{-18}-7.44\times10^{-19} =8.91\times10^{-19}\ \mathrm{J}, \]

so

\[ \lambda_{e,1} =\frac{6.6256\times10^{-34}} {\sqrt{2(9.1091\times10^{-31})(8.91\times10^{-19})}} =\boxed{5.20\times10^{-10}\ \mathrm{m}}. \]

For irradiation by the sixth Lyman line, retaining unrounded intermediate values,

\[ K_{\max,6} =2.135510\ldots\times10^{-18}-7.44\times10^{-19} =1.391510\ldots\times10^{-18}\ \mathrm{J}, \]

and

\[ \lambda_{e,6} =\frac{6.6256\times10^{-34}} {\sqrt{2(9.1091\times10^{-31})(1.391510\ldots\times10^{-18})}} =\boxed{4.16\times10^{-10}\ \mathrm{m}}. \]

These are the minimum de Broglie wavelengths, corresponding to the maximum kinetic energies. Photoelectrons emitted with less kinetic energy have longer wavelengths.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.