Atomic and Molecular Physics Questions and Answers

Answer: C

The factor is \(\frac{4n^{2}}{n^{2} - 4}\), up to a constant.

The fringe width is \(\beta = \lambda D/d\). With the screen distance fixed, keeping \(\beta\) constant requires \(d \propto \lambda\). For the Balmer series,

\[ \begin{aligned} \frac{1}{\lambda} &= R\left(\frac{1}{4} - \frac{1}{n^{2}}\right) = R\,\frac{n^{2} - 4}{4n^{2}} \\ \lambda &= \frac{1}{R} \cdot \frac{4n^{2}}{n^{2} - 4} \end{aligned} \]

so \(d\) must be scaled by \(4n^{2}/(n^{2} - 4)\). The option \((n^{2} - 4)/4n^{2}\) is proportional to \(1/\lambda\), the inverse of what is needed, and the forms with \(n^{2} + 4\) do not come from the Rydberg formula.

Answer: A

The first ionization potential is 0.89 atomic units.

In neutral helium each electron sees the screened charge \(Z = 1.7\), so with both in \(n = 1\) the atom's energy is

\[ \begin{aligned} E_{\text{He}} &= 2 \times \left(-\frac{1.7^{2}}{2}\right) \\ &= -2.89 \end{aligned} \]

After one electron is removed, the remaining electron in He⁺ sees the full nuclear charge \(Z = 2\):

\[ \begin{aligned} E_{\text{He}^{+}} &= -\frac{2^{2}}{2} = -2 \\ \text{IP} &= -2 - (-2.89) = 0.89 \end{aligned} \]

0.89 a.u. is about 24.2 eV, close to the measured 24.6 eV. 1.78 doubles the answer, and 1.445 (\(1.7^{2}/2\)) would be the result of removing one screened electron without letting the other relax, which is not offered.

Answer: D

The shortest wavelength is 8205 Å (the paper's 'Paschen back series' means the Paschen series).

Paschen lines end on \( n = 3 \). The shortest wavelength is the series limit, an electron falling from \( n = \infty \):

\[ \begin{aligned} \frac{1}{\lambda} &= R_{H}\left(\frac{1}{3^{2}} - 0\right) \\ \lambda &= \frac{9}{10967757.6}\ \text{m} \approx 8206\ \text{\AA} \end{aligned} \]

The option gives this limit as 8205 Å. 7800 Å and 7349 Å are shorter than the limit, so they cannot be Paschen lines at all, and 9546 Å, longer than the limit, belongs to a transition from a finite upper level, not the shortest line.