A sample of hydrogen atoms, all in the ground state, is lit with light of wavelength 97.32 nm…

(a) The orbit. One photon carries $E = \dfrac{hc}{\lambda} = \dfrac{1240.8\ \text{eV nm}}{97.32\ \text{nm}} = 12.75\ \text{eV}$. So $13.6\left(1 - \dfrac{1}{n^2}\right) = 12.75$, which gives $n^2 = 16.0$: the atoms reach $n = 4$.

(b) Lines. Every pair of the levels 1, 2, 3 and 4 gives one line: $\dfrac{4 \times 3}{2} = 6$ lines.

(c) Longest wavelength. The smallest gap is from $n = 4$ to $n = 3$: $13.6\left(\dfrac19 - \dfrac1{16}\right) = 0.6611\ \text{eV}$, so $\lambda = \dfrac{1240.8}{0.6611}\ \text{nm} = 1877\ \text{nm}$, in the infrared.

(d) Visible lines. The lines ending on $n = 1$ are in the ultraviolet and the one from 4 to 3 is in the infrared. The two ending on $n = 2$ are at $487\ \text{nm}$ (from 4) and $657\ \text{nm}$ (from 3), both visible. So $2$ lines are visible.

(e) Photoelectrons. The shortest line, from 4 to 1, carries $12.75\ \text{eV}$. So $K_{\max} = 12.75 - 4.30 = 8.45\ \text{eV}$. Its momentum is $p = \sqrt{2 m_e K} = 1.570 \times 10^{-24}\ \text{kg m s}^{-1}$, and