A sample of Li^2+ ions all in the n=3 state de-excite to the ground state through all possible transitions. The emitted radiation falls on a metal…

For $\mathrm{Li^{2+}}$ ($Z=3$), $E_n=-13.6\times9/n^2=-122.4/n^2\ \mathrm{eV}$. The photon energies are $3\to1$: $108.8\ \mathrm{eV}$, $2\to1$: $91.8\ \mathrm{eV}$ and $3\to2$: $17.0\ \mathrm{eV}$. Since $17.0\ \mathrm{eV}<20\ \mathrm{eV}$, the $3\to2$ line cannot eject electrons, so only two lines can and (A) is wrong. The maximum kinetic energy comes from the most energetic photon, $108.8-20=88.8\ \mathrm{eV}$, so (B) is correct. For the $2\to1$ line the stopping potential is $(91.8-20)/e=71.8\ \mathrm{V}$, not $91.8\ \mathrm{V}$ (that forgets the work function), so (C) is wrong. The longest wavelength belongs to the smallest energy: $\lambda=1240/17.0=72.9\ \mathrm{nm}$; $17.0$ is an energy in eV, not a wavelength, so (D) is wrong. Correct option: B only.