A Li^2+ ion is excited to n=3 and then makes a transition to n=2. The emitted photon falls on a metal of work function 5.0 eV. Take the hydrogen…

For a hydrogen-like ion $E_n=-13.6Z^2/n^2\ \mathrm{eV}$ with $Z=3$ for $\mathrm{Li^{2+}}$. The photon energy is $E=13.6\times9\left(\tfrac14-\tfrac19\right)=13.6\times9\times\tfrac5{36}=17.0\ \mathrm{eV}$. The maximum kinetic energy of photoelectrons is $K_{\max}=17.0-5.0=12.0\ \mathrm{eV}$. The shortest de Broglie wavelength belongs to the fastest electrons: $\lambda=12.27/\sqrt{12}=3.54\ \text{\AA}$. The value $2.98\ \text{\AA}$ comes from wrongly using the full photon energy $17\ \mathrm{eV}$, and $5.49\ \text{\AA}$ from using the work function $5\ \mathrm{eV}$ as the kinetic energy. Note that the work function only reduces the energy of the most energetic electrons; electrons ejected from deeper levels have lower kinetic energy and therefore longer wavelengths, so the question asks for the minimum wavelength, corresponding to the maximum kinetic energy. The ion's own transition does not depend on the metal, and the stopping potential here would be $12\ \mathrm{V}$.