A photon emitted when a He^+ ion de-excites from n=4 to n=2 falls on a metal surface of work function 4.0 eV. Take En=-13.6 Z^2/n^2 eV…

For $\mathrm{He^+}$, $Z=2$, so the photon energy is $E=13.6\times4\left(\frac14-\frac1{16}\right)=10.2$ eV. By the photoelectric equation the maximum kinetic energy is $K=10.2-4.0=6.2$ eV. The momentum is $p=\sqrt{2mK}=\sqrt{2\times9.11\times10^{-31}\times6.2\times1.6\times10^{-19}}\approx1.345\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}$ and $\lambda=h/p\approx0.49$ nm (also $\lambda\approx1.227/\sqrt{6.2}$ nm). Using the full photon energy of 10.2 eV as the kinetic energy gives $0.38$ nm, and using $3.1$ eV gives $0.70$ nm; forgetting the factor $Z^2$ would give a photon energy below the work function and no emission at all. The wavelength follows from $\lambda=h/\sqrt{2mK}$, and only the maximum kinetic energy corresponds to the fastest electron, so deeper-lying electrons are slower and have longer wavelengths.