An electron in a He^+ ion makes a transition from n=4 to n=2, and the emitted photon falls on a metal surface of work function 4.0 eV. Use En=-13.6…

For $\mathrm{He^+}$, $Z=2$. The photon energy is $\Delta E=13.6\times4\left(\tfrac14-\tfrac1{16}\right)=13.6\times4\times0.1875=10.2\ \mathrm{eV}$. By the photoelectric equation the maximum kinetic energy is $K=10.2-4.0=6.2\ \mathrm{eV}=6.2\times1.602\times10^{-19}=9.93\times10^{-19}\ \mathrm{J}$. The de Broglie wavelength is $\lambda=h/\sqrt{2m_eK}$. Computing $\sqrt{2\times9.109\times10^{-31}\times9.93\times10^{-19}}=1.345\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}$, so $\lambda=6.626\times10^{-34}/1.345\times10^{-24}=4.93\times10^{-10}\ \mathrm{m}=4.93\ \mathrm{\AA}$. Note that the $n=4\to2$ line of $\mathrm{He^+}$ coincides with the energy of hydrogen $n=2\to1$ ($10.2$ eV) because $Z^2=4$ scales every level fourfold. A common slip is to use $Z=1$ and obtain only $2.55$ eV, which is below the work function so that no photoelectron would be emitted at all; the check on the sign of $K$ is therefore an essential step before applying the de Broglie relation, and the final unit conversion from metres to angstrom is a factor $10^{10}$.