Photons emitted when a hydrogen-like Li^2+ ion makes the transition n=3→ n=2 fall on a metal surface of work function 4.0 eV. Take the hydrogen…

For a hydrogen-like ion the photon energy is $E=13.6\,Z^2\left(\dfrac1{n_f^2}-\dfrac1{n_i^2}\right)\ \mathrm{eV}$. With $Z=3$, $n_f=2$, $n_i=3$ this is $13.6\times 9\times\dfrac{5}{36}=17.0\ \mathrm{eV}$. The maximum kinetic energy of photoelectrons is $K_{max}=h\nu-\phi=17.0-4.0=13.0\ \mathrm{eV}$. The corresponding de Broglie wavelength is $\lambda=1.226/\sqrt{13.0}=0.340\ \mathrm{nm}$. The value $0.30\ \mathrm{nm}$ results if the work function is forgotten, $0.65\ \mathrm{nm}$ results from using $Z=2$ (giving $K=3.56\ \mathrm{eV}$), and $0.48\ \mathrm{nm}$ from using half the correct kinetic energy. Hence the correct choice is $0.34\ \mathrm{nm}$, the only value following from the full chain: ion transition energy, Einstein's photoelectric equation, then the de Broglie relation.