A hydrogen-like ion X^Z-1)+ contains a single electron. Photons emitted when its electron falls from n=2 to n=1 strike isolated hydrogen atoms in…

The $n=2\to1$ photon energy is $E=13.6Z^2\left(1-\tfrac14\right)=10.2Z^2\ \mathrm{eV}$. It ejects an electron from hydrogen, so the kinetic energy is $K=10.2Z^2-13.6\ \mathrm{eV}$. The de Broglie relation $\lambda=h/\sqrt{2m_eK}$ gives $K=\dfrac{h^2}{2m_e\lambda^2}$. With $\lambda=0.139\ \mathrm{nm}$, $K=\dfrac{(6.626\times10^{-34})^2}{2\times9.109\times10^{-31}\times(1.39\times10^{-10})^2}=1.247\times10^{-17}\ \mathrm{J}=77.8\ \mathrm{eV}$. Then $10.2Z^2=77.8+13.6=91.4$, so $Z^2=8.96$ and $Z=2.99\approx3$. The check with $Z=3$ gives $K=91.8-13.6=78.2\ \mathrm{eV}$ and $\lambda=0.1387\ \mathrm{nm}$, so the ion is $\mathrm{Li^{2+}}$. Since only one value of $Z$ gives a photoelectron wavelength near $0.139\ \mathrm{nm}$, and $Z=2$ gives $0.235\ \mathrm{nm}$ while $Z=4$ gives $0.100\ \mathrm{nm}$, the integer is unambiguous. Students should also remember that the ejected electron's energy is the photon energy minus the ionisation energy of hydrogen, and not the photon energy itself, otherwise they would obtain $Z=\sqrt{77.8/10.2}=2.76$, which is not an integer.