Physics - Modern Physics Question with Solution | TestHub

The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen-like species is given by $E_n=-13.6\frac{Z^2}{n^2}\text{ eV}$.

For the hydrogen atom ($Z=1$) in the $2^{\text{nd}}$ orbit ($n=2$):
$E_{\text{H}}=-13.6\times \frac{1^2}{2^2}=-13.6\times \frac{1}{4}\text{ eV}$

For the helium ion (${\text{He}}^{+}$) ($Z=2$) in the $2^{\text{nd}}$ orbit ($n=2$):
$E_{{\text{He}}^{+}}=-13.6\times \frac{2^2}{2^2}=-13.6\times 1\text{ eV}$

The ratio of their total energies is:
$\frac{E_{\text{H}}}{E_{{\text{He}}^{+}}}=\frac{-13.6\times \frac{1}{4}}{-13.6}=\frac{1}{4}$

Answer: $\frac{1}{4}$