Physics - Modern Physics Question with Solution | TestHub

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

For the hydrogen atom ($Z=1$) in the third orbit ($n=3$):

$E=k\left(\frac{1^2}{3^2}\right)=\frac{k}{9}\Rightarrow k=9E$

For the helium ion ($Z=2$) in the fifth orbit ($n=5$):

$E^{\prime }=k\left(\frac{2^2}{5^2}\right)=\frac{4k}{25}$

Substituting $k=9E$ into the above equation:

$E^{\prime }=\frac{4(9E)}{25}=\frac{36E}{25}$

Answer: $\frac{36\mathrm{E}}{25}$