What is the ratio of the potential energy of an electron in the...

For a hydrogen-like ion, the potential energy of an electron in the $n^{\text{th}}$ orbit is given by $U \propto \dfrac{Z^2}{n^2}$, where $Z$ is the atomic number.

For the first excited state of $\text{He}^{+}$ ion:
$Z_1 = 2$, $n_1 = 2$
$U_1 \propto \dfrac{2^2}{2^2} = 1$

For the second excited state of $\text{Be}^{3+}$ ion:
$Z_2 = 4$, $n_2 = 3$
$U_2 \propto \dfrac{4^2}{3^2} = \dfrac{16}{9}$

The ratio of their potential energies is:
$\dfrac{U_1}{U_2} = \dfrac{1}{16/9} = \dfrac{9}{16}$

Thus, the required ratio is $9:16$.

Answer: $9:16$