According to Bohr's theory for hydrogen-like species, the energy of an electron in the $n$-th orbit is given by the expression: $E_n = E_0 \times \frac{Z^2}{n^2}$ where $E_0$ is the ground state energy of hydrogen ($-2.18 \times 10^{-18}$ J) and $Z$ is the atomic number.

1. For Helium ion ($He^+$): Atomic number $Z = 2$. Ground state $n = 1$.

• Given Energy $E = -x$ J. $E_{He^+} \propto \frac{2^2}{1^2} = 4$ So, $-x \propto 4$.

2. For Beryllium ion ($Be^{3+}$): Atomic number $Z = 4$. Excited state $n = 2$. $E_{Be^{3+}} \propto \frac{4^2}{2^2} = \frac{16}{4} = 4$

3. Comparison: Since the value proportional to the energy depends on $\frac{Z^2}{n^2}$, and for both cases this ratio is 4, the energies are equal. $E_{Be^{3+}} = E_{He^+} = -x \text{ J}$