The energy of an electron in the ground state ($n=1$) for the $\mathrm{He}{}^{+}$ ion is $-x$ J. Then the energy of an electron in the $n=2$ state for the $\mathrm{Be}{}^{3+}$ ion, in J, is:

Solution & Explanation

\textbf{Answer:} (D) $-x$ \textbf{Solution:} $E_n\propto -\frac{Z^2}{n^2}$. For $\mathrm{He}{}^{+}$ ($Z=2$, $n=1$): $\frac{Z^2}{n^2}=\frac{4}{1}=4\Rightarrow E=-x$ J. For $\mathrm{Be}{}^{3+}$ ($Z=4$, $n=2$): $\frac{Z^2}{n^2}=\frac{16}{4}=4$, the same value, so $E=-x$ J.