The ratio of the energy of the electron in ground state of hydrogen to the electron in first excited state of ${\mathrm{Be}}^{3+}\mathrm{is}$

$\begin{array}{l}{\mathrm{E}}_{\mathrm{n}}=-\frac{13.6{\mathrm{Z}}^2}{{\mathrm{n}}^2}\mathrm{eV},\mathrm{where}\\ \mathrm{Z}=\mathrm{atomic}\mathrm{number}=4\\ \mathrm{N}=\mathrm{number}\mathrm{of}\mathrm{orbits}\\ {\mathrm{Be}}^{3+}\left(\mathrm{n}=2\right)\left(\mathrm{first}\mathrm{excited}\mathrm{state}\right)\\ {\mathrm{E}}_1=-13.6\mathrm{eV},{\mathrm{E}}_2=-13.6\times 16/4\end{array}$

${\mathrm{E}}_1:{\mathrm{E}}_2=4:16 =1:4$