The energy of an electron in first Bohr orbit of H -atom is -13.6 eV. The magnitude of energy value of electron in the first excited state of $\mathrm{Be}^{3+}$ is _________ eV (nearest integer value)
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Solution
$$
\mathrm{E}_{\mathrm{T}}=-13.6 \frac{\mathrm{z}^2}{\mathrm{n}^2} \mathrm{ev}
$$
For energy of H -atom, energy of $1^{\text {st }}$ Bohr orbit
$$
\mathrm{E}_1=-13.6 \mathrm{eV}[\mathrm{z}=1, \mathrm{n}=1]
$$
For $\mathrm{Be}^{+3}$ ion, energy of $\mathrm{I}^{\text {st }}$ E.S. $[\mathrm{z}=4, \mathrm{n}=2]$
$$
\begin{aligned}
& \frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{Be}^{+3}}}=\frac{\mathrm{z}_1^2}{\mathrm{n}_1^2} \times \frac{\mathrm{n}_2^2}{\mathrm{z}_2^2} \\
& \frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{Be}^{+3}}}=\frac{1}{1} \times \frac{4}{16} \\
& \mathrm{E}_{\mathrm{Be}^{+3}}=-13.6 \times 4=-54.4 \mathrm{eV} \\
& \left|\mathrm{E}_{\mathrm{Bc}^{+3}}\right|=54.4 \mathrm{eV}
\end{aligned}
$$
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