What is the electron separation energy (in $e V$ ) for $B e^{3 +}$ in the first excited state.

13.6 e V

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27.2 e V

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40.8 e V

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54.4 e V

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Solution

The correct option is D $54.4 e V$
Electron separation energy is similar to ionization energy.
For $B e^{3 +}, E_{\infty} - E_{2} = + 13.6 \frac{z^{2}}{n^{2}}$
$= 13.6 \times \frac{4^{2}}{2^{2}} = 54.4 e V$

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