A) 27.2eV
B) 13.6eV
C) 6.8eV
D) 3.4eV
Correct Answer: D
Solution :
Energy of electron in n th orbit of H-atom is given by \[{{E}_{n}}=\frac{13.6}{{{n}^{2}}}=eV/atom\] For \[n=2\] \[{{E}_{2}}=\frac{-13.6}{{{(2)}^{2}}}=-3.4\,eV\]
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