A) -3.4 eV
B) -6.8 eV
C) -27.2 eV
D) -52.4 eV
Correct Answer: A
Solution :
Given: Energy of the ground electronic state of hydrogen atom E = -13.6 eV. We know that energy of the first excited state for second orbit (where n = 2) \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}=-\frac{13.6}{{{(2)}^{2}}}=-\,3.4eV.\]
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