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Q.

The ground state energy of hydrogen atom is $- 13.6 e V .$ When its electron is in the first excited state, its excitation energy is

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a

$10.2 e V$

b

$6.8 e V$

c

$13.6 e V$

d

$3.4 e V$

answer is A.

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Detailed Solution

First excited state means: $n = 2$

Energy for excitation: $E = \frac{13.6}{n^{2}} = \frac{13.6}{{(2)}^{2}} = 3.4 e V$

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