Q. The total energy of an electron in the $n^{t h}$ stationary orbit of the hydrogen atom can be obtained by

9059 208 NEETNEET 2020Atoms Report Error

A

$E_{n} = \frac{13.6}{n ^{2}} e V$

2%

B

$E_{n} = - \frac{13.6}{n ^{2}} e V$

90%

C

$E_{n} = - \frac{1.36}{n ^{2}} e V$

4%

D

$E_{n} = - 13.6 \times n^{2} e V$

3%

Solution:

For hydrogen
$E = - \frac{13.6}{n ^{2}} e V$