The energy of an electron present in Bohr’s second orbit of hydrogen atom is

The correct option is A
$E_{n} = \frac{- 1312}{n^{2}} k J m o l e^{- 1} = \frac{- 1312}{2^{2}} = - 328 k J m o l^{- 1}$

Chemistry

Suggest Corrections

The correct option is A
$E_{n} = \frac{- 1312}{n^{2}} k J m o l e^{- 1} = \frac{- 1312}{2^{2}} = - 328 k J m o l^{- 1}$

Chemistry