Energy `E` of a hydrogen atom with principle quantum number `n` is given by `E = (-13.6)/(n^(2)) eV`. The energy of a photon ejected when the electron jumps from `n = 3` state to `n = 2` state of hydrogen is approximately

Options

A. `1.9 eV`
B. `2.3 eV`
C. `3.4 eV`
D. `4.5 eV`

Answer

A

Explanation

`13.6 ((1)/(2^(2)) - (1)/(3^(2))) eV = 1.9 eV`