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. `1.5eV`
B. `0.85eV`
C. `3.3eV`
D. `1.9eV`

Answer

D

Explanation

Energy of ejected photon
`E=Rhc(1/(n_(1)^(2))-1/(n_(2)^(2)))`
`=13.6(1/(2^(2))-1/3^(2))eV`
`=13.6xx5/36=1.9eV`