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. 3.4 eV
D. 1.9 eV

Answer

D

Explanation

`E_(3)- E_(2)= -13.6 [(1)/((3)^(2))- (1)/((2)^(2))] eV= (13.6 xx 5)/(36)= 1.89 eV ~~ 1.9eV`