Light emitted during the de excitation of electron from n = 3 to n = 2, when incident on a metal, photoelectrons are just emitted from that metal. In which of the following deexcitations photoelectric effect is not possible ?

Options

A. From n =2 to n=1
B. From n=3 to n=1
C. From n=5 to n=2
D. From n=4 to n=3

Answer

D

Explanation

According to Bohr's theory energy of the electron in `n^(th)` state is
`E_(n) = (13.6)/(n^(2))eV`
`:. E_(1) = (-13.6)/(1^(2)) = -13.6eV E_(2) = (-13.6)/(2^(2)) = -3.4eV`
`E_(3) = (-13.6)/(3^(2)) = -1.5eV E_(4) = (-13.6)/(4^(2)) = -0.85eV`
`E_(5) = (-13.6)/(5^(2)) = -0.54eV`
For photoelectric effect energy should be more than
`E_(3) - E_(2) = -1.5eV -(-3.4eV) = 1.9eV`
It is not possible in option(d)
`:. E_(4) -E_(3) = -0.85eV-(-1.5eV) = 0.65eV`
It is possible in all other options