An electron in the hydrogen atom jumps from excited state `n` to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function `2.75eV`. If the stopping potential of the photoelectron is `10V`, the value of `n` is

Options

A. 5
B. 2
C. 3
D. 4

Answer

D

Explanation

`k_("max") = hv - phi_0 =E-phi_0`
`implies E = 10 eV +2.75 eV = 12.75 eV`
This is the energy difference between n = 4 and n = 1.
As `DeltaE = (-13.6)/n_1^2-((-13.6)/(n_2^2))`