If the electron in hydrogen atom jumps from second Bohr orbit to ground state and difference between energies of the two states is radiated in the form of photons. If the work function of the material is `4.2 eV`, then stopping potential is
[Energy of electron in nth orbit `= - (13.6)/(n^(2)) eV`]

Options

A. 2V
B. 4V
C. 6V
D. 8V

Answer

C

Explanation

Energy radiated when the electron jumps from n = 2 to n=1 is given by
`E = E_2 -E_1`
`- (13.6)/(2^2) - (-(13.6)/(1^2))`
`=(13.6)/(1^2) - (13.6)/(4) = 13.6 [ 1-1/4]`
`=13.6 xx 3/4 = 10.2 eV`
This energy is the energy of the photon
`therefore E_2 - E_1 = hv therefore hv = 10.2 eV " " ...(1)`
From Einstein's photoelectric equation,
`hv = hv_0 + W_0`
and If `V_0` is the stopping potential , then `hv_0 = ev_0 " " ..(2)`
`therefore` Form (1) and (2) , hv`=eV_0 + W_0`
`therefore 10.2eV = eV_0 + 4.2eV`
`therefore 6eV = eV_0 therefore V_0 = 6V`