Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be `3.57V`. The threshold frequency of the material is

Options

A. `4 xx 10^(15)Hz`
B. `5 xx 10^(15)Hz`
C. `1.6 xx 10^(15)Hz`
D. `2.5 xx 10^(15)Hz`

Answer

C

Explanation

For hydrogen atom , `E_n = -(13.6)/n^2 eV`
For ground state, n =1
`:. E_1 =-(13.6)/1^2 =-13.6eV`
For first excited state, n = 2
`:. E_2 =-(13.6)/2^2=-3.4 eV`
The energy of the emitted photon when an electron jumps from first excited state to ground state is
`hupsilon=E_2 -E_1 =-3.4 eV -(-13.6 eV) = 10.2 eV`
Maximum kinetic energy ,
`K_(max) = eV_s = e xx 3.57 V = 3.57 eV`
According to Einstein.s photoelectric equation
`K_(max) = hupsilon - phi_0`
where `phi_0` is the work function and `hupsilon` is the incident energy .
`phi_0=hupsilon-K_(max)=10.2 eV - 3.57 eV = 6.63 eV`
Threshold frequency , `upsilon_0/h=(6.63 xx1.6 xx10^(-19)J)/(6.63 xx10^(-34) Js)`
`= 1.6 xx10^(15) Hz`