The work function of a surface of a photosensitive material is `6.2 eV`. The wavelength of the incident radiation for which the stopping potential is `5 V` lies in the

Options

A. visible region
B. infrared region
C. X- ray region
D. ultraviolet region

Answer

D

Explanation

`hv_0` = work function `phi_0=6.2eV`
Stopping potential , `V_0=5`
`K.E._(max)=eV_0=5eV`
From Einstein.s photoelectric equation
`hv=hv_0+K.E._(max)`
`hv=6.2eV+5.0eV`
`(hc)/lamda=11.2 eV`
`lamda=(4.136xx10^(-15)xx3xx10^8(eV)m)/(11.2eV)`
`lamda=1.11xx10^(-7)m`
This wavelength lies in region of ultraviolet rays.