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

Answer

c

Explanation

Here, `phi_(0)=6.2eV, V_(0)=5V`
As, `eV_(0)=(hc)/lambda-phi_(0)` or `(hc)/lambda=eV_(0)+phi_(0)`
`=exx5V+6.2eV=11.2eV`
or `lambda=(hc)/(11.2eV)=(12400 eVÅ)/(11.2 eV)=1107Å`
This wavelength is in the ultraviolet region.