Question

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

Answer 1

Correct Answer - A
According to laws of photoelectric effect
`KE_(max) = E - phi`
where `phi` is work function and `KE_(max)` , is maximum , kinetic energy of photoelectron.
`:. Hv = eV_(0) + phi`
or `hv = 5 eV + 6.2 eV = 11.2 eV`
`:. Lambda = ((12400)/(11.2)) Å ~~ 1000 Å`
Hence , the radiation lies in ultraviolet region.