1.

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 theA. ultraviolet regionB. visible regionC. infrared regionD. X - ray region

Answer» 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.


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