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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 |
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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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