Power of a source of 2mW and wavelength 400nm light is directed on a photoelectric cell.If the current in the cell is 0.40 microampere,then the percentage of the incident photons that produce photoelectrons is Please solve this question and explain it. Share with your friends Share 0 Anshu Agrawal answered this Solution:P =2mW =2×10-3WEnergy of each photon =hcλ=6.62×10-34×3×1084×10-7=4.96×10-19JNumber of photons incident per second =PEnergy of each photon=2×10-34.96×10-19 ≃4×1015 Let n percentage of the incident photons that produce photoelectrons The number of photo electrons produced=n%×4×1015 Current i =n%×4×1015×1.6×10-190.40×10-6 =n%×4×1015×1.6×10-19n% =4×10-74×1015×1.6×10-19=0.625×10-3or n =0.06% 0 View Full Answer
