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Question:

A beam of light of wavelength 400nm and power 1.55 mW is directed at the cathode of a photoelectric cell. If only 10% of the incident photons effectively produce photoelectrons, then find the current due to these electrons. (given, hc=1240 eV-nm, e=1.6×10^-19C)

5µA

114µA

40µA

50µA

Solution:

Correct option is C. 50µA
Energy of incident photons = hc/λ = 1240/400 eV = 3.1 eV = 4.96 × 10^-19 J
Hence the frequency of incident photons is, ν = Power/Energy = (1.55 × 10^-3)/(4.96 × 10^-19) = 3.125 × 10^15 Hz
Hence the photocurrent = eν = 50 × 10^-6 A = 50µA