Q.
Find the number of electrons emitted per second by a 24W source of monochromatic light of wavelength 6600A˚, assuming 3% efficiency for photoelectric effect (take h=6.6×10−34Js )
2322
217
Dual Nature of Radiation and Matter
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Solution:
Energy per photon =hf or λhc
Number of photon's per second =hf24
Number of electrons emitted =1003×hc24×λ