Ultraviolet light of wavelength 300 nm and intensity 1 .0 W m - 2 falls on the surface on photoelectric metal. If one percent of incident photons produce photoelectrons, then the number of photoelectr

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Ultraviolet light of wavelength 300 nm and intensity 1 .0 W m - 2 falls on the surface on photoelectric metal. If one percent of incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1 .0 cm 2 of the surface is nearly

TestHub · Accepted Answer

Energy of each photon = h c λ = 6.6 × 10 - 34 × 3 × 10 8 300 × 10 - 9 = 6 .6 × 10 - 19 J Power of source is given by p = intensity × area = 1 .0 × 1 .0 × 10 - 4 W = 10 - 4 W No. of photons per second = p e = 10 - 4 6.6 × 10 - 19 Now the number of electrons emitted = 1 100 × 10 - 4 6.6 × 10 - 19 = 1.5 × 10 12 s - 1

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