Light rays of wavelength 6000 Å and intensity 39.6 W m - 2 is incident on a metal surface. If only one percent of photons incident on surface emit photoelectrons, then the number of electrons emitted

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Light rays of wavelength 6000 Å and intensity 39.6 W m - 2 is incident on a metal surface. If only one percent of photons incident on surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be

TestHub · Accepted Answer

λ = 6000 × 1 0 - 10 m Ι = 39.6 W m -2 1 % of photons emit photoelectrons. I = n h c A t λ 39.6 = n × 6.6 × 1 0 - 34 × 3 × 1 0 8 1 × 1 × 6000 × 1 0 - 10 39.6 = n × 1.1 × 1 0 - 24 × 3 × 1 0 5 n p = 12 × 1 0 19 No. of photoelectrons = 1% of photons n e = 1 100 × 12 × 1 0 19 n e = 12 × 1 0 17

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