1Thanks Govind.
Light rays of wavelength 6000 A and of photon intensity 39.6 W/m2 is incident on a metal surface. If only 1% of the photons incident on surface emit photoelectrons, then the number of electrons emmited per second per unit area from the surface will be
(h = 6.64 x 10-34 J s, c = 3 x 108 m/s )
A) 12 x 1018 B) 10 x 1018 C) 12 x 1017 D) 12 x 1016
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2 Answers
29the answer is C
Find the energy per photon by E= hc/λ
that comes out to be 3.3*10-19
let x photoelectrons are emitted in the 100% efficient process
so 3.3*10-19 *x = 39.6
so x = 12*1019
now the efficiency of the process is only 1%
so no. of photoelectrons emitted = 12*1017