When a beam of 10.6 eV photon of intensity 2 w/m2 falls on a platinum surface of area 104 m2 and work function 5.6 eV, 0.53% of incident photons eject photoelectrons. The number of photoelectrons emitted per second is 125 n×1017. Calculate n
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answer is 0005.00.
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Detailed Solution
Power =2×104 WattsNo of photons falling per second = MM=2×10410.6×1.6×10−19 No of electrons ejected =N=0.53100×M=0.53100×2×10410.6×1.6×10−19N=625×1017=n×125×1017n=5