Light of frequency 8×1015Hzis incident on a substance of photoelectric work function 6.125 eV. The maximum kinetic energy of the emitted photoelectrons is
A
17 eV
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B
22 eV
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C
27 eV
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D
37 eV
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Solution
The correct option is C 27 eV Energy of incident photon E=hv=6.6×10−34×8×1015=5.28×10−18J=33eV. From E=W0+Kmax⇒Kmax=E−W0=33−6.125=26.87eV≈27eV.