Light of frequency 8×1015Hz is incident on a substance of photoelectric work function 6.125 eV. The maximum kinetic energy of the emitted photoelectrons is
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a
17 eV
b
22 eV
c
27 eV
d
37 eV
answer is C.
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Detailed Solution
E=hv=6.6×10−34×8×1015=5.28×10−18J=33eVBy using E=Wo+Kmax⇒Kmax=E−Wo=33−6.125≈27eV
Light of frequency 8×1015Hz is incident on a substance of photoelectric work function 6.125 eV. The maximum kinetic energy of the emitted photoelectrons is