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Q.

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
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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