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Q.

If a metal is irradiated with light of frequency 3×1019 sec−1 , electron is emitted with kinetic energy of 6.625×10−15J . The threshold frequency of the metal is

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a

2×1019 sec−1

b

1.25×1019 sec−1

c

6.625×1035 sec−1

d

6.625×1019 sec−1

answer is A.

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Detailed Solution

hϑ=hϑ0+K.E K.E=h(ϑ−ϑ0) ϑ−ϑ0=6.625×10−156.6×10−34=1019 ϑ0=3×1019−1019       =1019(3−1)       =1019×2 ϑ0=2×1019 sec−1
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