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Question

Photoelectric emission is observed from a metallic surface for frequencies v1 and v2 and of the incident light rays (v1>v2). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is


A

v1v2k1

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B

kv1v2k1

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C

kv2v1k1

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D

v2v1k

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Solution

The correct option is B

kv1v2k1


By using hvhv0=Kmass
h(v1v0)=K1(i)
And h(v2v0)=K2(ii)
v1v0v2v0=K1K2=1K,
Hence v0=Kv1v2K1


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