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Question

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

A
v1kv2k1
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B
v2kv1k1
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C
v1kv21k
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D
v1v2kk1
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Solution

The correct option is A v1kv2k1
Maximum Kinetic Energy, K.E.max=hvhvT

K.E.max(1)=hv1hvT.....1

K.E.max(2)=hv2hvT.....2
Dividing 1 by 2:

K.E.max(1)K.E.max(2)=hv1hvThv2hvT
1k=hv1hvThv2hvT
hv2hvT=hv1khvTk
hvT(k1)=hv1khv2
vT=v1kv2k1

So, the answer is (A).

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