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Questions  

Photo electric emission is observed from a metallic surface for frequencies v1 andv2   of the incident light rays(ν1>ν2).  If the maximum values of kinetic energy of the photo elections emitted in the two cases are in the ratio of 1:K, then the threshold frequency of the metallic surface is

a
ν2−ν1K−1
b
Kν1−ν2K−1
c
Kν2−ν1K−1
d
ν2−ν1K

detailed solution

Correct option is B

K.E=hv−hv0=h(v−v0) kE1kE2=v1−v0v2−v0=1k kv1−kv0=v2−v0⇒kv1−v2=kv0−v0 kv1−v2=v0(k−1)r0 v0=kv1−v2(k−1)

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