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Q.

Photoelectric emission is observed from a metallic surface for frequencies ν1 and ν2 of the incident light rays (ν1>ν2). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of I : k, then the threshold frequency of the metallic surface is

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a

v1−v2k−1

b

kv1−v2k−1

c

kv2−v1k−1

d

v2−v1k

answer is B.

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Detailed Solution

By  using  hv−hvo=Kmax⇒hv1−vo=K1   .    .     .    .    iand  hv1−vo=K2    .    .     .     ii⇒v1−vov2−vo=K1K2=1K,   Hence ,  vo=Kv1−v2K−1
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