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Question

When a monochromatic light of frequency v is incident on a metal, stopping potential is V0. Frequency of the incident light for which stopping potential becomes double is :

A
v
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B
v+eV0h
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C
2veV0h
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D
veV0h
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Solution

The correct option is B v+eV0h
hv=hv0+K.E

At stopping potential, K.E=eV

hv=hv0+eV

V=he(vv0)

Given for, V=V0;v=v.

V0=he(vv0)

For V=2V0;v1=?

2V0=(he(v1v0))

v1=2eV0h+v0
v1=2eV0h+veV0h

v1=eV0h+v

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