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Question

When a certain metal surface is illuminated with light of frequency ν, the stopping potential for photoelectric current is V0. When the same surface is illuminated by light of frequency ν2 the stopping potential is V04. The threshold frequency for photoelectric emission is :

A
ν/6
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B
ν/3
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C
2ν/3
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D
4ν/3
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Solution

The correct option is B ν/3
Incident frequency =ν

Stopping potential =V0

Using Photoelectric effect :

E=ϕ+(KE)max

hν=ϕ+eV0------(1)

and similarly ,

hν2=ϕ+eV04-------(2)

from (2)

hν=2ϕ+eV02-----(3)

from (1) and (3)

hν=2ϕ+12[hνϕ]

hν=2ϕ+hν2ϕ2

=(2ϕϕ2)+hν2

hν2=32ϕ

ϕ=hν3

frequency =ν3

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