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A stopping potential of 0.82 volt is required to stop the photo electrons emitted from a metallic surface by light of wavelength 4000˚A. The stopping potential for wavelength 3000˚A will be

A
1.1 volt
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B
1.85 volt
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C
2.03 volt
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D
2.5 volt
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Solution

The correct option is B 1.85 volt
Here, a stopping potential of 0.82 volt is required to stop the photo electrons emitted from a metallic surface by light of wavelength 4000A, hence
eV=hνϕ
where, V is the stopping potential, e is the electronic charge,
ν is the frequency of incident light and ϕ is photoelectric work function of metal, h is Plank's constant.

eV=hcλϕ

1.6×0.82×1019=6.6×1034×3×1084000×1010ϕ

1.312×1019=4.95×1019ϕ
ϕ=(4.951.312)×1019
ϕ=3.638eV
The stopping potential for wavelength 3000A

V=hcλϕe

V=6.6×1034×3×1083000×10103.638×10191.6×1019

V=2.9621.6

V=1.85V
So, the answer is option (B).

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