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Question

When a certain metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential for photoelectric current is 3V0 and when the same surface is illuminated with light of wavelength 2λ, the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is

A
6λ
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B
4λ/3
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C
4λ
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D
8λ
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Solution

The correct option is B 4λ
Let threshold wavelength be λ0

Therefore work function for the surface will be W=hcλ0

Stopping potential V=EWe where E is the energy of the incident beam (hcλ) and e is the charge on an electron.
Given

3eV0=hcλhcλ0

eV0=hc2λhcλ0

Multiplying the second eqaution by 3 and subtracting it from the first we get

2hcλ0hc2λ=0

Solving we get λ0=4λ

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