When a metallic surface is illuminate with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V4. The threshold wavelength for the metallic surface is:
4λ
5λ
52λ
3λ
A
5λ
B
4λ
C
3λ
D
52λ
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Solution
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We know that:
eV=hcλ−ϕ ........ (1)
eV4=hc2λ−ϕ ........ (2)
From equations (1) and (2), we get:
⇒4=1λ−1λ012λ−1λ0
⇒λ0=3λ
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