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When a certain metal surface is illuminated with a light of wavelength λ, the stopping potential is V, When the same surface is illuminated by light of wavelength 2λ, the stopping potential is (V/3). The threshold wavelength for the surface is
  • a)
    8λ/3
  • b)
    4λ/3
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
When a certain metal surface is illuminated with a light of wavelengt...
Given for a metal, the wavelength of light used = λ,
Stopping potential = V
If λ0 be the threshold wavelength, then the maximum kinetic energy of emitted electrons.
Again, the wavelength of used light
λ' = 2λ
Stopping potential V' = V/3
Then
From Eqs (i) and (ii), we have
⇒ λ0 = 4λ
So, the threshold wavelength is 4 times of wavelength of light.
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