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Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential is
  • a)
    Greater with visible light
  • b)
    Equal in both cases
  • c)
    Greater with ultraviolet light
  • d)
    Infinite in both cases
Correct answer is 'C'. Can you explain this answer?
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Verified Answer
Sodium surface is illuminated by ultraviolet and visible radiation suc...
λ for U.V is less than λ for visible light 
ν for U.V is greater than ν for visible light 
∴ potential is greater for U.V light.
as K.Eα1​/λ
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Sodium surface is illuminated by ultraviolet and visible radiation suc...
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Sodium surface is illuminated by ultraviolet and visible radiation suc...
I think answer is A because stopping potential ∝ wavelength λ(visible) greater than λ(UV)
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According to wave theory of light, the light of any frequency can emit electrons from metallic surface provided the intensity of light be sufficient to provide necessary energy for emission of electrons, but according to experimental observations, the light of frequency less than threshold frequency can not emit electrons; whatever be the intensity of incident light. Einstein also proposed that electromagnetic radiation is quantised.If photoelectrons are ejected from a surface when light of wavelength λ1 = 550 nm is incident on it. The stopping potential for such electrons is Vs =0.19 . Suppose the radiation of wavelength λ2 = 190 nm is incident on the surface.Q. Photoelectric effect supports quantum nature of light because(i) there is a minimum frequency of light below which no photoelectrons are emitted.(ii) the maximum K.E. of photoelectric depends only on the frequency of light and not on its intensity.(iii) even when the metal surface is faintly illuminated, the photo electrons leave the surface immediately.(iv) electric charge of the photoelectrons is quantized.

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Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential isa)Greater with visible lightb)Equal in both casesc)Greater with ultraviolet lightd)Infinite in both casesCorrect answer is 'C'. Can you explain this answer?
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Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential isa)Greater with visible lightb)Equal in both casesc)Greater with ultraviolet lightd)Infinite in both casesCorrect answer is 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential isa)Greater with visible lightb)Equal in both casesc)Greater with ultraviolet lightd)Infinite in both casesCorrect answer is 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential isa)Greater with visible lightb)Equal in both casesc)Greater with ultraviolet lightd)Infinite in both casesCorrect answer is 'C'. Can you explain this answer?.
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