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Wave Optics

Question
CBSEENPH12039781

When a metallic surface is illuminated with radiation of wavelength straight lambda, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2 straight lambda, the stopping potential is V/4. The threshold wavelength for the metallic surface is,

  • 5 straight lambda
  • 5 over 2 lambda
  • 3straight lambda

  • 4straight lambda

Solution

C.

3straight lambda

When a metallic surface is illuminated with radiation of wavelength straight lambda, the stopping potential is V.
Photoelectric equation can be written as,
eV equals space space hc over straight lambda minus hc over straight lambda subscript straight o        ... (i)
Now, when the same surface is illuminated with radiation of wavelength 2straight lambda, the stopping potential is V/4. So, photoelectric equation can be written as,
space space space space space space eV over 4 equals fraction numerator h c over denominator 2 lambda end fraction minus fraction numerator h c over denominator lambda subscript o end fraction
rightwards double arrow space e V space equals space fraction numerator 4 h c over denominator 2 lambda end fraction minus fraction numerator italic 4 h c over denominator lambda subscript o end fraction space space space space... space left parenthesis i i right parenthesis
From equations (i) and (ii), we get
space space space hc over straight lambda minus hc over straight lambda subscript straight o equals fraction numerator 4 h c over denominator 2 lambda end fraction minus fraction numerator 4 h c over denominator straight lambda subscript straight o end fraction
rightwards double arrow space 1 over straight lambda minus 1 over straight lambda subscript straight o equals italic 2 over lambda minus 4 over straight lambda subscript straight o
rightwards double arrow space space straight lambda subscript straight o space equals space 3 straight lambda

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