When a metallic surface of threshold wavelength 4 λ is illuminated with light of wavelength λ , the stopping potential is V . When the same surface is illuminated by the light of wavelength 2 λ , stop

Question

When a metallic surface of threshold wavelength 4 λ is illuminated with light of wavelength λ , the stopping potential is V . When the same surface is illuminated by the light of wavelength 2 λ , stopping potential is V 3 . Threshold wavelength for the metallic surface is

TestHub · Accepted Answer

When a metallic surface is illuminated with light of wavelength $λ$ ,

$e V = h c (\frac{1}{λ} - \frac{1}{4 λ})$

$e V = \frac{3 h c}{4 λ}$ ..............(1)

When a metallic surface is illuminated with light of wavelength $2 λ$ ,

$e V^{'} = h c (\frac{1}{2 λ} - \frac{1}{4 λ})$

$e V^{'} = \frac{h c}{4 λ}$ .............(2)

$\frac{V^{'}}{V} = \frac{1}{3}$

$V^{'} = \frac{V}{3}$

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