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When a metallic surface is illuminated with monochromatic light of wavelength $λ$ stopping potential for photoelectric current is $3 V_{0}$ . When the same metallic surface is illuminated with a light of wavelength $2 λ$ the stopping potential is $V_{0}$ . The threshold wavelength for the surface is

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$6 λ$

$4 λ$

$\frac{4 λ}{3}$

$8 λ$

answer is B.

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Detailed Solution

$\frac{h c}{λ} = \frac{h c}{λ_{0}} + 3 e V_{0} - - - (1) \frac{h c}{2 λ} = \frac{h c}{λ_{0}} + e V_{0} - - - - (2) s o l v i n g w e g e t λ_{0} = 4 λ$

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