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A metallic surface is illuminated with monochromatic light of wavelength $λ$ and stopping potential for photoelectric current is $5 v_{o}$ when the same metallic surface is illuminated with a light of wavelength $2 λ$ , the stopping potential is $V_{o}$ . What is the threshold wavelength for the surface ?

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$\frac{8 λ}{3}$

$8 λ$

$5 λ$

$3 λ$

answer is A.

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Detailed Solution

$\frac{hc}{λ} = ϕ + e 5 V_{o} . .. (1)$

$\frac{hc}{2 λ} = ϕ + {eV}_{o} . .. (2)$

$∴ 2 (ϕ + {eV}_{o}) = ϕ + 5 {eV}_{o} \Rightarrow ϕ = 3 {eV}_{o} \Rightarrow \frac{hc}{λ_{o}} = 3 {eV}_{o} \Rightarrow λ_{o} = \frac{hc}{3 {eV}_{o}}$

$From (1) \frac{hc}{λ} = 3 {eV}_{o} + 5 {eV}_{o} \Rightarrow \frac{hc}{3 {eV}_{o}} = \frac{8 λ}{3}$

$∴ λ_{o} = \frac{8 λ}{3}$

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