A certain metallic surface is illuminated with monochromatic light of wavelength $λ$ . The stopping potential for photoelectric current for this light is $3 V_{0}$ . If the same surface is illuminated with light of wavelength $2 λ$ , The stopping potential is $V_{0}$ . The Threshold wavelength for this surface for photoelectric effect is

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$4 λ$

$\frac{λ}{4}$

$\frac{λ}{6}$

$6 λ$

answer is A.

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Detailed Solution

${eV}_{s} = \frac{hc}{λ} - Ψ$

$3 {eV}_{o} = \frac{hc}{λ} - Ψ$

${eV}_{o} = \frac{hc}{2 λ} - Ψ$

$\frac{3 {eV}_{o}}{2} = \frac{3 hc}{2 λ} - 3 Ψ$

substructing both the equations

$Ψ = \frac{hc}{4 λ}$

So $λ_{th} = \frac{hc}{Ψ} = \frac{hc}{hc / 4 λ} = 4 λ$

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