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Q.

When a surface 1 cm thick is illuminated with light of wavelength $λ$ the stopping potential is V₀ but when the same surface is illuminated by light of wavelength 3 $λ$ the stopping potential is V₀ /6. the threshold wavelength for metallic surface is

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a

$4 λ$

b

$5 λ$

c

$3 λ$

d

$2 λ$

answer is B.

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Detailed Solution

Here ${eV}_{0} = hc [\frac{1}{λ} - \frac{1}{λ_{0}}]$
and $\frac{{eV}_{0}}{6} = hc [\frac{1}{3 λ} - \frac{1}{λ_{0}}]$
$Dividing eq. (1) by eq. (2), we get 6 = \frac{(\frac{1}{λ} - \frac{1}{λ_{0}})}{(\frac{1}{3 λ} - \frac{1}{λ_{0}})} or \frac{6}{3 λ} - \frac{6}{λ_{0}} = \frac{1}{λ} - \frac{1}{λ_{0}} or \frac{6}{3 λ} - \frac{1}{λ} = \frac{6}{λ_{0}} - \frac{1}{λ_{0}} or \frac{1}{λ} = \frac{5}{λ_{0}} or λ = \frac{λ_{0}}{5} or λ_{0} = 5 λ$

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