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 photo-electric effect is

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$\frac{λ}{6}$

$4 λ$

$6 λ$

$\frac{π}{4}$

answer is D.

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Detailed Solution

According to Einstein's photoelectric equation

$e V_{s} = \frac{h c}{λ} - \frac{h c}{λ_{0}}$

$where V_{s} = Stopping potential$

$λ = Incident wavelength$

$λ_{0} = Threshold wavelength$

$or V_{s} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ_{0}})$

for the first case

$3 V_{0} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ_{0}})$

for the second case

$V_{0} = \frac{h c}{e} (\frac{1}{2 λ} - \frac{1}{λ_{0}})$

Divide eqn. (i) by (ii), we get

$3 = \frac{(\frac{1}{λ} - \frac{1}{λ_{0}})}{(\frac{1}{2 λ} - \frac{1}{λ_{0}})} 3 (\frac{1}{2 λ} - \frac{1}{λ_{0}}) = (\frac{1}{λ} - \frac{1}{λ_{0}}) \frac{3}{2 λ} - \frac{3}{λ_{0}} = \frac{1}{λ} - \frac{1}{λ_{0}} \frac{1}{2 λ} = \frac{2}{λ_{0}} or λ_{0} = 4 λ$

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