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When 1cm thick surface is illuminated with light of wavelength $λ,$ the stopping potential is V. When the same surface is illuminated by light of wavelength $2 λ$ . the stopping potential is V/3. Threshold wavelength for metallic surface is

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$4 λ / 3$

$4 λ$

$6 λ$

$8 λ / 3$

answer is B.

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Detailed Solution

According to the question

$e V = h c (\frac{1}{λ} - \frac{1}{λ_{0}})$ ........(i)

and $\frac{e V}{3} = h c (\frac{1}{2 λ} - \frac{1}{λ_{0}})$ .......(ii)

Dividing Eq.(i) by Eq.(ii), we get

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

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

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

$or \frac{1}{2 λ} = \frac{2}{λ_{0}}$

Threshold wavelength for metallic

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