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When radiation of wavelength $λ$ is incident on a metallic surface, the stopping potential is 4.8 volts. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1. 6 volts. Then the threshold wavelength for the surface is:

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$2 λ$

$4 λ$

$6 λ$

$8 λ$

answer is B.

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Detailed Solution

$By using \frac{hc}{c} (\frac{1}{λ} - \frac{1}{λ_{0}}) = V_{0}$

$\Rightarrow \frac{hc}{e} (\frac{1}{λ} - \frac{1}{λ_{0}}) = 4 .8 \dots ... (i)$

$and \frac{hc}{e} (\frac{1}{2 λ} - \frac{1}{λ_{0}}) = 1 .6 \dots (ii)$

From equation (i) by (ii),

$\frac{(\frac{1}{λ} - \frac{1}{λ_{0}})}{(\frac{1}{2 λ} - \frac{1}{λ_{0}})} = \frac{4 .8}{1 .6} \Rightarrow λ_{0} = 4 λ$

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