When a certain metallic surface is illuminated with mono-chromatic light of wave length $λ$ , the stopping potential for photo electric current is $6 V_{0}$ . When the same surface is illuminated with light of wave length $2 λ$ , the stopping potential is $2 V_{0}$ . The threshold wavelength of this surface for photoelectric effect is:-

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$4 λ / 3$

$8 λ$

$6 λ$

$4 λ$

answer is C.

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Detailed Solution

$K_{\max} = \frac{h c}{λ} - \frac{h c}{λ_{0}} o r e V_{0} = \frac{h c}{λ} - \frac{h c}{λ_{0}}$
$\frac{h c}{λ} - \frac{h c}{λ_{0}} = 6 e V_{0} ...... (i)$
$\frac{h c}{2 λ} - \frac{h c}{λ_{0}} = 2 e V_{0} ...... (i i)$
Dividing (i) and (ii), we get
$\frac{h c}{λ} - \frac{h c}{λ_{0}} = \frac{3 h c}{2 λ} - \frac{3 h c}{λ_{0}} o r \frac{3}{2 λ} - \frac{1}{λ} = \frac{3}{λ} - \frac{1}{λ_{0}}$
$\frac{1}{2 λ} = \frac{2}{λ_{0}} o r λ_{0} = 4 λ$