Q: When a centimeter thick surface is illuminated with light of wavelength λ , the stopping potential is V. When the same surface is illuminated by light of wavelength 2λ , the stopping potential is V/3. Threshold wavelength for the metallic surface is

(A) 4λ/3

(B) 4λ

(C) 6λ

(D) 8λ/3

**Solution :**

$\large \frac{hc}{\lambda} = \phi + e V $

$\large \frac{hc}{\lambda} – \phi = e V $ …(i)

$\large \frac{hc}{2 \lambda} = \phi + e \frac{V}{3} $

$\large \frac{hc}{2 \lambda} – \phi = e \frac{V}{3} $ …(ii)

On dividing (i) by (ii)

$\large \frac{\frac{hc}{\lambda} – \phi }{\frac{hc}{2 \lambda} – \phi } = 3 $

$\large \frac{hc}{\lambda} – \phi = 3 (\frac{hc}{2 \lambda} – \phi) $

$\large 3 \phi – \phi = \frac{3 hc}{2 \lambda} – \frac{hc}{\lambda} $

$\large 2 \phi = \frac{hc}{2 \lambda} $

$\large \phi = \frac{hc}{4 \lambda} $

$\large \frac{hc}{ \lambda_o} = \frac{hc}{4 \lambda} $

$\large \lambda_o = 4 \lambda$

Correct option is (B)