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)