Dual Nature Of Matter Question 8
Question 8 - 2024 (31 Jan Shift 1)
When a metal surface is illuminated by light of wavelength $\lambda$, the stopping potential is $8 \mathrm{~V}$. When the same surface is illuminated by light of wavelength $3 \lambda$, stopping potential is $2 \mathrm{~V}$. The threshold wavelength for this surface is :
(1) $5 \lambda$
(2) $3 \lambda$
(3) $9 \lambda$
(4) $4.5 \lambda$
Show Answer
Answer: (3)
Solution:
$\mathrm{E}=\phi+\mathrm{K}_{\text {max }}$
$\phi=\frac{\mathrm{hc}}{\lambda_{0}}$
$\mathrm{K}{\text {max }}=\mathrm{eV}{0}$
$8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}} \ldots$ (i)
$2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_{0}} \ldots$ (ii)
on solving (i) $\backslash &$ (ii)
$\lambda_{0}=9 \lambda$
