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 $8 V$. When the same surface is illuminated by light of wavelength $3 \lambda$, stopping potential is $2 V$. 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:
$E=\phi+K _{\text {max }}$
$\phi=\frac{hc}{\lambda _0}$
$K _{\text {max }}=eV _0$
$8 e=\frac{hc}{\lambda}-\frac{hc}{\lambda _0} \ldots$ (i)
$2 e=\frac{hc}{3 \lambda}-\frac{hc}{\lambda _0} \ldots$ (ii)
on solving (i) $ \&$ (ii)
$\lambda _0=9 \lambda$
