Dual Nature of Radiation and Matter - Result Question 23
25. When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2 \lambda$, the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is :
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======= ####25. When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2 \lambda$, the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is :
c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (a) $4 \lambda$
(b) $5 \lambda$
(c) $\frac{5}{2} \lambda$
(d) $3 \lambda$
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Answer:
Correct Answer: 25. (d)
Solution:
- (d) According to Einstein’s photoelectric effect,
$eV=\frac{hc}{\lambda}-\frac{hc}{\lambda_0}$
$eV / 4=\frac{hc}{2 \lambda}-\frac{hc}{\lambda_0}$
Dividing equation (i) by (ii) by
$\Rightarrow 4=\frac{\frac{1}{\lambda}-\frac{1}{\lambda_0}}{\frac{1}{2 \lambda}-\frac{1}{\lambda_0}}$ on solving we get,
$\lambda_0=3 \lambda$
Threshold wavelength $(\lambda_0)$ is the maximum wavelength of incident radiations required to eject the electrons from a metallic surface.
If incident wavelength $\lambda>\lambda_0$
$\Rightarrow$ No photoelectron emission occur
