Question 322: Dual Nature of Matter and Radiation

Question

When a metallic surface is illuminated with a radiation of wavelength ' $\lambda$ ', the stopping potential is ' $V$ '. If the same surface is illuminated with radiation of wavelength ' $3 \lambda$ ', the stopping potential is ' $\left(\frac{\mathrm{V}}{6}\right)$ '. The threshold wavelength for the surface is

Accepted Answer

$$\begin{aligned}
& 	ext { From Einstein's equation, } \
& \mathrm{hv}=\mathrm{e} \mathrm{V}_0+\mathrm{h} v_0 \
& 	herefore \quad \frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}=e \mathrm{~V}_0 \
& 	ext { case (i) } \lambda=\lambda ; \mathrm{V}_0=\mathrm{V} \
& \frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}=\mathrm{eV}\quad	ext{... (i)} \
& 	ext { case (ii) } \lambda=3 \lambda ; \mathrm{V}_0=\frac{\mathrm{V}}{6} \
& \frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0}=\frac{\mathrm{eV}}{6} \quad	ext{... (ii)}
\end{aligned}$$
$$\begin{aligned}
	herefore\quad & \frac{\left(\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}ight)}{\left(\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0}ight)}=6 \
	herefore \quad & \frac{1}{\lambda}-\frac{1}{\lambda_0}=6\left(\frac{1}{3 \lambda}-\frac{1}{\lambda_0}ight)
\end{aligned}$$
$$\begin{array}{ll}
	herefore & \frac{1}{\lambda}-\frac{1}{\lambda_0}=\frac{2}{\lambda}-\frac{6}{\lambda_0} \
	herefore & \frac{-1}{\lambda_0}+\frac{6}{\lambda_0}=\frac{2}{\lambda}-\frac{1}{\lambda} \
	herefore & \frac{5}{\lambda_0}=\frac{1}{\lambda} \Rightarrow \lambda_0=5 \lambda
\end{array}$$

Answer

$3 \lambda$

Answer

$4 \lambda$

Answer

$5 \lambda$

Answer

$6 \lambda$

Dual Nature of Matter and Radiation

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