Question 313: Dual Nature of Matter and Radiation

Question

When a metallic surface is illuminated with radiation of wavelength '$\lambda$', the stopping potential is '$\mathrm{V}$'. If the same surface is illuminated with radiation of wavelength '$2 \lambda$', the stopping potential is '$\left(\frac{\mathrm{v}}{4}\right)$'. The threshold wavelength for the metallic surface is

Accepted Answer

For stopping potential $\mathrm{V}, \mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}$
For stopping potential $\frac{\mathrm{V}}{4}, \frac{\mathrm{eV}}{4}=\frac{\mathrm{hc}}{2 \lambda}-\frac{\mathrm{hc}}{\lambda_0}$
Taking the ratio, we get
$\begin{aligned}
& 4=\frac{\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}ight)}{\frac{1}{2 \lambda}-\frac{1}{\lambda_0}} \
& 4\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}ight)=\frac{1}{\lambda}-\frac{1}{\lambda_0} \
& \left(\frac{2}{\lambda}-\frac{4}{\lambda_0}ight)=\frac{1}{\lambda}-\frac{1}{\lambda_0} \
& \left(\frac{2}{\lambda}-\frac{1}{\lambda}ight)=-\frac{1}{\lambda_0}+\frac{4}{\lambda_0} \
& \quad\left(\frac{1}{\lambda}ight)=\frac{3}{\lambda_0} \
& 	herefore \quad \lambda_0=3 \lambda
\end{aligned}$

Answer

$\frac{5}{2} \lambda$

Answer

$3 \lambda$

Answer

$4 \lambda$

Answer

$5 \lambda$

Dual Nature of Matter and Radiation

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