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:
Solution
<p>$$\begin{aligned}
& \mathrm{E}=\phi+\mathrm{K}_{\max } \\
& \phi=\frac{\mathrm{hc}}{\lambda_0} \\
& \mathrm{~K}_{\max }=\mathrm{eV}_0 \\
& 8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) } \\
& 2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text { (ii) } \\
& \text { on solving (i) & (ii) } \\
& \lambda_0=9 \lambda
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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