JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 9)
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:
3$$\lambda$$
9$$\lambda$$
5$$\lambda$$
4.5$$\lambda$$
Explanation
$$\begin{aligned}
& \mathrm{E}=\phi+\mathrm{K}_{\max } \\
& \phi=\frac{\mathrm{hc}}{\lambda_0} \\
& \mathrm{~K}_{\max }=\mathrm{eV}_0 \\
& 8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) } \\
& 2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . \text { (ii) } \\
& \text { on solving (i) & (ii) } \\
& \lambda_0=9 \lambda
\end{aligned}$$
Comments (0)
