Question 77: Dual Nature of Matter and Radiation

Question

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:

Accepted Answer

$\begin{aligned}
& \mathrm{E}=\phi+\mathrm{K}_{\max } \
& \phi=\frac{\mathrm{hc}}{\lambda_0} \
& \mathrm{~K}_{\max }=\mathrm{eV}_0 \
& 8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . 	ext { (i) } \
& 2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . . 	ext { (ii) } \
& 	ext { on solving (i) & (ii) } \
& \lambda_0=9 \lambda
\end{aligned}$

Answer

3$\lambda$

Answer

9$\lambda$

Answer

5$\lambda$

Answer

4.5$\lambda$

Dual Nature of Matter and Radiation

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:

Choose the correct answer: