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JEE MAINS 2024
31.01.24 S1
Question
When a metal surface is illuminated by light of wavelength λ, the stopping potential is 8" " V. When the same surface is illuminated by light of wavelength 3λ, stopping potential is 2" " V. The threshold wavelength for this surface is
Select the correct option:
A
B
C
D
4.5λ
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol. $\quad \mathrm{E}=\phi+\mathrm{K}_{\max }$ $$ \begin{aligned} & \phi=\frac{\mathrm{hc}}{\lambda_0} \\ & \mathrm{~K}_{\max }=\mathrm{eV}_0 \\ & 8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \\ & 2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \end{aligned} $$ on solving (i) \& (ii) $$ \lambda_0=9 \lambda $$
Question Tags
JEE Main
Physics
Medium
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