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JEE MAIN 2025
29-01-2025 SHIFT-2
Question
In an experiment with photoelectric effect, the stopping potential
Select the correct option:
A
increases with increase in the intensity of the incident light
B
is $\left(\frac{1}{\mathrm{e}}\right)$ times the maximum kinetic energy of the emitted photoelectrons
C
decreases with increase in the intensity of the incident light
D
increases with increase in the wavelength of the incident light
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\frac{\mathrm{hC}}{\lambda}=\mathrm{W}+\mathrm{eV}_{\mathrm{s}}$
$\frac{\mathrm{hC}}{\lambda}=\mathrm{W}+\left(\mathrm{K}_{\text {max }}\right)$
$\therefore \mathrm{V}_{\mathrm{s}}=\frac{\mathrm{K}_{\text {max }}}{\mathrm{e}}$
Question Tags
JEE Main
Physics
Medium
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