Physics · Atomic Physics

JEE Main 2025 — 29 January, Evening Shift — Question 20

In an experiment with photoelectric effect, the stopping potential.

Option A:
increases with increase in the wavelength of the incident light
Option B:
increases with increase in the intensity of the incident light
Option C:
is $\left(\frac{1}{e}\right)$ times the maximum kinetic energy of the emitted photoelectrons
Correct
Option D:
decreases with increase in the intensity of the incident light

Answer: C

Step-by-step solution

$\frac{h \mathrm{C}}{\lambda}=\mathrm{W}+\mathrm{eV}_{\mathrm{S}}$ $\frac{\mathrm{hC}}{\lambda}=\mathrm{W}+\left(\mathrm{K}_{\text {max }}\right)$ $\therefore \mathrm{V}_{\mathrm{S}}=\frac{\mathrm{K}_{\text {max }}}{\mathrm{e}}$

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Photoelectric Effect

The number of spectral lines emitted by atomic hydrogen that is in the 4^ th energy level, is

Answer: C

Step-by-step solution

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