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JEE Main 2024 (30 Jan, Shift 2) · Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from JEE Main 2024 (30 Jan, Shift 2)

Q35JEE Main 2024 · 30 JAN · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

For the photoelectric effect, the maximum kinetic energy $(E_k)$ of the photoelectrons is plotted against the frequency $(v)$ of the incident photons as shown in figure. The slope of the graph gives

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Answer: (a)

$E_k = h\nu - \phi$ , so $\frac{dE_k}{d\nu} = h$ .
Slope $\tan\theta = h$ .
So the answer is (a) Planck's constant.

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