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JEE Main 2024 (4 Apr, Shift 1) · Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from JEE Main 2024 (4 Apr, Shift 1)

Q29JEE Main 2024 · 4 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Which figure shows the correct variation of applied potential difference (V) with photoelectric current (I) at two different intensities of light $(\mathrm{I_1} < \mathrm{I_2})$ of same wavelengths:

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Answer: (c)

Same wavelength (frequency) means the same stopping potential, so both curves meet the V axis at the same $-V_0$ .
Saturation current $\propto$ intensity, so the $I_2$ curve is higher since $I_2 > I_1$ .
Only graph (c) shows a common $-V_0$ with $I_2$ above $I_1$ .
So the answer is (c).

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