In an experiment with photoelectric effect, the stopping potential, (a) is (1/e) times the maximum kinetic energy of the emitted photoelectrons (b) increases with increase in the intensity of the incident light (c) decreases with increase in the intensity of the incident light (d) increases with increase in the wavelength of the incident light

Answer: (a). eV₀ = Kmax ⇒ V₀ = 1/eKmax. V₀ does not depend on intensity, and it decreases as wavelength increases ( Kmax = hc/λ - φ). So the answer is (a).

In an experiment with photoelectric effect, the stopping potential, | JEE Main 2025 (29 Jan, Shift 2)

Q15JEE Main 2025 · 29 JAN · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

In an experiment with photoelectric effect, the stopping potential,

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Answer: (a)

$eV_0 = K_{max} \Rightarrow V_0 = \dfrac{1}{e}K_{max}$ .
$V_0$ does not depend on intensity, and it decreases as wavelength increases ( $K_{max} = hc/\lambda - \phi$ ).
So the answer is (a).

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