In photoelectric experiment keeping the frequency of incident radiation and accelerating potential fixed, if the intensity of incident light is increased, (a) photoelectric current decreases (b) kinetic energy of emitted photoelectrons decreases (c) photoelectric current increases (d) kinetic energy of emitted photoelectrons increases

Answer: (c). More intensity at fixed frequency means more photons per second, so more electrons per second. Kmax = hν - φ is unchanged because ν is unchanged. So the answer is (c) photoelectric current increases.

In photoelectric experiment keeping the frequency of incident… | MHT CET 2021 (24 Sep, Shift 1)

Q48MHT CET 2021 · 24 SEP · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

In photoelectric experiment keeping the frequency of incident radiation and accelerating potential fixed, if the intensity of incident light is increased,

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Answer: (c)

More intensity at fixed frequency means more photons per second, so more electrons per second.
$K_{max} = h\nu - \phi$ is unchanged because $\nu$ is unchanged.
So the answer is (c) photoelectric current increases.

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