In a photoelectric experiment, increasing the intensity of incident light: (a) increases the number of photons incident and also increases the K.E. of the ejected electrons (b) increases the frequency of photons incident and increases the K.E. of the ejected electrons (c) increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged (d) increases the number of photons incident and the K.E. of the ejected electrons remains unchanged

Answer: (d). Intensity I = n hν per unit area per second; at fixed ν, more intensity means more photons n. Frequency is unchanged, so Kmax = hν - φ is unchanged. So the answer is (d).

In a photoelectric experiment, increasing the intensity of incident… | JEE Main 2021 (27 Aug, Shift 1)

Q50JEE Main 2021 · 27 AUG · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

In a photoelectric experiment, increasing the intensity of incident light:

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Answer: (d)

Intensity $I = n h\nu$ per unit area per second; at fixed $\nu$ , more intensity means more photons $n$ .
Frequency is unchanged, so $K_{max} = h\nu - \phi$ is unchanged.
So the answer is (d).

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