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MHT CET 2019 (2 May, Shift 2) · Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from MHT CET 2019 (2 May, Shift 2)

Q61MHT CET 2019 · 2 MAY · SHIFT 2Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum KE of the emitted photoelectrons would be

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Answer: (d)

$K_{max} = h\nu - \phi$ depends only on frequency and work function.
Reducing intensity to $I/4$ only reduces the number of photons per second.
So the answer is (d) unchanged.

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