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MHT CET 2026 (16 Apr, Shift 1) · Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from MHT CET 2026 (16 Apr, Shift 1)

Q7MHT CET 2026 · 16 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Using Einstein's photoelectric equation, the graphical representation between the kinetic energy (E) of emitted Photoelectrons and the frequency of incident radiation ( $\nu$ ) is show correctly in figure

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Answer: (c)

$E = h\nu - \phi$ : a straight line of positive slope $h$ .
$E = 0$ at $\nu = \nu_0 = \phi/h > 0$ , so the line starts on the $\nu$ axis at the threshold frequency, not at the origin.
Graph C shows this.
So the answer is (c) C.

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