This question has Statement - 1 and Statement - 2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement - 1: A metallic surface is irradiated by a monochromatic light of frequency v > v₀ (the threshold frequency). The maximum kinetic energy and the stopping potential are Kmax and V₀ respectively. If the frequency incident on the surface is doubled, both the Kmax and V₀ are also doubled. Statement - 2: The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light. (a) Statement - 1 is true, Statement - 2 is true, Statement - 2 is the correct explanation of Statement - 1. (b) Statement - 1 is true, Statement - 2 is true, Statement - 2 is not the correct explanation of Statement - 1. (c) Statement - 1 is false, Statement - 2 is true. (d) Statement - 1 is true, Statement - 2 is false.

Answer: (c). Kmax = hν - φ and eV₀ = hν - φ. Doubling ν: K' = 2hν - φ = 2Kmax + φ ≠ 2Kmax. Statement 1 is false. Both Kmax and V₀ are linear (not proportional) in ν. Statement 2 is true. So the answer is (c).

This question has Statement - 1 and Statement - 2. Of the four… | JEE Main 2011 (AIEEE)

Q69JEE Main 2011 · AIEEEMedium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

This question has Statement - $1$ and Statement - $2$ . Of the four choices given after the statements, choose the one that best describes the two statements.

Statement - $1$ : A metallic surface is irradiated by a monochromatic light of frequency $v > v_0$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $K_\mathrm{max}$ and $V_0$ respectively. If the frequency incident on the surface is doubled, both the $K_\mathrm{max}$ and $V_0$ are also doubled.

Statement - $2$ : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.

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Answer: (c)

$K_{max} = h\nu - \phi$ and $eV_0 = h\nu - \phi$ .
Doubling $\nu$ : $K' = 2h\nu - \phi = 2K_{max} + \phi \neq 2K_{max}$ . Statement 1 is false.
Both $K_{max}$ and $V_0$ are linear (not proportional) in $\nu$ . Statement 2 is true.
So the answer is (c).

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Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from JEE Main 2011 (AIEEE)

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