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JEE Main 2024 (5 Apr, Shift 1) · Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from JEE Main 2024 (5 Apr, Shift 1)

Q34JEE Main 2024 · 5 APR · SHIFT 1Easy

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Given below are two statements:

Figure

Statement I: Figure shows the variation of stopping potential with frequency (v)(v) for the two photosensitive materials M1M_1 and M2M_2. The slope gives value of he\frac{h}{e}, where hh is Planck's constant, e is the charge of electron.

Statement II: M2\mathrm{M}_2 will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency.

In the light of the above statements, choose the most appropriate answer from the options given below.

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Answer: (a)

  1. V0=heν−ϕeV_0 = \frac{h}{e}\nu - \frac{\phi}{e}, so the slope of each line is h/eh/e: Statement I is correct.
  2. M2M_2 has the higher threshold frequency, so larger work function ϕ\phi.
  3. For the same ν\nu, KEmax=hν−ϕKE_{max} = h\nu - \phi is smaller for M2M_2: Statement II is incorrect.
  4. So the answer is (a).
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