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JEE Main 2023 (10 Apr, Shift 2) · Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from JEE Main 2023 (10 Apr, Shift 2)

Q46JEE Main 2023 · 10 APR · SHIFT 2Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

The variation of stopping potential (V0)(\mathrm{V}_0) as a function of the frequency (v)(v) of the incident light for a metal is shown in figure. The work function of the surface is

Figure

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Answer: (d)

  1. The line cuts the ν\nu-axis at ν0=5×1014\nu_0 = 5 \times 10^{14} Hz.
  2. ϕ=hν0=6.63×10−34×5×1014=3.32×10−19\phi = h\nu_0 = 6.63 \times 10^{-34} \times 5 \times 10^{14} = 3.32 \times 10^{-19} J
  3. ϕ=3.32×10−191.6×10−19≈2.07\phi = \frac{3.32 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.07 eV
  4. So the answer is (d) 2.07 eV.
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