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JEE Main 2019 (12 Apr, Shift 1) · Photoelectric Experiment: Intensity, Stopping Potential & Graphs

Photoelectric Experiment: Intensity, Stopping Potential & Graphs question from JEE Main 2019 (12 Apr, Shift 1)

Q63JEE Main 2019 · 12 APR · SHIFT 1Medium

Photoelectric Experiment: Intensity, Stopping Potential & Graphs

The stopping potential V0 (in volt) as a function of frequency (vv) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be: (Given: Planck’s constant (h) = 6.63 × 10–34 Js, electron charges e = 1.6 × 10–19 C)

Figure

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Answer: (d)

  1. From the graph, V0=0V_0 = 0 at ν0=4×1014\nu_0 = 4 \times 10^{14} Hz.
  2. ϕ=hν0=6.63×10−34×4×1014=2.65×10−19\phi = h\nu_0 = 6.63 \times 10^{-34} \times 4 \times 10^{14} = 2.65 \times 10^{-19} J
  3. ϕ=2.65×10−191.6×10−19≈1.66\phi = \frac{2.65 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.66 eV
  4. So the answer is (d) 1.66 eV.
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