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MHT CET 2025 (19 Apr, Shift 2) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from MHT CET 2025 (19 Apr, Shift 2)

Q34MHT CET 2025 · 19 APR · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation

The maximum velocity of the photoelectrons emitted by a metal surface is 9×105 m/s9 \times 10^{5}\text{ m/s}. The value of ratio of charge (e) to mass (m) of the photoelectron is 1.8×1011 C/kg1.8 \times 10^{11}\text{ C/kg}. The value of stopping potential in volt is

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Answer: (b)

  1. eV0=12mvmax2⇒V0=vmax22(e/m)eV_0 = \frac{1}{2}mv_{max}^2 \Rightarrow V_0 = \frac{v_{max}^2}{2(e/m)}
  2. V0=(9×105)22×1.8×1011=8.1×10113.6×1011V_0 = \frac{(9 \times 10^5)^2}{2 \times 1.8 \times 10^{11}} = \frac{8.1 \times 10^{11}}{3.6 \times 10^{11}}
  3. V0=2.25 VV_0 = 2.25\text{ V}, so the answer is (b) 2.25.
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