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JEE Main 2026 (5 Apr, Shift 1) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2026 (5 Apr, Shift 1)

Q5JEE Main 2026 · 5 APR · SHIFT 1Easy

Photoelectric Effect: Einstein's Equation

Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is α×10−19 J\alpha \times 10^{-19}\text{ J}. The value of α\alpha is ______.

(h=6.62×10−34 J s, e=1.6×10−19 C\left(h = 6.62 \times 10^{-34}\text{ J s},\ e = 1.6 \times 10^{-19}\text{ C}\right. and c=3×108 m/s)\left.c = 3 \times 10^{8}\text{ m/s}\right)

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Answer: (c)

  1. ϕ=hcλ−eV0\phi = \frac{hc}{\lambda} - eV_0
  2. hcλ=6.62×10−34×3×108331×10−9=6.0×10−19 J\frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{331 \times 10^{-9}} = 6.0 \times 10^{-19}\text{ J}
  3. eV0=1.6×10−19×0.2=0.32×10−19 JeV_0 = 1.6 \times 10^{-19} \times 0.2 = 0.32 \times 10^{-19}\text{ J}
  4. ϕ=(6.0−0.32)×10−19=5.68×10−19 J\phi = (6.0 - 0.32) \times 10^{-19} = 5.68 \times 10^{-19}\text{ J}, so the answer is (c) 5.68.
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