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JEE Main 2019 (9 Apr, Shift 1) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2019 (9 Apr, Shift 1)

Q131JEE Main 2019 · 9 APR · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation

The electric field of light wave is given as

E⃗=10−3cos⁡(2πx5×10−7−2π×6×1014t)x^ NC\vec{E} = 10^{-3}\cos\left(\frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14}t\right)\hat{x}\,\frac{\text{N}}{\text{C}}

This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is:

Given, E (in eV) = 12375/λ\lambda(inÅ)

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Answer: (b)

  1. From the wave, k=2πλk = \frac{2\pi}{\lambda} with λ=5×10−7\lambda = 5 \times 10^{-7} m =5000= 5000 Å.
  2. E=123755000=2.475E = \frac{12375}{5000} = 2.475 eV
  3. eV0=E−ϕ=2.475−2=0.475eV_0 = E - \phi = 2.475 - 2 = 0.475 eV
  4. So the answer is (b) 0.48 V.
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