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JEE Main 2019 (9 Jan, Shift 2) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2019 (9 Jan, Shift 2)

Q135JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Photoelectric Effect: Einstein's Equation

The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 ×\times 107)ct + sin(6.28 ×\times 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons?

(Take c = 3 ×\times 108 ms−1{}^{-1}, h = 6.6 ×\times 10−-34J-s)

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Answer: (d)

  1. ω=2πν\omega = 2\pi\nu: ν1=3.14×107×3×1082π=1.5×1015\nu_1 = \frac{3.14 \times 10^7 \times 3 \times 10^8}{2\pi} = 1.5 \times 10^{15} Hz and ν2=3×1015\nu_2 = 3 \times 10^{15} Hz.
  2. Maximum KE comes from the higher frequency: hν2=6.6×10−34×3×1015=1.98×10−18h\nu_2 = 6.6 \times 10^{-34} \times 3 \times 10^{15} = 1.98 \times 10^{-18} J ≈12.4\approx 12.4 eV.
  3. Kmax=12.4−4.7≈7.7K_{max} = 12.4 - 4.7 \approx 7.7 eV
  4. So the answer is (d) 7.72 eV.
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