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JEE Main 2026 (21 Jan, Shift 1) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2026 (21 Jan, Shift 1)

Q10JEE Main 2026 · 21 JAN · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation

A light wave described by E=60[sin⁡(3×1015)t+sin⁡(12×1015)t]E = 60\left[\sin\left(3 \times 10^{15}\right)t + \sin\left(12 \times 10^{15}\right)t\right] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately)

______ eV. (h=6.6×10−34\left(h = 6.6 \times 10^{-34}\right. J.s. and e=1.6×10−19 C)\left.e = 1.6 \times 10^{-19}\text{ C}\right)

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Answer: (d)

  1. The wave has two angular frequencies; the larger one, ω=12×1015\omega = 12 \times 10^{15} rad/s, gives the maximum KE.
  2. E=hω2π=6.6×10−34×12×10152π=1.26×10−18E = \frac{h\omega}{2\pi} = \frac{6.6 \times 10^{-34} \times 12 \times 10^{15}}{2\pi} = 1.26 \times 10^{-18} J =7.88= 7.88 eV
  3. Kmax=7.88−2.8≈5.1K_{max} = 7.88 - 2.8 \approx 5.1 eV
  4. So the answer is (d) 5.1 eV.
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