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JEE Main 2026 (22 Jan, Shift 2) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2026 (22 Jan, Shift 2)

Q3JEE Main 2026 · 22 JAN · SHIFT 2Easy

Photoelectric Effect: Einstein's Equation

Light is incident on a metallic plate having work function 110×10−20 J110 \times 10^{-20}\text{ J}. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ______ rad/s. (h=6.63×10−34 J.s)(h = 6.63 \times 10^{-34}\text{ J.s}).

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Answer: (d)

  1. Zero KE means hν=ϕh\nu = \phi, so ν=ϕh=110×10−206.63×10−34≈1.66×1015 Hz\nu = \dfrac{\phi}{h} = \dfrac{110\times10^{-20}}{6.63\times10^{-34}} \approx 1.66\times10^{15}\text{ Hz}
  2. Angular frequency ω=2πν=2π×1.66×1015\omega = 2\pi\nu = 2\pi\times1.66\times10^{15}
  3. ω≈1.04×1016 rad/s\omega \approx 1.04\times10^{16}\text{ rad/s}
  4. So the answer is (d) 1.04×10161.04\times10^{16}.
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