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JEE Main 2022 · 29 Jun · Shift 2, Physics question 36

The electric field at a point associated with a light wave is given by
E=200[sin⁡(6×1015)t+sin⁡(9×1015)t]Vm−1E=200\left[\sin\left(6\times10^{15}\right)t+\sin\left(9\times10^{15}\right)t\right]\text{Vm}^{-1}
Given : h=4.14×10−15h=4.14\times10^{-15} eVs

If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

Choose one option
Answer and worked solution

Official answer: D

Highest frequency. f=9×10152π≈1.43×1015f=\dfrac{9\times10^{15}}{2\pi}\approx1.43\times10^{15} Hz, photon energy 4.14×10−15×1.43×1015≈5.934.14\times10^{-15}\times1.43\times10^{15}\approx5.93 eV.

Kinetic energy 5.93−2.50≈3.425.93-2.50\approx3.42 eV.

(D)

Topics:Photoelectric effect
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