The electric field at a point associated with a light wave is given… | JEE Main 2022 (29 Jun, Shift 2)

Q93JEE Main 2022 · 29 JUN · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation

The electric field at a point associated with a light wave is given by

E = 200 [sin (6 $\times$ 1015)t + sin (9 $\times$ 1015)t] Vm ${}^{-1}$

Given: h = 4.14 $\times$ 10 $-$ 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

Show answer and solution

Answer: (d)

Highest angular frequency $\omega = 9 \times 10^{15}$ rad/s gives the maximum KE: $\nu = \frac{\omega}{2\pi} = 1.43 \times 10^{15}$ Hz.
$E = h\nu = 4.14 \times 10^{-15} \times 1.43 \times 10^{15} = 5.93$ eV
$K_{max} = 5.93 - 2.50 \approx 3.42$ eV
So the answer is (d) 3.42 eV.

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