The electric field at a point associated with a light wave is given by E=200[\sin (6 \times 10^{15}) t+\sin (9 \times 10^{15}) t] V^{-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 [29-Jun-2022-Shift-2]

JEE Main 29 June 2022 Shift 2 Solved Paper

Section: Physics

Question : 36 of 90

Marks: +1, -0

The electric field at a point associated with a light wave is given by
E=200[sin(6×1015)‌t+sin(9×1015)‌t]V−1
Given : h=4.14×10−15‌eVs
If this light falls on a metal surface having a work function of 2.50‌eV, the maximum kinetic energy of the photoelectrons will be

[29-Jun-2022-Shift-2]

1.90‌eV
3.27‌eV
3.60‌eV
3.42‌eV

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