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Light wave given as a sum of two sine terms falls on a metal: maximum kinetic energy of photoelectrons

The same question, asked 2 times: 29 Jun 2022*, 21 Jan 2026 S1. Every wording below, oldest first.

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  1. Question 1

    29 Jun 2022

    96
    The electric field at the point associated with a light wave is given by
    E=200[sin⁡(6×1015)t+sin⁡(9×1015)t]Vm−1E = 200\left\lbrack \sin\left( 6 \times10^{15} \right)t + \sin\left( 9 \times10^{15} \right)t \right\rbrack Vm^{- 1}
    Given: h=4.14×10−15eVsh = 4.14 \times10^{- 15}eVs If this light falls on a metal surface having a work function of 2.50eV2.50eV, the maximum kinetic energy of the photoelectrons will be:

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  2. Question 2

    21 Jan 2026 S1

    12
    A light wave described by E=60[sin⁡(3×1015t)+sin⁡(12×1015t)]E = 60\left[\sin\left(3 \times 10^{15}t\right) + \sin\left(12 \times 10^{15}t\right)\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 J sh = 6.6 \times 10^{-34}\ J\,s and e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ C)

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