The electric field at a point associated with a light wave is given by
$E = 200 [\sin (6 \times 10^{15}) t + \sin (9 \times 10^{15}) t] {Vm}^{- 1}$

Question

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

TestHub · Accepted Answer

Given: E = 200 sin 6 × 10 15 t + sin 9 × 10 15 t V m - 1 Here, angular velocity, ω 1 = 6 × 10 15 and ω 2 = 9 × 10 15 Maximum kinetic energy of the photoelectron is KE max . = E - ϕ = h f - ϕ = h ω 2 π - ϕ = 4 . 14 × 10 - 15 × 9 × 10 15 2 × 3 . 14 - 2 . 5 = 5 . 92 - 2 . 50 = 3 . 42 eV

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