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The electric field associated with a light wave is given by E = E 0 sin ⁡ 1 .57 × 10 7 x - ct (where x and t are in metre and second respectively). This light is used in an experiment on photoelectric effect with the emitter having work function ϕ = 1.9 e V . What is the maximum kinetic energy (in eV ) of the ejected photoelectron? [Take π = 3 . 14 , hc = 12400 eV A ∘ ]

Correct answer

1.2

Step-by-step solution

Given, E = E 0 sin ⁡ 1 .57 × 10 7 x - ct ...(i) ∴ E = E 0 sin ⁡ ω c x - ct ...(ii) On comparing equations (i) and (ii), we get ω c = 1 .57 × 10 7 = π 2 × 10 7 λ = 4 × 10 - 7 m = 400 n m As, ϕ 0 = hc λ - E K 1.9 = 1240 400 - E K E K = 3.1 - 1.9 eV = 1.2 e V

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