Let K 1 and K 2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ 1 and λ 2 , respectively are incident on a metallic surface. If λ 1 = 3 λ 2 then :

Question

Let K 1 and K 2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ 1 and λ 2 , respectively are incident on a metallic surface. If λ 1 = 3 λ 2 then :, K 1 > K 2 3, K 1 K 2 3, K 1 = K 2 3, K 2 = K 1 3

MARKS App · Accepted Answer

According to Einstein's photoelectric equation h ⁡ c λ = ϕ 0 + KE max , where ϕ 0 is a work function of a metal ∴ K ⁡ 1 = h ⁡ c λ 1 - ϕ 0 … i K ⁡ 2 = h ⁡ c λ 2 - ϕ 0 … i i or K 1 - K 2 = h c 1 λ 1 - 1 λ 2 = h c 1 3 λ 2 - 1 λ 2 = - 2 h c 3 λ 2 Given λ 1 = 3 λ 2 = - 2 3 K ⁡ 2 + ϕ 0 Using (ii) or ⁡ K ⁡ 1 = K ⁡ 2 - 2 3 K ⁡ 2 - 2 3 ϕ 0 = K ⁡ 2 3 - 2 3 ϕ 0 or K ⁡ 1 K ⁡ 2 3

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