If K 1 and K 2 are the maximum kinetic energies of photo electrons emitted when lights of wavelength λ 1 and λ 2 respectively are incident on a metallic surface and λ 1 = 3 λ 2 . Then

Question

If K 1 and K 2 are the maximum kinetic energies of photo electrons emitted when lights of wavelength λ 1 and λ 2 respectively are incident on a metallic surface and λ 1 = 3 λ 2 . Then, K 1 > K 2 / 3, K 1 K 2 / 3, K 1 = 3 K 2, K 2 = 3 K 1

MARKS App · Accepted Answer

Maximum kinetic energies of photo electrons for light of wavelength K 1 = h c λ 1 - f , Maximum kinetic energies of photo electrons for light of wavelength K 2 = h c λ 2 - f Also given, λ 1 = 3 λ 2 K 2 K 1 = 3 h c λ 1 - ϕ h c λ 1 - ϕ = 3 h c λ 1 - ϕ 3 h c λ 1 - ϕ Therefore K 2 K 1 > 3

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