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If K 1 and K 2 are maximum kinetic energies of photoelectrons emitted when lights of wavelength λ 1 and λ 2 respectively incident on a metallic surface. If λ 1 = 3 λ 2 , then

Options

  1. AK 1 > K 2 3
  2. BK 1 < K 2 3
  3. CK 1 = 3 K 2
  4. DK 2 = 3 K 1

Correct answer

B. K 1 < K 2 3

Step-by-step solution

Eienstein's photoelectric equation is given as, K 1 = h c 1 λ 1 - 1 λ 0 ...(i) and K 2 = h c 1 λ 2 - 1 λ 0 ...(ii) Dividing Equation (i) by Equation (ii), we get K 1 K 2 = 1 λ 1 - 1 λ 0 1 λ 2 - 1 λ 0 Since, λ 1 = 3 λ 2 K 1 K 2 = 1 3 λ 2 - 1 λ 0 1 λ 2 - 1 λ 0 = 1 3 λ 2 1 - 3 λ 0 / λ 2 1 λ 2 1 - 1 λ 0 K 1 K 2 = 1 3 1 - 3 λ 0 / λ 2 1 - 1 λ 0 / λ 2 3 K 1 K 2 = 1 - 3 / λ 0 / λ 2 1 - 1 / λ 0 λ 2 < 1 ∴ 3 K 1 K 2 < 1 ⇒ K 1 < K 2 3

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