Question:

Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ1 and λ2, respectively are incident on a metallic surface. If λ= 3λ2 then:

Updated On: Jan 31, 2026
  • \(K_1> \frac{K_2}{3}\)

  • \(K_1< \frac{K_2}{3}\)

  • \(K_1=\frac{K_2}{3}\)

  • \(K_2<=\frac{K_1}{3}\)

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The Correct Option is B

Solution and Explanation

K1=\(\frac{ℎc}{λ_1}−Φ\)=\(\frac{ℎc}{3λ_2}−Φ\)….(i)
and
K2=\(\frac{ℎc}{λ_2}−Φ\)….(ii)
from (i) and (ii) we can say
3K1 = K2 – 2φ
\(K1<\frac{K2}{3}\)

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