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 λ₁ = 3λ₂ then :

Solution

Physics sir · 4 steps

Do not just divide by 3: the work function is subtracted in full, so K₁ falls below K₂/3.

1. Einstein's equation. For each beam, the maximum kinetic energy is the photon energy minus the work function: K = hc/λ - φ. The same metal means the same φ for both beams.
2. Write both energies. Let E = hc/λ₂. Then K₂ = E - φ. Since λ₁ = 3λ₂, the first photon carries E/3, so K₁ = E/3 - φ.

   K₂/3 = E/3 - φ/3 ⇒ K₁ = K₂/3 - 2φ/3

3. Compare. Dividing K₂ by 3 also divides φ by 3, but K₁ has the full φ subtracted. Because φ > 0, K₁ is smaller than K₂/3 by exactly 2φ/3.
4. So. K₁ < K₂/3, which is option (b).

The idea to keep. Photon energy scales as 1/λ, but kinetic energy does not, because the same work function is subtracted every time. Write K = hc/λ - φ for both beams and compare.

Answer from the NTA final answer key. Solution by PrepWiser.