Question

Wavelength λA and λB are incident on two identical metal plates and photo electrons are emitted. If λ_A = 2λ_B, the maximum kinetic energy of photo electrons is………

(A) 2K_A = K_B

(B) K_A < (K_B / 2)

(C) K_A = 2K_B

(D) K_A > (K_B / 2)

Answer 1

The correct option is (B) K_A < (K_B / 2).

Explanation:

λ = (h/p) = {h / √(2mE)}

 i.e. 2mE = (h² / λ²)

∴ Energy = {h² / (2mλ²)} i.e. Energy ∝ (1 / λ²)

Let energies corresponding to λ_A, λ_B are K_A, K_B respectively.

∴ (K_A / K_B) = (λ_B / λ_A)²

∴ (K_A / K_B) = (1/2)² = (1/4)

∴ K_B = 4K_A

∴ (K_B / 2) = 2K_A

hence K_A < (K_B / 2)