Light of wavelength λ_A and λ_B falls on two identical metal plates A and B respectively. The maximum kinetic energy of photoelectrons is K_A and K_B respectively, then which one of the following relations is true (λ_A = 2λ_B)

Concept:

The energy of a photon is given by E = hν = hc/λ. The maximum kinetic energy of the photoelectrons is given by the photoelectric equation:

K.E. = hf - φ

where φ is the work function of the metal.

Calculation:

Light of wavelength λA and λB falls on two identical metal plates A and B respectively. The maximum kinetic energy of photoelectrons is KA and KB respectively, with λA = 2λB.

For plate A, with wavelength λ_A:

⇒ K_A = hc/λ_A - φ

For plate B, with wavelength λ_B:

⇒ K_B = hc/λ_B - φ

Given λ_A = 2λ_B, substitute λ_A:

⇒ K_A = hc/(2λ_B) - φ

Since λ_A = 2λ_B, we have:

⇒ K_A = (1/2)hc/λ_B - φ

⇒ K_A = (1/2)(hc/λ_B - 2φ) + φ

From K_B equation:

⇒ K_B = hc/λ_B - φ

Since hc/λ_B - φ = K_B:

⇒ K_A = (1/2)(K_B + φ)

Since φ is the same for both metals:

⇒ K_A = (1/2)K_B + (1/2)φ

Therefore, the correct relation is:

∴ K_A = (1/2)K_B + (1/2)φ, indicating K_A is less than K_B.