Question

Photoelectric effect on surface is found for frequencies 5.5 × 10⁸ MHz and 4.5 × 10⁸ MHz If ratio of maximum kinetic energies of emitted photo electrons is 1:5, threshold frequency for metal surface is………… 

(A) 7.55 × 10⁸ MHz 

(B) 4.57 × 10⁸ MHz 

(C) 9.35 × 10⁸ MHz 

(D) 5.75 × 10⁸ MHz

Answer 1

The correct option is (D) 5.75 × 10⁸ MHz.

Explanation:

Kinetic energy = E = (1/2) mV² = hf – hf_o

∴ E = h(f – f_o)

 ∴ (E₁ / E₂) = [{h(f₁ – f_o)} / {h(f₂ – f_o)}] = {(f₁ – f_o) / (f₂ – f_o)} .....(1)

Given: f₁ = 5.5 × 10⁸ MHz = 5.5 × 10¹⁴ Hz

f₂ = 4.5 × 10⁸ MHz = 4.5 × 10¹⁴ Hz

also (E₁ / E₂) = (1/5)

∴ (1/5) = {(f₁ – f_o) / (f₂ – f_o)} ----- from (1)

∴ f₂ – f_o  = 5f₁ – 5f_o

∴ f₂ – 5f₁ = – 4 f_o

∴ f_o = Threshold frequexy = {(5f₁ – f₂) / 4} = [{5(5.5 × 10¹⁴)}  / 4] - 4.5 × 10¹⁴

∴ f_o = [{23 × 10¹⁴} / 4] = 5.75 × 10¹⁴ Hz

f_o = 5.75 × 10⁸ MHz