Question

Photons of energy 1 eV and 2.5 eV successively illuminate a metal, whose work function is 0.5 eV, the ratio of maximum speed of emitted election is………… 

(A) 1:2 

(B) 2:1 

(C) 3:1 

(D) 1:3

Answer 1

The correct option is (A) 1:2.

Explanation:

KE_max = hf – ɸ

Given: hf₁ = 1 eV, ɸ = 0.5 eV

hf₂ = 2.5 eV

∴ [{(KE_max)₁} / {(KE_max)₂}] = {(1 – 0.5) / (2.5 – 0.5)} = {(0.5) / 2} = (1/4)

 ∴ [{(1/2) mV_(1)max²} / {(1/2) mV_(2)max²}] = (1/4)

∴ {(V_1max) / (V_2max)} = 1:2