Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speed of emitted electrons will be

Options

(a) 1 : 4
(b) 1 : 2
(c) 1 : 1
(d) 1 : 5

Correct Answer:

1 : 2

Explanation:

The maximum kinetic energy of emitted electrons is given by
K.E = WorkFunction(₀) – WorkFunction(₁)
K.E₁ = 1 eV – 0.5 eV = 0.5 eV
K.E₂ = 2.5 eV – 0.5 eV = 2.0 eV
K.E₁ / K.E₂ = 0.5 eV / 2 eV = 1/4

KE = mv² /2
v₁ / v₂ = √1/4 = 1/2