Question

Radiations of two frequencies are incident on a metal surface of work function 2.0 eV one by one. The energies of their photons are 2.5 eV and 4.5 eV respectively. Find the ratio of the maximum speed of the electrons emitted in the two cases.

Answer 1

Given:

Work function of the metal (Φ) = 2.0 eV 

Photon energy in Case 1 (E₁) = 2.5 eV 

Photon energy in Case 2 (E₂) = 4.5 eV 

Using the photoelectric equation:

K_max = E − Φ

Maximum Kinetic Energy in Case 1 (K₁) = E₁ − Φ

= 2.5 − 2.0

= 0.5 eV

Maximum Kinetic Energy in Case 2 (K₂) = E₂ − Φ

= 4.5 − 2.0

= 2.5 eV

The maximum kinetic energy is related to the electron speed using:

K = `1/2 mv^2`

So,

v = `sqrt((2 K)/m)`

Taking the ratio of speeds in both cases:

`v_1/v_2 = sqrt(K_1/K_2`

= `sqrt(0.5/2.5)`

= `1/sqrt5`

∴ v₁ : v₂ = 1 : `sqrt 5`