Question

Light of wavelength 3000 Å falls on a metal surface having work function 2.3 eV. Calculate the maximum velocity of ejected electrons.

(Planck’s constant h = 6.63 x 10^–34 Js, Velocity of light c = 3 x 10⁸ m/s, mass of an electron = 9.1 x 10^–31 kg)

Answer 1

Given : λ = 3000 Å = 3 x 10^–7 m, me = 9.1 x 10^–31 kg,

ϕ_o = 2.3 eV, h = 6.63 x 10^–34 Js, c = 3 x 10⁸ m/s

To find: Maximum velocity (v_max)

Formula: (K.E)_max = \(\frac{hc}{λ}\) - ϕ_o

Calculation: From formula,

Ans: The maximum velocity of electron is 8.052 x 10⁵ m/s