Question

1. Find the kinetic energy of a photoelectron emitted from a rubidium (Rb) surface if the wavelength of the incident light is 400 nm (1 nm = 1·10^−9 m), and the binding (or threshold) energy of an electron is 3.6·10^−19J.

(A) 5.0·10−19 J

(B) 3.6·10−19J

(C) 1.4·10−19 J

(D) λthreshold > 400 nm, no emission of photoelectrons

(E) λthreshold < 400 nm, no emission of photoelectrons

Please show how you got the problem

Answer 1

Ans:

Let ‘E’ be the energy associated with the incident light

According to photoelectric effect,

E = BE + KE

Where, E = energy associated with the incident light = (h.c/ λ)

            h = plank’s constant = 6.626 × 10^-34 m² kg s^-1

c = speed of light = 3 x 10⁸ m/s

            λ = wavelength of incident light = 400 x 10^-9 m

            BE = binding energy of electron = 3.6 x 10^-19 J

            KE = kinetic energy of emitted electron

KE = E – BE =

KE = {(6.626 × 10^-34 m² kg s^-1 x 3 x 10⁸ m s^-1)/ 400 x 10^-9 m} - 3.6 x 10^-19 J

{Note: 1 J = 1 kg m²s^-2 }

KE = 1.3695 x 10^-19 J

Kinetic energy of a photoelectron emitted from a rubidium (Rb) surface = 1.3695 x 10^-19 J