Question

What is the

(a) momentum,

(b) speed, and

(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

Answer 1

Kinetic energy of the electron, E_k = 120 eV

Planck’s constant, h = 6.6 × 10⁻³⁴ Js

Mass of an electron, m = 9.1 × 10⁻³¹ kg

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

(a) For the electron, we can write the relation for kinetic energy as:

Where,

v = Speed of the electron

Momentum of the electron, p = mv

= 9.1 × 10⁻³¹ × 6.496 × 10⁶

= 5.91 × 10⁻²⁴ kg m s⁻¹

Therefore, the momentum of the electron is 5.91 × 10⁻²⁴ kg m s⁻¹_.

(b) Speed of the electron, v = 6.496 × 10⁶ m/s

(c) De Broglie wavelength of an electron having a momentum p, is given as:

Therefore, the de Broglie wavelength of the electron is 0.112 nm.