# What is the

Question:

What is the

(a) momentum,

(b) speed, and

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

Solution:

Kinetic energy of the electron, Ek = 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:

$E_{k}=\frac{1}{2} m v^{2}$

Where,

v = Speed of the electron

$\therefore v^{2}=\sqrt{\frac{2 e E_{k}}{m}}$

$=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 120}{9.1 \times 10^{-31}}}$

$=\sqrt{42.198 \times 10^{12}}=6.496 \times 10^{6} \mathrm{~m} / \mathrm{s}$

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:

$\lambda=\frac{h}{p}$

$=\frac{6.6 \times 10^{-34}}{5.91 \times 10^{-24}}=1.116 \times 10^{-10} \mathrm{~m}$

$=0.112 \mathrm{~nm}$

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