Question 11.13: What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with the kinetic energy of 120 eV.

p= mv
Step 1: Find the Speed of the electron.

For the electron, we can write the relation for kinetic energy as, Ek=12mv2

Where, v = Speed of the electron


Step 2: Find the momentum of the electron.
Momentum of the electron, p = mv = 9.1 x 10-31 x 6.496 x 106
= 5.91 x 10-24 kg ms
Therefore, the momentum of the electron is 5.91x 10-24 kg m s-1
Hint: \(E_{k}=\frac{1}{2}mv^{2}\)

Speed of the electron, v = 6.496 x 106 m/s [from part (a)]
Hint: \(\lambda=\frac{h}{p}\)

Step: Find de Broglie wavelength of an electron.
De Broglie wavelength of an electron having a momentum P is given as:

λ=hp=6.6×10145.91×1024=1.116×1010m=0.112  nm.

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