Question 11.12: Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

At equilibrium, the kinetic energy of each electron is equal to the accelerating potential.
Step 1: Find the velocity of the electron.
we can write the relation for velocity (v) of each electron as:


Step 2: Find the momentum of the electron.
The momentum of each accelerated electron is given as:
p= mv = 9.1 x 10-31 x 4.44 x 106 = 4.04 x 10-24 ms-1

Hint: \(\lambda=\frac{12.27}{\sqrt{V}}\)

Step: Find the De Broglie wavelength of an electron.
De Broglie wavelength of an electron accelerating through a potential V is given by the relation:

λ=12.27VA=12.2756×1010m=0.1639  nm.

Therefore, the de Broglie wavelength of each electron is 0.1639 nm.