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

(a)
Hint:
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:

$\begin{array}{l}\frac{1}{2}m{v}^{2}=eV\\ {v}^{2}=\frac{2eV}{m}\\ \therefore v=\sqrt{\frac{2×1.6×{10}^{-19}×56}{9.1×{10}^{-31}}}\\ =\sqrt{19.69×{10}^{12}}=4.44×{10}^{6}m/s\end{array}$

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

(b)
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:

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