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.

(a)
Hint:
p= mv
Step 1: Find the Speed of the electron.

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

$\begin{array}{l}\therefore {v}^{2}=\sqrt{\frac{2e{E}_{k}}{m}}\\ =\sqrt{\frac{2×1.6×{10}^{-19}×120}{9.1×{10}^{-31}}}\\ =\sqrt{42.198×{10}^{12}}=6.496×{10}^{6}m/s\end{array}$

Step 2: Find the momentum of the electron.
Momentum of the electron, p = mv = 9.1 x ${10}^{-31}$ x 6.496 x ${10}^{6}$
= 5.91 x ${10}^{-24}$ kg ms
Therefore, the momentum of the electron is 5.91x ${10}^{-24}$ kg m ${\mathrm{s}}^{-1}$
(b)
Hint: $$E_{k}=\frac{1}{2}mv^{2}$$

Speed of the electron, v = 6.496 x ${10}^{6}$ m/s [from part (a)]
(c)
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:

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