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

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

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:

$\begin{array}{l}\lambda =\frac{h}{p}\\ =\frac{6.6\times {10}^{-14}}{5.91\times {10}^{-24}}=1.116\times {10}^{-10}m\\ =\mathrm{0.112 }n\mathrm{m.}\end{array}$

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