What is the $(a)$ momentum,$(b)$ speed,and $(c)$ de Broglie wavelength of an electron with kinetic energy of $120 \ eV$?

$(a)$ The momentum $p$ is given by $p = \sqrt{2mK}$. Given $K = 120 \ eV = 120 \times 1.6 \times 10^{-19} \ J = 1.92 \times 10^{-17} \ J$ and $m = 9.11 \times 10^{-31} \ kg$. Thus,$p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.92 \times 10^{-17}} \approx 5.91 \times 10^{-24} \ kg \ m/s$.
$(b)$ The speed $v$ is given by $K = \frac{1}{2}mv^2$,so $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 1.92 \times 10^{-17}}{9.11 \times 10^{-31}}} \approx 6.5 \times 10^6 \ m/s$.
$(c)$ The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{5.91 \times 10^{-24}} \approx 1.12 \times 10^{-10} \ m = 0.112 \ nm$.