Calculate de Broglie wavelength of an electron travelling at 1% of the speed of light

One percent of the speed of light is

$\begin{array}{l}\mathrm{v}=\left(\frac{1}{100}\right)\left(3.00\times {10}^8{\mathrm{ms}}^{-1}\right)\\ =3.00\times {10}^6{\mathrm{ms}}^{-1}\\ \mathrm{Momentum}\mathrm{of}\mathrm{the}\mathrm{electron}\left(\mathrm{p}\right)=\mathrm{mv}\\ =\left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(3.00\times {10}^6{\mathrm{ms}}^{-1}\right)\\ =2.73\times {10}^{-24}\mathrm{kg}{\mathrm{ms}}^{-1}\\ \mathrm{The}\mathrm{de}-\mathrm{Broglie}\mathrm{wavelength}\mathrm{of}\mathrm{this}\mathrm{electron}\mathrm{is}\\ \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}=\frac{6.626\times {10}^{-34}}{2.73\times {10}^{-24}{\mathrm{kgms}}^{-1}}\\ \mathrm{\lambda }=2.424\times {10}^{-10}\mathrm{m}.\end{array}$