The velocity of an electron is $1\%$ of the velocity of light. Calculate the de-Broglie wavelength of the electron.

The velocity of the electron is $v = (\frac{1}{100}) \times (3.00 \times 10^8 \ m \ s^{-1}) = 3.00 \times 10^6 \ m \ s^{-1}$.
The momentum of the electron is given by $p = m \times v$.
Substituting the mass of the electron $(m = 9.11 \times 10^{-31} \ kg)$:
$p = (9.11 \times 10^{-31} \ kg) \times (3.00 \times 10^6 \ m \ s^{-1}) = 2.733 \times 10^{-24} \ kg \ m \ s^{-1}$.
The de-Broglie wavelength is calculated using the formula $\lambda = \frac{h}{p}$,where $h = 6.626 \times 10^{-34} \ J \ s$.
$\lambda = \frac{6.626 \times 10^{-34} \ J \ s}{2.733 \times 10^{-24} \ kg \ m \ s^{-1}} = 2.424 \times 10^{-10} \ m$.