According to Heisenberg’s Uncertainty principle:

$∆\mathrm{x}.∆\mathrm{P} \ge \frac{\mathrm{h}}{4\mathrm{\pi }}$

$∆\mathrm{x}=$Uncertainty in position.

$∆\mathrm{P}=$Uncertainty in momentum$=\mathrm{m}∆\mathrm{V}$

$∆\mathrm{V}=$Uncertainty in velocity

$\mathrm{m}=9.11\times {10}^{-31} \mathrm{kg}$(mass of electron)

$\mathrm{h}=6.626\times {10}^{-34} \mathrm{Js}$

So, $\mathrm{\Delta V}\ge \frac{\mathrm{h}}{4\mathrm{\pi m\Delta x}}$ ​

$\mathrm{\Delta V}\ge \frac{6.626\times {10}^{-34}}{4\times 3.14\times 9.11\times {10}^{-31}\times {10}^{-12}}$

$\mathrm{\Delta V}\ge 5.79\times {10}^7\mathrm{m}/\mathrm{s}$