The minimum uncertainty in the speed of an electron in one...

Heisenberg's uncertainty principle

$\Delta x \times {\Delta p}_{x} \ge \dfrac{h}{4\pi}$

$⇒ 2{a}_{0} \times {m \Delta v}_{x} = \dfrac{h}{4\pi}$(minimum)

$⇒ {\Delta v}_{x} = \dfrac{h}{4\pi} \times \dfrac{1}{2{a}_{0}} \times \dfrac{1}{m}$

$= \dfrac{6 . 63 \times {10}^{- 34}}{4 \times 3 . 14 \times 2 \times 52 . 9 \times {10}^{- 12} \times 9 . 1 \times {10}^{- 31}}$

$= 548273{ms}^{- 1}$

$= 548 . 273km{s}^{- 1}$

$= 548km{s}^{- 1}$