The uncertainty in the position of an electron is $1 \, \mathring{A}$. Find the uncertainty in its velocity.

(N/A) According to Heisenberg's uncertainty principle:
$\Delta x \cdot \Delta p \geqslant \frac{h}{4 \pi}$
Since $\Delta p = m \cdot \Delta v$,we have:
$\Delta x \cdot m \cdot \Delta v \geqslant \frac{h}{4 \pi}$
Rearranging for $\Delta v$:
$\Delta v = \frac{h}{4 \pi \cdot m \cdot \Delta x}$
Given:
$\Delta x = 1 \, \mathring{A} = 10^{-10} \, m$
$m = 9.1 \times 10^{-31} \, kg$
$h = 6.626 \times 10^{-34} \, J \cdot s$
Substituting the values:
$\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 9.1 \times 10^{-31} \times 10^{-10}}$
$\Delta v = \frac{6.626 \times 10^{-34}}{114.35 \times 10^{-41}}$
$\Delta v \approx 5.797 \times 10^5 \, m/s$