$A$ microscope using suitable photons is employed to locate an electron in an atom within a distance of $0.1 \, \mathring{A}$. What is the uncertainty involved in the measurement of its velocity?

(N/A) According to Heisenberg's uncertainty principle: $\Delta x \Delta p = \frac{h}{4 \pi}$ or $\Delta x \, m \Delta v = \frac{h}{4 \pi}$.
Rearranging for velocity uncertainty: $\Delta v = \frac{h}{4 \pi \Delta x m}$.
Given values: $h = 6.626 \times 10^{-34} \, J \, s$,$\Delta x = 0.1 \, \mathring{A} = 0.1 \times 10^{-10} \, m$,$m = 9.11 \times 10^{-31} \, kg$.
Substituting the values: $\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 0.1 \times 10^{-10} \times 9.11 \times 10^{-31}}$.
Calculation: $\Delta v = 0.579 \times 10^{7} \, m \, s^{-1} = 5.79 \times 10^{6} \, m \, s^{-1}$.