The position of a particle is measured with an uncertainty of 6.6 × $$10^{-10}$$​ m. What is the minimum uncertainty in any simultaneous measurement of the momentum of the particle? ($$h$$​ = 6.63 × $$10^{-34}$$​ J s)

Correct option is A

​$$\begin{aligned}&\text{Given:} \\&\bullet \ \text{Position uncertainty of particle:} \quad \Delta x = 6.6 \times 10^{-10} \, m \\&\bullet \ \text{Planck's constant:} \quad h = 6.63 \times 10^{-34} \, Js \\[10pt]&\text{Heisenberg's uncertainty principle states:} \\&\Delta x \times \Delta p_x = \frac{h}{4 \pi}\end{aligned}$$

$$\begin{aligned}&\Delta p_x = \frac{h}{4 \pi \Delta x} \\[10pt]&\Delta p_x = \frac{6.63 \times 10^{-34}}{4 \pi \times 6.6 \times 10^{-10}} = 0.8 \times 10^{-25} \, kg \cdot m/s\end{aligned}$$​​​