​The momentum of an electron is measured as 4.80 × $$10^{-27}$$​ kg m/s with an uncertainty of 2.4 × $$10^{-29}$$​ kg m/s. What is the minimum uncertainty in the determination of its position? ($$h$$​ = 6.63 × $$10^{-34}$$​ J s) ​

Correct option is A

​$$\begin{aligned}&\Delta x \cdot \Delta p \geq \frac{h}{4\pi} \\\\&\textbf{Where:} \\&\quad \Delta x = \text{uncertainty in position} \\&\quad \Delta p = 2.4 \times 10^{-29} \, \text{kg} \cdot \text{m/s (uncertainty in momentum)} \\&\quad h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \\\\&\textbf{Calculation:} \\&\quad \Delta x \geq \frac{6.63 \times 10^{-34}}{4\pi \times 2.4 \times 10^{-29}} \\&\quad \Delta x \geq \frac{6.63 \times 10^{-34}}{30.16 \times 10^{-29}} = \frac{6.63}{30.16} \times 10^{-5} \\&\quad \Delta x \geq 0.22 \times 10^{-5} = 2.2 \times 10^{-6} \, \text{m}\end{aligned}$$​​