The uncertainty in the momentum of a moving electron is determined to be . What will be the minimum — Atomic Structure Chemistry Question

💡 Solution & Explanation

Applying $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$, we get $\Delta x \ge \frac{h}{4\pi \Delta p}$. Substituting the values: $\Delta x \ge \frac{6.62 \times 10^{-34}}{4 \times 3.14 \times 10^{-5}} = \frac{6.62 \times 10^{-29}}{12.56} \approx 0.527 \times 10^{-29} \text{ m}$, which simplifies to $5.27 \times 10^{-30} \text{ m}$.