According to Heisenberg's Uncertainty Principle, if the uncertainty in the position of a microscopic — Atomic Structure Chemistry Question

💡 Solution & Explanation

Heisenberg's Uncertainty Principle states $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$. If $\Delta x = \Delta p$, then $(\Delta p)^2 \ge \frac{h}{4\pi}$, which means $\Delta p \ge \frac{1}{2}\sqrt{\frac{h}{\pi}}$. Since $\Delta p = m\Delta v$, we have $m\Delta v \ge \frac{1}{2}\sqrt{\frac{h}{\pi}}$. Therefore, the uncertainty in velocity $\Delta v \ge \frac{1}{2m}\sqrt{\frac{h}{\pi}}$.