According to Heisenberg's Uncertainty Principle, if the uncertainty in position is numerically equal — Atomic Structure Chemistry Question

💡 Solution & Explanation

$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$. Minimum uncertainty occurs when equality holds. If $\Delta x = \Delta p$, then $(\Delta p)^2 = \frac{h}{4\pi} \implies \Delta p = \frac{1}{2} \sqrt{\frac{h}{\pi}}$. Since $\Delta p = m\Delta v$, then $\Delta v = \frac{1}{2m} \sqrt{\frac{h}{\pi}}$.