If uncertainty in position and momentum of a particle is numerically equal, then the minimum uncerta — Atomic Structure Chemistry Question

💡 Solution & Explanation

Given $\Delta x = \Delta p$. The minimum uncertainty principle is $\Delta x \cdot \Delta p = \frac{h}{4\pi}$. Substituting $\Delta x$ gives $(\Delta p)^2 = \frac{h}{4\pi} \implies \Delta p = \frac{1}{2}\sqrt{\frac{h}{\pi}}$. Since $\Delta p = m \Delta v$, the minimum uncertainty in speed is $\Delta v = \frac{\Delta p}{m} = \frac{1}{2m}\sqrt{\frac{h}{\pi}}$. Therefore, correct answer is B.