Uncertainty in measuring speed of a particle is numerically equal to uncertainty in measuring its po — Atomic Structure Chemistry Question

💡 Solution & Explanation

By Heisenberg's uncertainty principle, $\Delta x \cdot m\Delta v \ge \frac{h}{4\pi}$. Given $\Delta x = \Delta v$ numerically, $m(\Delta x)^2 \ge \frac{h}{4\pi} \implies (\Delta x)^2 \ge \frac{h}{4\pi m} \implies \Delta x \ge \sqrt{\frac{h}{4\pi m}}$. Thus, the value can be equal to or greater than the given expression. Therefore, correct answer is D.