Question

If the uncertainty in the velocity of a moving electron is found to be two times the uncertainty in its position, the maximum uncertainty in its momentum will be:

• A. $\sqrt{m\hbar}$
• B. $\frac{\hbar}{2m}$
• C. $m\hbar$
• D. $\frac{m}{\hbar}$

Answer 1

Answer: (A)

$\sqrt{m\hbar}$

Answer 2

Heisenberg's Uncertainty Principle: $\Delta x \cdot \Delta p \ge \frac{\hbar}{2}$. Given $\Delta v = 2 \Delta x$ and $\Delta p = m \Delta v$. Substituting $\Delta v = \frac{\Delta p}{m}$ into the given condition: $\frac{\Delta p}{m} = 2 \Delta x \implies \Delta x = \frac{\Delta p}{2m}$. Substituting $\Delta x$ into the uncertainty principle (assuming equality): $\frac{\Delta p}{2m} \cdot \Delta p = \frac{\hbar}{2} \implies \frac{(\Delta p)^2}{2m} = \frac{\hbar}{2} \implies (\Delta p)^2 = m\hbar \implies \Delta p = \sqrt{m\hbar}$.