Question

If \(\Delta x\) is the uncertainty in position and \(\Delta v\) is the uncertainty in velocity of a particle are equal, the correct expression for uncertainty in momentum for the same particle is

• A. \(\displaystyle \frac{1}{4}\sqrt{\frac{mh}{\pi}}\)
• B. \(\displaystyle \frac{1}{3}\sqrt{\frac{mh}{2\pi}}\)
• C. \(\displaystyle \frac{1}{2}\sqrt{\frac{mh}{\pi}}\)
• D. \(\displaystyle \frac{1}{2}\sqrt{\frac{h}{m\pi}}\)

Hint:

Remember that uncertainty in momentum and velocity are related by \(\Delta p=m\Delta v\), which is very useful in Heisenberg uncertainty problems.

Answer 1

The Correct Option is C

Step 1: Recall the uncertainty principle.
Heisenberg's principle links the uncertainty in position and momentum as $\Delta x \, \Delta p \geq \dfrac{h}{4\pi}$. We will work with the limiting equality to get the minimum value.
Step 2: Write momentum in terms of velocity.
Momentum is mass times velocity, so $p = mv$, which gives $\Delta p = m\,\Delta v$. Hence $\Delta v = \dfrac{\Delta p}{m}$.
Step 3: Apply the given condition.
The problem tells us the position uncertainty equals the velocity uncertainty, so $\Delta x = \Delta v = \dfrac{\Delta p}{m}$.
Step 4: Substitute into the relation.
Replacing $\Delta x$ in the uncertainty relation, \[ \left(\frac{\Delta p}{m}\right)\Delta p = \frac{h}{4\pi} \] so that \[ \frac{(\Delta p)^2}{m} = \frac{h}{4\pi}. \]
Step 5: Solve for the momentum uncertainty.
Multiplying by $m$ gives $(\Delta p)^2 = \dfrac{mh}{4\pi}$. Taking the square root, \[ \Delta p = \sqrt{\frac{mh}{4\pi}} = \frac{1}{2}\sqrt{\frac{mh}{\pi}}. \]
Step 6: Match with the options.
This matches option (3), so the uncertainty in momentum is \[ \boxed{\dfrac{1}{2}\sqrt{\dfrac{mh}{\pi}}} \]