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: Use Heisenberg uncertainty principle.
According to Heisenberg uncertainty principle,
\[ \Delta x\,\Delta p \geq \frac{h}{4\pi} \]
where \(\Delta p\) is the uncertainty in momentum.

Step 2: Relate uncertainty in momentum and velocity.
We know that
\[ \Delta p=m\Delta v \]
Given in the question,
\[ \Delta x=\Delta v \]
Substitute
\[ \Delta v=\frac{\Delta p}{m} \]
Thus,
\[ \Delta x=\frac{\Delta p}{m} \]

Step 3: Substitute into uncertainty relation.
\[ \left(\frac{\Delta p}{m}\right)\Delta p=\frac{h}{4\pi} \]
\[ \frac{(\Delta p)^2}{m}=\frac{h}{4\pi} \]
\[ (\Delta p)^2=\frac{mh}{4\pi} \]
Taking square root,
\[ \Delta p=\sqrt{\frac{mh}{4\pi}} \]
\[ \Delta p=\frac{1}{2}\sqrt{\frac{mh}{\pi}} \]

Step 4: Final conclusion.
Hence, the uncertainty in momentum is
\[ \boxed{\frac{1}{2}\sqrt{\frac{mh}{\pi}}} \]