If uncertainty in position and momentum are equal, then uncertainty in velocity is
- $\frac{1}{2 m} \sqrt{\frac{h}{\pi}}$
- $\sqrt{\frac{h}{2 \pi}}$
- $\frac{1}{m} \sqrt{\frac{h}{\pi}}$
- $\sqrt{\frac{h}{\pi}}$
Solution
$\begin{aligned}
& \Delta x \times \Delta p \geq \frac{h}{4 \pi} \\
& \because \Delta x \times \Delta p \geq \frac{h}{4 \pi}
\end{aligned}$
Here $\Delta x=\Delta p$ and $\Delta p=m \cdot \Delta v$
$\begin{aligned}
& \therefore \Delta v^2=\frac{h}{m^2 4 \pi} \\
& \text {or } \\
& \Delta v=\frac{1}{2 m} \sqrt{\frac{h}{\pi}}
\end{aligned}$
Note: The uncertainty principle in terms of energy and time is given as
$\Delta E \cdot \Delta t \geq \frac{h}{4 \pi}$
Asked in: NEET 2008 (Screening)