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If uncertainty in position and momentum are equal, then uncertainty in velocity is
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Verified Answer
The correct answer is:
$\frac{1}{2 m} \sqrt{\frac{h}{\pi}}$
Key Idea: According to Heisenberg's uncertainty principle, it is impossible to determine simultaneously the position and momentum of a moving microscopic particle, ie,
$\begin{aligned}
& \Delta x \times \Delta p \geq \frac{h}{4 \pi} \\
& \because \Delta x \times \Delta p \geq \frac{h}{4 \pi}
\end{aligned}$
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
$\Delta E \cdot \Delta t \geq \frac{h}{4 \pi}$
$\begin{aligned}
& \Delta x \times \Delta p \geq \frac{h}{4 \pi} \\
& \because \Delta x \times \Delta p \geq \frac{h}{4 \pi}
\end{aligned}$
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
$\Delta E \cdot \Delta t \geq \frac{h}{4 \pi}$
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