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Q. If uncertainty in position and velocity are equal, then uncertainty in momentum will be

ManipalManipal 2018

Solution:

$\Delta x=\Delta v$
$\Delta x \times \Delta p \geq \frac{h}{4 \pi}$
$\Delta x \times m \cdot \Delta v=\frac{h}{4 \pi}$
$(\Delta v)^{2}=\frac{h}{4 \pi m}(\because \Delta x=\Delta v)$
$\therefore \Delta p=m \cdot \Delta v$
$=m \sqrt{\frac{h}{4 \pi m}}=\sqrt{\frac{m h}{4 \pi}}$
$\Delta p=\frac{1}{2} \sqrt{\frac{m h}{\pi}}$