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If the uncertainty in velocity is $\frac{1}{2 \mathrm{~m}} \sqrt{\frac{\mathrm{h}}{\pi}}$, then the ratio of uncertainty in position and momentum is
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Verified Answer
The correct answer is:
$1: 1$
Uncertainty Principle,
$$
\Delta \mathrm{p} \cdot \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi}
$$
or, $\Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi \cdot(\mathrm{m} \Delta \mathrm{v})}=\frac{\mathrm{h}}{4 \pi \times \mathrm{m} \times \frac{1}{2 \mathrm{~m}} \sqrt{\frac{\mathrm{h}}{\pi}}}=\frac{1}{2} \sqrt{\frac{\mathrm{h}}{\pi}}$
$\because \quad$ Ratio of uncertainty in position $(\Delta \mathrm{x})$ and momentum
$$
(\Delta \mathrm{p})=\frac{\Delta \mathrm{x}}{\Delta \mathrm{p}}=\frac{\Delta \mathrm{x}}{\mathrm{m}(\Delta \mathrm{v})}=\frac{\frac{1}{2} \sqrt{\mathrm{h} / \pi}}{\mathrm{m} \times \frac{1}{2 \mathrm{~m}} \sqrt{\frac{\mathrm{h}}{\pi}}}=\frac{1}{1}
$$
$$
\therefore \quad \Delta \mathrm{x}: \Delta \mathrm{p}=1: 1
$$
$$
\Delta \mathrm{p} \cdot \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi}
$$
or, $\Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi \cdot(\mathrm{m} \Delta \mathrm{v})}=\frac{\mathrm{h}}{4 \pi \times \mathrm{m} \times \frac{1}{2 \mathrm{~m}} \sqrt{\frac{\mathrm{h}}{\pi}}}=\frac{1}{2} \sqrt{\frac{\mathrm{h}}{\pi}}$
$\because \quad$ Ratio of uncertainty in position $(\Delta \mathrm{x})$ and momentum
$$
(\Delta \mathrm{p})=\frac{\Delta \mathrm{x}}{\Delta \mathrm{p}}=\frac{\Delta \mathrm{x}}{\mathrm{m}(\Delta \mathrm{v})}=\frac{\frac{1}{2} \sqrt{\mathrm{h} / \pi}}{\mathrm{m} \times \frac{1}{2 \mathrm{~m}} \sqrt{\frac{\mathrm{h}}{\pi}}}=\frac{1}{1}
$$
$$
\therefore \quad \Delta \mathrm{x}: \Delta \mathrm{p}=1: 1
$$
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