If the uncertainty in velocity is frac{1}{2m} | vedclass

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(C) According to Heisenberg's Uncertainty Principle: $\Delta p \cdot \Delta x = \frac{h}{4\pi}$.Given $\Delta v = \frac{1}{2m} \sqrt{\frac{h}{\pi}}$.F

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$1$

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(C) According to Heisenberg's Uncertainty Principle: $\Delta p \cdot \Delta x = \frac{h}{4\pi}$.Given $\Delta v = \frac{1}{2m} \sqrt{\frac{h}{\pi}}$.First, calculate uncertainty in position $\Delta x$:$\Delta x = \frac{h}{4\pi \cdot m \cdot \Delta v} = \frac{h}{4\pi \cdot m \cdot (\frac{1}{2m} \sqrt{\frac{h}{\pi}})} = \frac{h}{2\pi \sqrt{\frac{h}{\pi}}} = \frac{1}{2} \sqrt{\frac{h}{\pi}}$.Next, calculate uncertainty in momentum $\Delta p$:$\Delta p = m \cdot \Delta v = m \cdot (\frac{1}{2m} \sqrt{\frac{h}{\pi}}) = \frac{1}{2} \sqrt{\frac{h}{\pi}}$.Finally, the ratio of uncertainty in position to momentum is:$\frac{\Delta x}{\Delta p} = \frac{\frac{1}{2} \sqrt{\frac{h}{\pi}}}{\frac{1}{2} \sqrt{\frac{h}{\pi}}} = \frac{1}{1}$.Thus, the ratio is $1: 1$.

If the uncertainty in velocity is $\frac{1}{2m} \sqrt{\frac{h}{\pi}}$, then the ratio of uncertainty in position and momentum is (in $: 1$)

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