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(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geqslant \frac{h}{4\pi}$.Given that the uncertainty in position $(\Delt

Vedclass · Accepted Answer

$\frac{1}{2m}\sqrt{\frac{h}{\pi}}$

Vedclass · Answer

(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geqslant \frac{h}{4\pi}$.Given that the uncertainty in position $(\Delta x)$ is equal to the uncertainty in momentum $(\Delta p)$,we have $\Delta x = \Delta p$.Substituting this into the equation: $(\Delta p)^2 = \frac{h}{4\pi}$.Taking the square root: $\Delta p = \frac{1}{2}\sqrt{\frac{h}{\pi}}$.Since $\Delta p = m \cdot \Delta v$,we have $m \cdot \Delta v = \frac{1}{2}\sqrt{\frac{h}{\pi}}$.Therefore,the uncertainty in velocity is $\Delta v = \frac{1}{2m}\sqrt{\frac{h}{\pi}}$.

If the uncertainty in position and momentum are equal,then the uncertainty in velocity is .......

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