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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

Uncertainty in position and momentum are equal. Uncertainty in velocity is :-

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