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Question

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

Vedclass · Accepted Answer

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

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(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$.Given that the uncertainty in position and momentum are equal,i.e.,$\Delta x = \Delta p$.Substituting this into the equation: $(\Delta p)^2 = \frac{h}{4\pi}$.Taking the square root on both sides: $\Delta p = \sqrt{\frac{h}{4\pi}} = \frac{1}{2} \sqrt{\frac{h}{\pi}}$.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 $\Delta v = \frac{1}{2m} \sqrt{\frac{h}{\pi}}$.

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

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