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(A) According to Heisenberg's uncertainty principle: $\Delta x 	imes \Delta p \ge h / (4 \pi)$.Given that the uncertainty in position and momentum ar

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$1 / (2 m) \sqrt{h / \pi}$

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(A) According to Heisenberg's uncertainty principle: $\Delta x 	imes \Delta p \ge h / (4 \pi)$.Given that the uncertainty in position and momentum are equal: $\Delta x = \Delta p$.Substituting this into the equation: $(\Delta p)^2 = h / (4 \pi)$.Taking the square root on both sides: $\Delta p = \sqrt{h / (4 \pi)} = 1 / 2 \sqrt{h / \pi}$.Since $\Delta p = m \Delta v$,we have $m \Delta v = 1 / 2 \sqrt{h / \pi}$.Therefore,the uncertainty in velocity is $\Delta v = 1 / (2 m) \sqrt{h / \pi}$.

If uncertainty in position and momentum of an electron are equal,then uncertainty in its velocity is

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