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(B) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given that $\Delta x = \Delta p$,we substitute this

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$\sqrt{\frac{h}{4 \pi}} 	imes \frac{1}{m}$

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(B) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given that $\Delta x = \Delta p$,we substitute this into the equation: $(\Delta p)^2 = \frac{h}{4\pi}$.Since $\Delta p = m \Delta v$,we have $(m \Delta v)^2 = \frac{h}{4\pi}$.Expanding this gives $m^2 \Delta v^2 = \frac{h}{4\pi}$.Solving for $\Delta v^2$,we get $\Delta v^2 = \frac{h}{4\pi m^2}$.Taking the square root of both sides,we obtain $\Delta v = \frac{1}{m} \sqrt{\frac{h}{4\pi}}$.

If uncertainties in the measurement of position and momentum of a microscopic object of mass $m$ are equal,then the uncertainty in the measurement of velocity is given by the expression:

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