If Delta x is the uncertainty in position and | vedclass

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

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

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

If $\Delta x$ is the uncertainty in position and $\Delta v$ is the uncertainty in velocity of a particle are equal,the correct expression for uncertainty in momentum for the same particle is

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