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(B) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta P \geq \frac{h}{4 \pi}$.Given that the uncertainty in momentum $(\Delta P)$

Vedclass · Accepted Answer

$\Delta v \geq \frac{1}{2 m} \sqrt{\frac{h}{\pi}}$

Vedclass · Answer

(B) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta P \geq \frac{h}{4 \pi}$.Given that the uncertainty in momentum $(\Delta P)$ is equal to the uncertainty in position $(\Delta x)$,i.e.,$\Delta P = \Delta x$.Substituting $\Delta x = \Delta P$ in the uncertainty principle equation:$(\Delta P)^2 \geq \frac{h}{4 \pi}$.Since $\Delta P = m \cdot \Delta v$,we have $(m \cdot \Delta v)^2 \geq \frac{h}{4 \pi}$.$m^2 \cdot (\Delta v)^2 \geq \frac{h}{4 \pi}$.$(\Delta v)^2 \geq \frac{h}{4 \pi m^2}$.Taking the square root on both sides:$\Delta v \geq \sqrt{\frac{h}{4 \pi m^2}} = \frac{1}{2 m} \sqrt{\frac{h}{\pi}}$.

If the uncertainty in momentum and uncertainty in the position of a particle are equal,then the uncertainty in its velocity would be given by

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