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(D) 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

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

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(D) 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 twice the uncertainty in position $(\Delta x)$,we have $\Delta p = 2\Delta x$.Substituting this into the equation: $\Delta x \cdot (2\Delta x) = \frac{h}{4\pi}$.$2(\Delta x)^2 = \frac{h}{4\pi}$.$(\Delta x)^2 = \frac{h}{8\pi}$.$\Delta x = \sqrt{\frac{h}{8\pi}} = \sqrt{\frac{h}{4 	imes 2\pi}} = \frac{1}{2}\sqrt{\frac{h}{2\pi}}$.Thus,the correct option is $D$.

If the uncertainty in momentum of a particle is twice the uncertainty in its position,what will be the uncertainty in the position of the particle?

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