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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) 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 momentum $(\Delta p)$,i.e.,$\Delta x = \Delta p$.Substituting $\Delta x$ with $\Delta p$ in the equation: $(\Delta p)^2 = \frac{h}{4\pi}$.Taking the square root on both sides: $\Delta p = \sqrt{\frac{h}{4\pi}} = \frac{1}{2}\sqrt{\frac{h}{\pi}}$.

If the uncertainty in position and momentum of a particle are equal,then what will be the uncertainty in the momentum of the particle?

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