If uncertainty in the measurement of position | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \ge \frac{h}{4 \pi}$.Given that the uncertainty in position $(\Delta x)$ is equal to the uncertainty in momentum $(\Delta p)$: $\Delta x = \Delta p$.Substituting this into the equation: $(\Delta p)^2 = \frac{h}{4 \pi}$.Therefore,$\Delta p = \sqrt{\frac{h}{4 \pi}}$.Since momentum uncertainty is related to velocity uncertainty by $\Delta p = m \Delta v$,we have $m \Delta v = \sqrt{\frac{h}{4 \pi}}$.Solving for $\Delta v$: $\Delta v = \frac{1}{m} \sqrt{\frac{h}{4 \pi}} = \frac{1}{m} \cdot \frac{1}{2} \sqrt{\frac{h}{\pi}} = \frac{1}{2m} \sqrt{\frac{h}{\pi}}$.

If uncertainty 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:

Similar Questions

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The uncertainty in the position of an electron and a helium atom is the same. If the uncertainty in the momentum of the electron is $32 \times 10^5$,then the uncertainty in the momentum of the helium atom is:

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Reason $(R)$: Subatomic particles possess both particle and wave nature.

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