Find the uncertainty in the position of an el | vedclass

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(D) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$,where $\Delta p = m \Delta v$. Thus,$\Delta x \geq

Vedclass · Accepted Answer

$12.11 \ mm$

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(D) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$,where $\Delta p = m \Delta v$. Thus,$\Delta x \geq \frac{h}{4 \pi m \Delta v}$. Given velocity $v = 2.99 	imes 10^4 \ cm \ s^{-1}$ and accuracy $0.0016 \%$. Uncertainty in velocity $\Delta v = v 	imes \frac{0.0016}{100} = 2.99 	imes 10^4 	imes 1.6 	imes 10^{-5} = 0.4784 \ cm \ s^{-1}$. Substituting the values: $\Delta x = \frac{6.626 	imes 10^{-27}}{4 	imes 3.1416 	imes 9.1 	imes 10^{-28} 	imes 0.4784}$. $\Delta x = \frac{6.626 	imes 10^{-27}}{5.475 	imes 10^{-27}} \approx 1.21 \ cm$. Converting to $mm$: $1.21 \ cm = 12.1 \ mm$.

Find the uncertainty in the position of an electron which is moving with a velocity of $2.99 \times 10^4 \ cm \ s^{-1}$,accurate up to $0.0016 \%$. (Given,$m_e = 9.1 \times 10^{-28} \ g, h = 6.626 \times 10^{-27} \ erg \cdot s$)

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