The uncertainty in the position of an electro | vedclass

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(B) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given velocity $v = 300 \, ms^{-1}$ and accuracy $=

Vedclass · Accepted Answer

$1.92 	imes 10^{-2} \, m$

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(B) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given velocity $v = 300 \, ms^{-1}$ and accuracy $= 0.001 \%$.Uncertainty in velocity $\Delta v = \frac{0.001}{100} 	imes 300 = 3 	imes 10^{-3} \, ms^{-1}$.Using $\Delta x = \frac{h}{4\pi m \Delta v}$:$\Delta x = \frac{6.63 	imes 10^{-34}}{4 	imes 3.14 	imes 9.1 	imes 10^{-31} 	imes 3 	imes 10^{-3}}$.$\Delta x = \frac{6.63 	imes 10^{-34}}{342.31 	imes 10^{-34}} \approx 0.01937 \, m \approx 1.92 	imes 10^{-2} \, m$.

The uncertainty in the position of an electron (mass = $9.1 \times 10^{-31} \, kg$) moving with a velocity of $300 \, ms^{-1}$ with an accuracy of $0.001 \%$ is ...... $(h = 6.63 \times 10^{-34} \, Js)$

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