Uncertainty in the position of an electron (m | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot m \Delta v \geq \frac{h}{4 \pi}$.Given velocity $v = 300 \ ms^{-1}$ and accuracy is $0.001 \ \%$,so uncertainty in velocity $\Delta v = 300 	imes \frac{0.001}{100} = 3 	imes 10^{-3} \ ms^{-1}$.Substituting the values: $\Delta x = \frac{h}{4 \pi m \Delta v} = \frac{6.63 	imes 10^{-34}}{4 	imes 3.1416 	imes 9.1 	imes 10^{-31} 	imes 3 	imes 10^{-3}}$.$\Delta x = \frac{6.63 	imes 10^{-34}}{34.21 	imes 10^{-33}} \approx 0.1938 	imes 10^{-1} \ m = 1.938 	imes 10^{-2} \ m$.The closest value is $1.92 	imes 10^{-2} \ m$.

Uncertainty in the position of an electron (mass $= 9.1 \times 10^{-31} \ kg$) moving with a velocity $300 \ ms^{-1}$ accurate up to $0.001 \ \%$ will be $(h = 6.63 \times 10^{-34} \ J \cdot s)$.

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