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(C) Given: $m = 9.1 	imes 10^{-28} \ g$,$v = 3 	imes 10^4 \ cm \ sec^{-1}$,$\Delta v = 0.011 \% 	ext{ of } v = \frac{0.011}{100} 	imes 3 	imes 10

Vedclass · Accepted Answer

$0.175$

Vedclass · Answer

(C) Given: $m = 9.1 	imes 10^{-28} \ g$,$v = 3 	imes 10^4 \ cm \ sec^{-1}$,$\Delta v = 0.011 \% 	ext{ of } v = \frac{0.011}{100} 	imes 3 	imes 10^4 = 3.3 \ cm \ sec^{-1}$.Using Heisenberg's uncertainty principle: $\Delta x \cdot \Delta v \geq \frac{h}{4 \pi m}$.Taking $h = 6.626 	imes 10^{-27} \ g \ cm^2 \ sec^{-1}$ and $\pi = 3.1416$:$\Delta x = \frac{h}{4 \pi m \Delta v} = \frac{6.626 	imes 10^{-27}}{4 	imes 3.1416 	imes 9.1 	imes 10^{-28} 	imes 3.3}$.$\Delta x = \frac{6.626 	imes 10^{-27}}{37.83} \approx 0.175 \ cm$.

Calculate the uncertainty in the position of an electron $(mass = 9.1 \times 10^{-28} \ g)$ moving with a velocity of $3 \times 10^4 \ cm \ sec^{-1}$,if the uncertainty in velocity is $0.011 \%$ (in $cm$)?

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