Uncertainty in the position of an electron (m | vedclass

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(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$ or $\Delta x \cdot m \cdot \Delta v = \frac{h}{4 \

Vedclass · Accepted Answer

$1.93 	imes 10^{-2} \, m$

Vedclass · Answer

(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$ or $\Delta x \cdot m \cdot \Delta v = \frac{h}{4 \pi}$.Given: $m = 9.1 	imes 10^{-31} \, kg$,$v = 300 \, ms^{-1}$,uncertainty in velocity $\Delta v = 0.001\% 	ext{ of } 300 = \frac{0.001}{100} 	imes 300 = 3 	imes 10^{-3} \, ms^{-1}$.Substituting the values: $\Delta x = \frac{h}{4 \pi m \Delta v} = \frac{6.63 	imes 10^{-34}}{4 	imes 3.14 	imes 9.1 	imes 10^{-31} 	imes 3 	imes 10^{-3}}$.$\Delta x = \frac{6.63 	imes 10^{-34}}{342.588 	imes 10^{-34}} \approx 0.0193 \, m = 1.93 	imes 10^{-2} \, m$.

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

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