Uncertainty in the position of an electron (m | vedclass

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Question

(A) The percentage error in velocity is given as $0.001\%$.$\frac{\Delta V}{V} 	imes 100 = 0.001$$\Delta V = \frac{0.001 	imes 300}{100} = 3 	imes

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The percentage error in velocity is given as $0.001\%$.$\frac{\Delta V}{V} 	imes 100 = 0.001$$\Delta V = \frac{0.001 	imes 300}{100} = 3 	imes 10^{-3} \ ms^{-1}$According to Heisenberg's uncertainty principle:$\Delta x \cdot m \Delta V \geq \frac{h}{4 \pi}$$\Delta x = \frac{h}{4 \pi m \Delta V}$Substituting the values ($h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 9.1 	imes 10^{-31} \ kg$,$\Delta V = 3 	imes 10^{-3} \ ms^{-1}$):$\Delta x = \frac{6.63 	imes 10^{-34}}{4 	imes 3.14159 	imes 9.1 	imes 10^{-31} 	imes 3 	imes 10^{-3}}$$\Delta x \approx 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:

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