If the position of the electron was measured with an accuracy of +0.002 nm . the uncertainty in the momentum of it would be (in $\left.\mathrm{kg} \mathrm{ms}^{-1} ight)\left(\mathrm{h}=6.626 imes 10^{-34} \mathrm{~J} \mathrm{~s} ight)$

Question

If the position of the electron was measured with an accuracy of +0.002 nm . the uncertainty in the momentum of it would be (in $\left.\mathrm{kg} \mathrm{ms}^{-1}ight)\left(\mathrm{h}=6.626 	imes 10^{-34} \mathrm{~J} \mathrm{~s}ight)$

Aicharya · Accepted Answer

Heisenberg uncertainty principle, $\Delta x \cdot \Delta p=\frac{h}{4 \pi}$ $\begin{aligned} & \Delta \mathrm{x}=2 	imes 10^{-3} \mathrm{~nm} \ & \Delta \mathrm{x}=2 	imes 10^{-3} 	imes 10^{-9} \mathrm{~m} \quad\left[1 \mathrm{~nm}=10^{-9} \mathrm{~m}ight] \ & 	herefore \quad \Delta \mathrm{x}=2 	imes 10^{-12} \mathrm{~m} \ & \mathrm{~h}=6.626 	imes 10^{-34} \mathrm{Js} \ & {\left[1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2}ight]} \ & 	herefore \quad \Delta x \cdot \Delta p=\frac{1}{4 \pi} \ & \Delta \mathrm{p}=\frac{\mathrm{h}}{4 \pi . \Delta \mathrm{x}}=\frac{6.626 	imes 10^{-34} \mathrm{Js}}{4 	imes 3.14 	imes 2 	imes 10^{-12} \mathrm{~m}} \ & =0.2637 	imes 10^{-22} \mathrm{~kg} \mathrm{~ms}^{-1} \ & =2.637 	imes 10^{-23} \mathrm{~kg} \mathrm{~ms}^{-1} \end{aligned}$

If the position of the electron was measured with an accuracy of +0.002 nm . the uncertainty in the momentum…

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