If the uncertainty in velocity of electron $(\Delta \mathrm{v})$ is $0.1 \mathrm{~m} / \mathrm{s}$, the uncertainty in its position $(\Delta x)$ is (given: $m_e=9.1 imes 10^{-31} \mathrm{~kg}$ )

Question

If the uncertainty in velocity of electron $(\Delta \mathrm{v})$ is $0.1 \mathrm{~m} / \mathrm{s}$, the uncertainty in its position $(\Delta x)$ is (given: $m_e=9.1 	imes 10^{-31} \mathrm{~kg}$ )

Aicharya · Accepted Answer

$\Delta x \cdot \Delta \mathrm{P}=\Delta \mathrm{x} \cdot \mathrm{m} \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi}=\frac{6.626 	imes 10^{-34}}{4 	imes 3.14}$ $\begin{aligned} & =5.27 	imes 10^{-35} \ & \Rightarrow \Delta x=\frac{5.27 	imes 10^{-35}}{\left(9.1 	imes 10^{-31}ight)(0.1)}=5.79 	imes 10^{-4} \mathrm{~m}\end{aligned}$

If the uncertainty in velocity of electron $(\Delta \mathrm{v})$ is $0.1 \mathrm{~m} / \mathrm{s}$, the…

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