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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: $m_e=9.1 \times 10^{-31} \mathrm{~kg}$ )
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(given: $m_e=9.1 \times 10^{-31} \mathrm{~kg}$ )
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Verified Answer
The correct answer is:
$5.79 \times 10^{-4} \mathrm{~m}$
$\Delta x \cdot \Delta \mathrm{P}=\Delta \mathrm{x} \cdot \mathrm{m} \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi}=\frac{6.626 \times 10^{-34}}{4 \times 3.14}$
$\begin{aligned} & =5.27 \times 10^{-35} \\ & \Rightarrow \Delta x=\frac{5.27 \times 10^{-35}}{\left(9.1 \times 10^{-31}\right)(0.1)}=5.79 \times 10^{-4} \mathrm{~m}\end{aligned}$
$\begin{aligned} & =5.27 \times 10^{-35} \\ & \Rightarrow \Delta x=\frac{5.27 \times 10^{-35}}{\left(9.1 \times 10^{-31}\right)(0.1)}=5.79 \times 10^{-4} \mathrm{~m}\end{aligned}$
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