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If a proton is accelerated to a velocity of $3 \times 10^7 \mathrm{~ms}^{-1}$ which is accurate up to $\pm 0.5 \%$, then the uncertainty in its position will be $\ldots \ldots \ldots$ [mass of proton $=1.66 \times 10^{-27} \mathrm{~kg}$, $\left.h=6.6 \times 10^{-34} \mathrm{Js}\right]$
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$2.11 \times 10^{-13} \mathrm{~m}$
$\begin{aligned} \Delta v & =\frac{0.5}{100} \times 3 \times 10^7 \mathrm{~ms}^{-1} \\ & =1.5 \times 10^5 \mathrm{~ms}^{-1} \\ \Delta x & =\frac{h}{4 \pi m \Delta v} \\ & =\frac{6.62 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}}{4 \times 3.14 \times 1.66 \times 10^{-27} \mathrm{~kg} \times 1.5 \times 10^5 \mathrm{~ms}^{-1}} \\ & =2.11 \times 10^{-13} \mathrm{~m} .\end{aligned}$
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