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The measurement of the electron position is associated with an uncertainty in momentum, which is equal to $1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1}$. The uncertainty in electron velocity is, (mass of an electron is $9 \times 10^{-28} \mathrm{~g}$ )
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The correct answer is:
$1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}$
$$
\begin{aligned}
& \Delta p=1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1} \\
& \text { or } \Delta p=\mathrm{m} \Delta \mathrm{v} \\
& \text { or } \Delta v=\frac{\Delta p}{\mathrm{~m}}=\frac{1 \times 10^{-18}}{9 \times 10^{-28}} \simeq 1 \times 10^9 \mathrm{~cm} / \text { second }
\end{aligned}
$$
\begin{aligned}
& \Delta p=1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1} \\
& \text { or } \Delta p=\mathrm{m} \Delta \mathrm{v} \\
& \text { or } \Delta v=\frac{\Delta p}{\mathrm{~m}}=\frac{1 \times 10^{-18}}{9 \times 10^{-28}} \simeq 1 \times 10^9 \mathrm{~cm} / \text { second }
\end{aligned}
$$
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