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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:
$1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}$
Key Idea : $\Delta x \times \Delta p \geq \frac{h}{4 \pi}$
where, $\Delta p=$ uncertainty in momentum and $\Delta v=$ uncertainty in velocity
Here, $\Delta p=1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1}$
$\begin{aligned}
m & =9 \times 10^{-28} \mathrm{~g} \\
\Delta y & =\frac{\Delta p}{m}=\frac{1 \times 10^{-18}}{9 \times 10^{-28}} \\
& =1.1 \times 10^9 \\
& \approx 1.0 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}
\end{aligned}$
where, $\Delta p=$ uncertainty in momentum and $\Delta v=$ uncertainty in velocity
Here, $\Delta p=1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1}$
$\begin{aligned}
m & =9 \times 10^{-28} \mathrm{~g} \\
\Delta y & =\frac{\Delta p}{m}=\frac{1 \times 10^{-18}}{9 \times 10^{-28}} \\
& =1.1 \times 10^9 \\
& \approx 1.0 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}
\end{aligned}$
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