The minimum uncertainty in the velocity of an electron whose position is known to within $\Delta x = 0.1\,nm$ is approximately
A$5.8 \times 10^7\,m\,s^{-1}$
B$5.8 \times 10^3\,m\,s^{-1}$
C$5.8 \times 10^5\,m\,s^{-1}$
D$3 \times 10^8\,m\,s^{-1}$
Answer & Solution
Correct answer: C. $5.8 \times 10^5\,m\,s^{-1}$
$\Delta p \ge \hbar/(2\Delta x) = (1.05 \times 10^{-34})/(2 \times 10^{-10}) \approx 5.3 \times 10^{-25}\,kg\,m\,s^{-1}$. $\Delta v = \Delta p/m_e = 5.3 \times 10^{-25}/9.11 \times 10^{-31} \approx 5.8 \times 10^5\,m\,s^{-1}$.
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