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The Bohr orbit velocity of the ground-state electron in hydrogen is about
A$2.18 \times 10^{6}\,m\,s^{-1}$
B$2.18 \times 10^{8}\,m\,s^{-1}$
C$2.18 \times 10^{3}\,m\,s^{-1}$
D$3 \times 10^{8}\,m\,s^{-1}$
Answer & Solution
Correct answer: A. $2.18 \times 10^{6}\,m\,s^{-1}$
$v_n = (Z/n) \cdot (2.18 \times 10^6)\,m\,s^{-1}$. For H ($Z=1, n=1$): $2.18 \times 10^6\,m\,s^{-1}$. About 1/137 of $c$ — the fine-structure constant.
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