alleryThe velocity of the electron in the $n$th Bohr orbit of a hydrogen-like atom is proportional to
A$n^2/Z$
B$Z^2/n^2$
C$Z/n$
D$n/Z$
Answer & Solution
Correct answer: C. $Z/n$
From $L = mvr = n\hbar$ and $r \propto n^2/Z$: $v = n\hbar/(mr) \propto Z/n$.
Ground-state H electron travels at about $c/137$ — the fine structure constant — much slower than light, justifying non-relativistic Bohr.
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