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For a hydrogen-like atom of nuclear charge $Z$, the radius of the $n$th Bohr orbit scales as
A$r_n \propto nZ$
B$r_n \propto n^2/Z$
C$r_n \propto Z/n^2$
D$r_n \propto n/Z^2$
Answer & Solution
Correct answer: B. $r_n \propto n^2/Z$
$r_n = (0.529\,Å)\,n^2/Z$. So radius grows as $n^2$ but shrinks as $Z$ — explaining why He$^+$ has a smaller $n=1$ orbit than H.
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