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The radius of the $n$-th Bohr orbit of a hydrogen-like atom of atomic number $Z$ varies as:

A$r \propto n / Z^2$
B$r \propto Z / n^2$
C$r \propto n^2 / Z$
D$r \propto Z n^2$
Answer & Solution
Correct answer: C. $r \propto n^2 / Z$
From Bohr's derivation, $r_n = \dfrac{n^2 h^2}{4\pi^2 m k Z e^2}$, hence $r_n \propto n^2 / Z$. Therefore higher orbits are larger ($n^2$) and higher-$Z$ atoms hold their electrons in tighter orbits ($1/Z$).
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