Bohr's atom model

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Question

Which energy state of the triply ionized beryllium (Be⁺⁺⁺) has the same electron orbital radius as that of the ground state of hydrogen? Given Z for beryllium = 4.

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Solution

$As r_{n} \propto \frac{n^{2}}{Z} we use$
$\Rightarrow \frac{n_{Be}^{2}}{Z_{Be}} = \frac{n_{H}^{2}}{Z_{H}} \Rightarrow n_{Be}^{2} = \frac{(1)^{2}}{1} \times 4 \Rightarrow n_{Be} = 2$

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