In a beryllium atom, if $a_{0}$ be the radius of the first orbit, then the radius of the second orbit will be in general

n a_{0}

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a_{0}

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n^{2} a_{0}

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\frac{a_{0}}{n^{2}}

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Solution

The correct option is C $n^{2} a_{0}$
$r α n^{2} \Rightarrow r_{n} = n^{2} a_{0} (∵ r_{1} = a_{0})$

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