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The radius of the second Bohr orbit, in terms of $a_0$, in Li$^{2+}$ is:
(1). ($\frac{2a_0}{3}$)
(2). ($\frac{4a_0}{9}$)
(3). ($\frac{4a_0}{3}$)
(4). ($\frac{2a_0}{9}$)
$r = \frac{a_0 n^2}{Z} = \frac{a_0 \times 4}{3} = \frac{4a_0}{3}$.
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