The radius of the second Bohr orbit, in terms of the Bohr radius, a0, in Li2+ is :

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Question

The radius of the second Bohr orbit, in terms of the Bohr radius, a₀, in Li²⁺ is :

$\frac{2 a_{0}}{3}$

$\frac{2 a_{0}}{9}$

$\frac{4 a_{0}}{9}$

$\frac{4 a_{0}}{3}$

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$r = \frac{a_{0} n^{2}}{Z}$

$For {Li}^{2 +} r = \frac{a_{0} {(2)}^{2}}{3} = \frac{4 a_{0}}{3}$