The radius of the second Bohr orbit, in terms of the Bohr radius, $a_{0}, in L i^{2 +}$ is:

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

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

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

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

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Solution

The correct option is D $\frac{4 a_{0}}{3}$
$r = a_{0} \frac{n^{2}}{Z} Å$
Bohr's radius of $L i^{2 +}$ ion for n = 2
$= a_{0} \frac{n^{2}}{Z}$
$= \frac{4 a_{0}}{3}$
Hence, option (d) is the correct answer.

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