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Q.

The radius of the second Bohr orbit, in terms of the Bohr radius, a₀, in ${Li}^{2 +}$ is:

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a

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

b

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

c

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

d

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

answer is D.

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Detailed Solution

$r_{2},_{{Li}^{+ 2}} = \frac{a_{0} \times 4}{z} = \frac{a_{0} \times 4}{3} = \frac{4 a_{0}}{3}$

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