Question:
The radius of the second Bohr orbit, in terms of the Bohr radius, $a_{0}$, in $\mathrm{Li}^{2+}$ is:
Correct Option: , 3
Solution:
$r=\frac{a_{0} n^{2}}{\mathrm{Z}}$
For $\mathrm{Li}^{2+}, r=\frac{a_{0}(2)^{2}}{3}=\frac{4 a_{0}}{3}$