Question

The radius of the second Bohr orbit, in terms of the Bohr radius, $a_0,$ in $L{i}^{2+}$ is :

• A. $\frac{2a_0}{3}$
• B. $\frac{4a_0}{3}$
• C. $\frac{4a_0}{9}$
• D. $\frac{2a_0}{9}$

Answer 1

Answer: (B)

$r_n=\frac{n^2\times a_0}{z}$
For $2^{nd}$ Bohr orbit of $L{i}^{+2}$
$n=2$
$z=3$
$\Rightarrow r_n=\frac{2^2\times a_0}{3}=\frac{4a_0}{3}$