Question

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

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

Answer 1

The correct option is D $\frac{4a_0}{3}$

$r=a_0\frac{n^2}{Z}\AA$
Bohr's radius of $L{i}^{2+}$ ion for n = 2
$=a_0\frac{n^2}{Z}$
$=\frac{4a_0}{3}$
Hence, option (d) is the correct answer.