Question

The ratio of the radius of second orbit of $L{i}^{2+}$ to that of third orbit of $B{e}^{3+}$ is

• A. $\frac{9}{8}$
• B. $\frac{8}{9}$
• C. $\frac{27}{16}$
• D. $\frac{16}{27}$

Answer 1

Answer: (D)

Radius of an electron $\left(r_n\right)$ in any orbit can be calculated as follows :

$r_n=\frac{0.52\times 10^{-10}{n}^2}{Z}m$

where, $n=$ number of orbit,

$Z=$ atomic number

For $L{i}^{2+}-n=2,Z=3$

For $B{e}^{3+}-n=3,Z=4$

Therefore, $\frac{r_{\left(L{i}^{2+}\right)}}{r_{\left(B{e}^{3+}\right)}}={\left[\frac{n^2}{Z}\right]}_{\left(L{i}^{2+}\right)}\times {\left[\frac{Z}{n^2}\right]}_{\left(B{e}^{3+}\right)}$

$=\left[\frac{2^2}{3}\right]\times \left[\frac{4}{3^2}\right]=\frac{16}{27}$