Question

The ratio of the radius of second orbit of $Li ^{2+}$ to that of third orbit of $Be ^{3+}$ 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 $Li ^{2+}-n=2, Z=3$

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

Therefore, $ \frac{r_{\left( Li ^{2+}\right)}}{r_{\left( Be ^{3+}\right)}} =\left[\frac{n^{2}}{Z}\right]_{\left( Li ^{2+}\right)} \times\left[\frac{Z}{n^{2}}\right]_{\left( Be ^{3+}\right)} $

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