# The difference between the radii of

Question:

The difference between the radii of $3^{\mathrm{rd}}$ and $4^{\text {th }}$ orbits of $\mathrm{Li}^{2+}$ is $\Delta \mathrm{R}_{1}$. The difference between the radii of $3^{\mathrm{rd}}$ and $4^{\text {th }}$ orbits of $\mathrm{He}^{+}$is $\Delta \mathrm{R}_{2}$. Ratio $\Delta \mathrm{R}_{1}: \Delta \mathrm{R}_{2}$ is :

1. 8 : 3

2. 3 : 8

3. 2 : 3

4. 3 : 2

Correct Option: , 3

Solution:

(c) $r=0.529 \frac{n^{2}}{Z} \AA$

For $\mathrm{Li}^{2+}$

$\left(r_{\mathrm{Li}^{2+}}\right)_{n=4}-\left(r_{\mathrm{Li}^{2+}}\right)_{n=3}=\frac{0.529}{3}\left[4^{2}-3^{2}\right]=\Delta R_{1}$

For $\mathrm{He}^{+}$,

$\left(r_{\mathrm{He}^{+}}\right)_{n=4}-\left(r_{\mathrm{He}^{+}}\right)_{n=3}=\frac{0.529}{2}\left[4^{2}-3^{2}\right]=\Delta R_{2}$

$\frac{\Delta R_{1}}{\Delta R_{2}}=\frac{2}{3}$