Question

The ratio of the radii of nuclei of ${}_{13}A{l}^{27}$ and ${}_{52}T{e}^{125}$ is approximately :

• A. 6: 10
• B. 13:52
• C. 40:177
• D. 14: 73

Answer 1

The correct option is A 6: 10

As we know, $r=1.33\times {10}^{-13}\times A^{1/3}cm$, where A is atomic mass.

$\therefore \frac{r_{Al}}{r_{Te}}={\left[\frac{27}{125}\right]}^{1/3}=\frac{3}{5}=0.6$