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Question: The ratio of radii of nuclei \({}_{13}^{27}Al\) and \({}_{52}^{125}Te\) is approximately...

The ratio of radii of nuclei 1327Al{}_{13}^{27}Al and 52125Te{}_{52}^{125}Te is approximately

A

6 : 10

B

13 : 52

C

40 : 177

D

14 : 7

Answer

6 : 10

Explanation

Solution

By using r∝A1/3r \propto A^{1/3} ⇒ r1r2=(A1A2)1/3=(27125)1/3=85=610\frac{r_{1}}{r_{2}} = \left( \frac{A_{1}}{A_{2}} \right)^{1/3} = \left( \frac{27}{125} \right)^{1/3} = \frac{8}{5} = \frac{6}{10}