Question: If the nucleus of $13Al^{27}$ has a nuclear radius of about $3.6$ fm then $52Te^{125}$ would h...

Question

If the nucleus of $13Al^{27}$ has a nuclear radius of about $3.6$ fm then $52Te^{125}$ would have its radius approximately as

Answer 1

: Here, $A_{1} = 27,A_{2} = 125$

$R_{1} = 3.6fm$

As $\frac{R_{2}}{R_{1}} = \left( \frac{A_{2}}{A_{1}} \right)^{1/3} = \left( \frac{125}{27} \right)^{1/3} = \frac{5}{3}$

$\therefore R_{2} = \frac{5}{3}R_{1} = \frac{5}{3} \times 3.6 = 6fm$

Question: If the nucleus of \(13Al^{27}\) has a nuclear radius of about \(3.6\) fm then \(52Te^{125}\) would h...