PHXII13:NUCLEI
363872
If the nucleus \({}_{13}^{27}Al\) has a nuclear radius of about \(3.6\,fm\), then \({}_{52}^{125}Te\) would have its radius approximately as
1 \(6.0\,fm\)
2 \(12.0\,fm\)
3 \(9.6\,fm\)
4 \(4.8\,fm\)
Explanation:
\(R = {R_0}{A^{1/3}}\)
Therefore, \(\frac{{{R_{Al}}}}{{{R_{Te}}}} = \frac{{{R_0}{{({A_{Al}})}^{1/3}}}}{{{R_0}{{({A_{Te}})}^{1/3}}}}\)
\(\frac{{{R_{Al}}}}{{{R_{Te}}}} = \frac{{{{({A_{Al}})}^{1/3}}}}{{{{({A_{Te}})}^{1/3}}}} = \frac{{{{(27)}^{1/3}}}}{{{{(125)}^{1/3}}}} = \frac{3}{5}\)
\( \Rightarrow {R_{Te}} = \frac{5}{3} \times {R_{Al}} = \frac{5}{3} \times 3.6 = 6\,fm\)