Question

If the radius of a ${}_{13}^{27}Al$ nucleus is estimated to be $3.6$ fermi, find the radius of a ${}_{52}^{125}Te$ nucleus.

Answer 1

Radius of ${}_{13}^{27}Al=3.6fermi=3.6\times {10}^{-13}m$

Mass number of ${}_{13}^{23}Al=27$

Using radius, Moc-number formula,

$R=R_{\circ }{\left(A\right)}^{\frac{1}{3}}=3.6\times {10}^{-13}={\left(27\right)}^{\frac{1}{3}}{R}_{\circ }\Rightarrow R_{\circ }=1.2\times {10}^{-13}m$

Now, for ${}_{32}^{123}Te$, radius is given by,

$R=R_{\circ }{\left(A\right)}^{\frac{1}{2}}=1.2\times {10}^{-13}\times {\left(123\right)}^{\frac{1}{3}}=6\times {10}^{-13}m$

Radius of ${}_{32}^{123}Te$nucleus$=6fermi$