Question

If radius of the $^{27}_{13} Al$ nucleus is estimated to be $3.6$ fermi, then the radius of $^{125}_{52} Te$ nucleus be nearly:

• A. 6 fermi
• B. 8 fermi
• C. 4 fermi
• D. 5 fermi

Answer 1

The Correct Option is A

$R=R_{0}\left(A\right)^{1/3}$ $\frac{R_{Al}}{R_{Te}}=\frac{R_{0}\left(Al\right)^{1/3}}{R_{0}\left(A_{Te}\right)^{1/3}}$ $\frac{R_{Al}}{R_{Te}}=\frac{\left(A_{Al}\right)^{1/3}}{\left(A_{Te}\right)^{1/3}}$ $=\frac{\left(27\right)^{1/3}}{\left(125\right)^{1/3}}=\frac{3}{5}$ $\therefore R_{Te}=\frac{5}{3}\times3.6$ $R_{Te}=6$ fermi