Question

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

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

Answer 1

Answer: (A)

$R=R_0{\left(A\right)}^{1/3}$
$\frac{R_{Al}}{R_{Te}}=\frac{R_0{\left(Al\right)}^{1/3}}{R_0{\left(A_{Te}\right)}^{1/3}}$
$\frac{R_{Al}}{R_{Te}}=\frac{{\left(A_{Al}\right)}^{1/3}}{{\left(A_{Te}\right)}^{1/3}}$
$=\frac{{\left(27\right)}^{1/3}}{{\left(125\right)}^{1/3}}=\frac{3}{5}$
$\therefore R_{Te}=\frac{5}{3}\times 3.6$
$R_{Te}=6$ fermi