Question

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

• A. 4.8 fm
• B. 6.0 fm
• C. 9.6 fm
• D. 12.0 fm

Answer 1

Answer: (B)

6.0 fm

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