Question

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

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

Answer 1

Answer: (D)

Nuclear radii $R=\left(R_0\right)A^{1/3}$
Where $A$ is the mass number
$\therefore \frac{R_{Te}}{R_A}={\left(\frac{A_{Tr}}{A_{Al}}\right)}^{1/3}$
$={\left(\frac{125}{27}\right)}^{1/3}=\left(\frac{5}{3}\right)$ or,
$R_{Te}=\frac{5}{3}\times R_{Al}=\frac{5}{3}\times 3.6=6fm$.
$($ Given $R_{Al}=3.6fm)$