1. Tardigrade
  2. Question
  3. Physics
  4. If the nucleus 1327 A l has a nuclear radius of about 3.6 fm, then 52125 Te would have its radius approximately is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Solution:

Nuclear radii $R=\left(R_{0}\right) A^{1 / 3}$
Where $A$ is the mass number
$\therefore \frac{R_{T e}}{R_{A}}=\left(\frac{A_{T r}}{A_{A l}}\right)^{1 / 3}$
$=\left(\frac{125}{27}\right)^{1 / 3}=\left(\frac{5}{3}\right)$ or,
$R_{T e}=\frac{5}{3} \times R_{A l}=\frac{5}{3} \times 3.6=6\, fm$.
$\left(\right.$ Given $\left.R_{A l}=3.6\, fm \right)$