Question

If the nuclear radius of ${}^{27}Al$ is $3.6$ Fermi, the approximate nuclear radius of ${}^{64}Cu$ in Fermi is

• A. $4.8$
• B. $3.6$
• C. $2.4$
• D. $1.2$

Answer 1

The correct option is A $4.8$

We know that, $r=r_0{A}^{1/3}$

So, $\frac{r_{Cu}}{r_{Al}}=\frac{A_{Cu}^{1/3}}{A_{Al}^{1/3}}=\frac{{64}^{1/3}}{{27}^{1/3}}=4/3$

or, $r_{Cu}=4/3\times 3.6=4.8$ Fermi