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

Answer: (A)

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