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)

4.8

Answer 2

Nuclear radius, $r \propto A^{1/3}$
where A is mass number
$\frac{r_{Cu}}{r_{Al}} = \left(\frac{A_{Cu}}{A_{Al}}\right)^{1/3} = \left(\frac{64}{27}\right)^{1/3}$
$\Rightarrow r_{Cu} = 3.6 \left(\frac{4}{3}\right) = 4.8$ Fermi