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  4. If the nuclear radius of 27 Al is 3.6 Fermi, the approximate nuclear radius of 64 Cu in Fermi is

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

Nuclei

Solution:

Nuclear radius $r \propto A^{1 / 3}$, where $A$ is mass number
$r =r_{0} A^{1 / 3}=r_{0}(27)^{1 / 3}=3 r_{0}$
$r_{0} =\frac{3.6}{3}=1.2 fm$
For ${ }^{64} Cu , r=r_{0} A^{1 / 3}=1.2\, fm (64)^{1 / 3}$
$=4.8\, fm$