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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

VITEEEVITEEE 2016

Solution:

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