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

AIPMTAIPMT 2012Atoms

Solution:

Nuclear radius, $R =R_{0} A^{1 / 3}$
where $R_{0}$ is a constant and $A$ is the mass number
$\therefore \frac{R_{A} I}{R_{C} u}=\frac{(27)^{1 / 3}}{(64)^{1 / 3}}=\frac{3}{4}$
or $R_{C u}=\frac{4}{3} \times R_{A l}=\frac{4}{3} \times 3.6$ fermi $=4.8$ fermi