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

Question

Dr. Abubacker · Accepted Answer

(c) $R = R_{0}(A)^{1/3}$
$\frac{R_{2}}{R_{1}} = \left( \frac{A_{2}}{A_{1}} ight)^{1/3} = \left( \frac{64}{27} ight)^{1/3} = \frac{4}{3}$
$R_{2} = 3.6 	imes \frac{4}{3} = 4.8$

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

Explanation

Reviewed by Dr. Abubacker

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