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If the nuclear radius of $^{27}Al$ is 3.6 Fermi, the approximate nuclear radius of $^{64}Cu$ in Fermi is:
A
3.6
B
2.4
C
1.2
D
4.8
Detailed Solution
$R = R_0A^{1/3}$, so $R \propto A^{1/3}$
$\frac{R_{Al}}{R_{Cu}} = \left(\frac{27}{64}\right)^{1/3} = \frac{3}{4}$
$\frac{3.6}{R_{Cu}} = \frac{3}{4} \Rightarrow R_{Cu} = 4.8$ Fermi
$\frac{R_{Al}}{R_{Cu}} = \left(\frac{27}{64}\right)^{1/3} = \frac{3}{4}$
$\frac{3.6}{R_{Cu}} = \frac{3}{4} \Rightarrow R_{Cu} = 4.8$ Fermi