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Q.
The radius of aluminium nucleus (in fermi) is nearly
AMUAMU 2003
Solution:
The radius $(R)$ of the nucleus of any metal is
$R=R_{0} A^{1 / 3}$
where $A$ is atomic number, and $R_{0}=1.5$ fermi
For aluminium $A=27$
$\therefore R =1.5 \times(27)^{1 / 3} $
$=1.5 \times 3=4.5$ fermi