Question

The radius of aluminium nucleus (in fermi) is nearly:

• A. $1.5$
• B. $2.5$
• C. $4.5$
• D. $13.5$

Answer 1

The correct option is C $4.5$

Radius of a nucleus, $R=R_0{A}^{1/3}$
A for aluminium = 27
i.e., $A^{1/3}={27}^{1/3}=|3^3{|}^{\frac{1}{3}}=3$
$R_0=1.5\times {10}^{-15}m$
$\therefore R_{Al}=1.5\times {10}^{-15}\times 3$
$=4.5\times {10}^{-15}m$
$=4.5f$