Q: If the nuclear radius of Al^{27} is 3.6 fermi , the approximate nuclear radius of Cu^{64} in Fermi is

(a) 2. 4

(b) 1. 2

(c) 4.8

(d) 3.6

Ans: (c)

Sol: $\large R = R_0 A^{1/3}$

$\large \frac{R_{Cu}}{R_{Al}} = (\frac{64}{27})^{1/3}$

$\large \frac{R_{Cu}}{R_{Al}} = \frac{4}{3}$

$\large R_{Cu} = \frac{4}{3} \times 3.6 $

= 4.8