If the nuclear radius of `.^27 A1` is `3.6` Fermi, the approximate nuclear radius of `64 Cu` in Fermi is :

Options

A. 2.4
B. 1.2
C. 4.8
D. 3.6

Answer

C

Explanation

`R= R_(0) A^((1)/(3)) rArr R prop A^((1)/(3))`
`(R_(Al))/(R_(Cu)) = ((27)/(64))^((1)/(3)) = (3)/(4) rArr R_(Cu) = (4)/(3) xx 3.6 = 4.8`