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)`
`(R_(Cu))/(R_(Al)) = ((64)/(27))^((1)/(3)) = (4)/(3)`
`(R_(Cu))/(3.6) = (4)/(3) rArr R_(Cu) = 4.8` fermi