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

Options

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

Answer

D

Explanation

`R_(Al)=R_(0)(27)^(1//2)`
`R_(Al) =3R_(0)`…………(i)
m `R_(Cu)=R_(0)(64)^(1//3)=4R_(0)` …………….(ii)
Divide equation (i) by (ii)
`R_(Cu)=4/3 R_(Al) = 4/3 xx 3.6` fermi
`R_(Cu) = 4.8` fermi