The radius of `""_(29) Cu^(64)` nucleus in Fermi is (given `R_(0) = 1.2 xx 10^(-15)` m )

Options

A. 4.8
B. 1.2
C. 7.7
D. 9.6

Answer

A

Explanation

Nuclear radius `R = R_(0) A^(1//3)` where `R_(0)` is a constant and A is the mass number
`R_(Cu) = (1 .2 xx 10^(-15) m) (64)^(1//3)`
= `(1.2 xx 10^(-15) m) (4^(3))^(1//3)`
`= 4.8 xx 10^(-15) m = 4.8 fm (because 1 fm = 10^(-15) m)`