The satble nucleus that has a radius half that of `Fe^(56)` is

Options

A. `S^16`
B. `Na^21`
C. `10^12`
D. `10^8`

Answer

B, C

Explanation

Nuclear radius `R prop (A)^(1//3)`
It is given that `R_2 = (R_1)/(2)`
`therefore R_2/R_1 = 1/2 " but " (R_2/R_1) = (A_2/A_1)^(1//3)`
`therefore (R_2/R_1)^3 = A_2/A_1 " As " A_1 = 56` = mass number of Fe
`therefore (1/2)^3 = (A_2)/(56) therefore (A_2)/(56) = 1/8 therefore A_2 = 56/8 = 7`
`therefore` Mass number of stable nucleus is 7
`therefore` It mass be `Li^+`