If radius of the `_(13)^(27) Al` necleus is estimated to be `3.6` fermi then the radius of `_(52)^(125)Te` nucleus be nearly

Options

A. 6.0 fm
B. 9.6 fm
C. 12.0 fm
D. 4.8 fm

Answer

A

Explanation

(a) If `R` is the radius of the nucleus, the corresponding volume `(4)/(3) pi R^3` has been found to be proportional to `A`.
This relationship is expressed in inverse form as
`R = R_0 A^(1//3)`
The value of `R_0` is `1.2 xx 10^-15 m`, i.e., `1.2 fm`
Therefore `= (R_(A1))/(R_(Te)) = (R_0(A_(A1))^(1//3))/(R_0(A_(Te))^(1//3))`
`(R_(A1))/(R_(Te)) = ((A_(A1))^(1//3))/((A_(Te))^(1//3)) = ((27)^(1//3))/((125)^(1//3)) = (3)/(5)`
or `R_(Te) = (3)/(5) xx R_(A1) = (3)/(5) xx 3.6 = 6 fm`.