If the nucleus of `._13Al^27` has a nuclear radius of about 3.6 fm, then `._52Te^125` would have its radius approximately as

Options

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

Answer

D

Explanation

Nuclear radius, `R = (R _(0)) A ^(1//3)`
Where A is the mass number:
`therefore ( R _(Te))/( R _(Al )) = ((A_(Te))/(A _(Al)))^(1//3) = ((125)/(27))^(1//3)= ((5)/(3))`
`or (R _(Te))/( R _(Al ))= 5/3 xx 3.6 = 6 fm`