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. 5.5
B. 6
C. 7.2
D. 8.4

Answer

B

Explanation

Nuclear radius `(R ) prop A^(1//3) " 1 fermi " = 10^(-15) m`
`therefore (R_(Te))/(R_(Al)) = ((A_(Te))/(A_(Al)))^(1//3) = (125/27)^(1//3) = 5/3`
`therefore (R_(Te))/(3.6) = 5/3 therefore R_(Te) = 5/3xx3.6 = 6` fermi