If radius of the `._13^27 A1` nucleus is taken to be `R_(A1)` then the radius of `._53^125 Te` nucleus is nearly.

Options

A. `((53)/(13))^(1//3) R_(A1)`
B. `(5)/(3) R_(A1)`
C. `(3)/(5) R_(A1)`
D. `((13)/(53))^(1//3) R_(A1)`

Answer

B

Explanation

(b) As we know, `R = R_0 (A)^(1//3)`
Where `A` = mass number
`R_(A1) = R_0 (27)^(1//3) = 3 R_0`
`R_(Te) = R_0 (125)^(1//3) = 5 R_0 = (5)/(3) R_(A1)`.