`""_(27)Co^(60)` is radioactive because

Options

A. its atomic number is high
B. it has high p/n ratio
C. it has high n/p ratio
D. None of these

Answer

C

Explanation

In `""_(27)Co^(60)=27, n=60 -27=33`
here , n/p =33/27=1.22
For a stable nucleide n/p ratio should be 1 (or slightly greater ) for lighter elements.