`(i)` The radioactive isotope `""_(27)^(60)Co` which is used in treatment of cancer can be mad by (n.p) reaction. What's the target nucleus for this reaction
`(A) ""_(28)^(59)Ni`
`(B) ""_(27)^(59)Co`
`(C )""_(28)^(60)Ni`
`(D) ""_(27)^(60)Co`
`(ii)` What is the product `P` in the nuclear reaction
`""_(92)^(235)U+_(0)^(1)n to ""_(27)^(60)Kr+3(""_(0)^(1)n)`

Explanation

`(i) ""_(28)^(60)Ni+_(0)^(1)n to _(27)^(60)Co+_(1)^(1)H`
`(ii)` Let symbol of element is `""_(z)^(m)P`
`92=Z+36`
`:.Z=56`
`235+1=M+92+3`
`:.M=141`
Thus element `P` will be `""_(56)^(14)Ba`