1.

(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)

Answer»

Solution :`(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`


Discussion

No Comment Found

Related InterviewSolutions