When a uranium isotope 23592U is bombarded with a neutron, it generates 8936Kr, three neutrons and :
- 9140Zr
- 10136Kr
- 10336Kr
- 14456Ba
Answer
(4) 14456Ba
Solution
23592U + 10n 8936Kr + 3 10n + AZX
Conservation of mass number: 235 + 1 = 89 + 3 + A
⇒ A = 144
Conservation of charge number: 92 + 0 = 36 + 0 + Z
⇒ Z = 56
∴ X = 14456Ba
Answer: (4)