29082X Y e+ Z - P e- Q In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are
- 280, 81
- 286, 80
- 288, 82
- 286, 81
Answer
(4) 286, 81
Solution
-emission lowers A by 4 and Z by 2, so Y: A = 286, Z = 80.
e+ ( +) emission lowers Z by 1, so Z: A = 286, Z = 79.
- emission raises Z by 1, so P: A = 286, Z = 80.
e- ( -) emission again raises Z by 1, so Q: A = 286, Z = 81.
Answer: (4)