In a hypothetical fission reaction 92X236 56Y141 + 36Z92 + 3R The identity of emitted particles (R) is :
- Proton
- Neutron
- Electron
- -radiations
Answer
(2) Neutron
Solution
Let each R be ZRA.
Mass number: 236 = 141 + 92 + 3A 3A = 3 A = 1
Charge number: 92 = 56 + 36 + 3Z Z = 0
A = 1, Z = 0 R is 0n1, a neutron
Answer: (2)