In a hypothetical fission reaction
₉₂ X²³⁶ → ₅₆ Y¹⁴¹+ ₃₆ Z⁹²+3 R
The identity of emitted particles (R) is :

Solution

Physics sir · 3 steps

Balance the charge numbers and the mass numbers separately, and R tells you who it is.

1. Two conservation laws. In any nuclear reaction the total charge number (bottom) and the total mass number (top) are the same on both sides. Let each R have charge number Z and mass number A.

   92 = 56 + 36 + 3Z 236 = 141 + 92 + 3A

2. Solve both. Charge: 56 + 36 = 92, so 3Z = 0 and Z = 0. Mass: 141 + 92 = 233, so 3A = 3 and A = 1.
3. So. A particle with no charge and mass number 1 is the neutron, ₀ n¹. The answer is (b).

The idea to keep. For any unknown particle in a nuclear equation, balance the bottom numbers and the top numbers separately. The pair (Z, A) names the particle at once.

Answer from the NTA final answer key. Solution by PrepWiser.