A nucleus with mass number 184 initially at rest emits an α-particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the α-particle.

Solution

Physics sir · 3 steps

The daughter nucleus recoils, so the alpha cannot take all 5.5 MeV. Split Q using momentum conservation.

1. Equal momenta. The parent nucleus is at rest, so the alpha particle (mass number 4) and the daughter nucleus (mass number 184 - 4 = 180) fly apart with equal and opposite momenta p.
2. Energy goes inversely. Since K = p²/2m and p is the same for both, kinetic energy is inversely proportional to mass. So K_α : K_d = 180 : 4, and the light alpha takes almost all of Q.

   K_α = Q × (A-4)/A = 5.5 × 180/184 MeV

3. So. K_α = 990/184 ≈ 5.38 MeV. The remaining 0.12 MeV is the recoil energy of the daughter. The answer is (c).

The idea to keep. In any decay from rest, the two fragments share momentum equally, so they share the Q value in inverse ratio of their masses. The lighter one gets the bigger share.

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