Given the masses of various atomic particles
mₚ = 1.0072 u, mₙ = 1.0087 u, mₑ = 0.000548 u,
m_(v̅) = 0, m d = 2.0141 u, where p ≡ proton, n ≡ neutron,
e ≡ electron, v̅ ≡ antineutrino and d ≡ deuteron. Which of the following process is allowed by momentum and energy conservation?

Solution

Physics sir · 3 steps

For each process ask two things: is there enough mass, and can momentum balance?

1. Two tests. Energy: a free particle can only decay on its own if the total rest mass of the products is not more than its own. Momentum: look from the frame where total momentum is zero. If only one particle comes out, it must sit at rest, so its rest mass must take up all the energy exactly.

   Process	Mass before (u)	Mass after (u)	Verdict
   (a) n + n → D atom	2 × 1.0087 = 2.0174	2.0141 + 0.000548 = 2.0146	Fails: a single product at rest cannot hold the extra energy
   (b) n + p → d + γ	1.0087 + 1.0072 = 2.0159	2.0141, plus a photon	Allowed: the extra mass becomes the photon, and two products can balance momentum
   (c) p → n + e⁺ + ν̄	1.0072	1.0087 + 0.000548 = 1.0092	Fails: products are heavier than a free proton
   (d) e⁺ + e⁻ → γ	2 × 0.000548 = 0.001096	one photon	Fails: one photon always carries momentum, so it cannot have zero

2. Check option (b). The mass defect is 2.0159 - 2.0141 = 0.0018 u, which is about 0.0018 × 931.5 ≈ 1.68 MeV. The photon and the recoiling deuteron share this energy, moving in opposite directions, so energy and momentum both balance.
3. So. Only (b), n + p → d + γ, is allowed by both energy and momentum conservation.

The idea to keep. Two particles fusing into one single product almost always fails momentum or energy balance. And a free particle can never decay into products heavier than itself.

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