1. Calculate Q-Value

Reaction: $^2\text{D} + ^3\text{T} \to ^4\text{He} + \text{n}$

The energy released ($Q$) is the difference between total binding energy of products and reactants.

Reactant Binding Energies (Given):

• Deuterium ($^2\text{D}$): $1.00 \text{ MeV/nucleon} \times 2 = 2.00 \text{ MeV}$
• Tritium ($^3\text{T}$): $2.80 \text{ MeV/nucleon} \times 3 = 8.40 \text{ MeV}$
• Total Initial BE $= 10.40 \text{ MeV}$

Product Binding Energies:

• Helium ($^4\text{He}$): $7.00 \text{ MeV/nucleon} \times 4 = 28.00 \text{ MeV}$
• Neutron: Free particle, BE = 0
• Total Final BE $= 28.00 \text{ MeV}$

Q-Value:

$$Q = \text{BE}_{\text{final}} – \text{BE}_{\text{initial}} = 28.00 – 10.40 = 17.6 \text{ MeV}$$

2. Energy Distribution

Assuming initial kinetic energy is negligible, the total energy $Q$ is shared as Kinetic Energy ($K$) between the products based on momentum conservation.

$$p_{\alpha} = p_n$$ $$\sqrt{2 m_{\alpha} K_{\alpha}} = \sqrt{2 m_n K_n}$$ $$m_{\alpha} K_{\alpha} = m_n K_n$$

The energy is distributed inversely proportional to mass:

$$\frac{K_n}{K_{\alpha}} = \frac{m_{\alpha}}{m_n} \approx \frac{4}{1}$$

Thus, the neutron takes $\frac{4}{5}$ of the energy, and the alpha particle takes $\frac{1}{5}$.

3. Calculation

Total Energy $Q = 17.6 \text{ MeV}$.

Neutron Energy:

$$K_n = \frac{4}{5} \times 17.6 = 0.8 \times 17.6 = 14.08 \text{ MeV}$$

Alpha Particle Energy:

$$K_{\alpha} = \frac{1}{5} \times 17.6 = 0.2 \times 17.6 = 3.52 \text{ MeV}$$