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**Worked Example** **Reaction**: 12C (target) + 4He (projectile) → 16O* (compound nucleus) → 12C + 4He (elastic scattering) or 16O + γ (gamma emission) **Step‑by‑Step** 1. **Identify the target and projectile**: The target nucleus is carbon‑12 (12C) and the projectile is an alpha particle (4He). 2. **Calculate the centre‑of‑mass energy**: If the alpha has kinetic energy 5 MeV in the lab frame and the target is at rest, the centre‑of‑mass energy E* ≈ 5 MeV (for light systems the difference is small). 3. **Form the compound nucleus**: The 12C and 4He fuse to form an excited 16O nucleus (16O*). The excitation energy is E* = E_cm + Q, where Q = (mass of 12C + mass of 4He – mass of 16O)c² ≈ 0 MeV for this reaction. 4. **Determine possible decay channels**: The compound nucleus can either re‑emit an alpha (elastic scattering) or emit a gamma ray and become a stable 16O nucleus. 5. **Compute cross‑section using a simple statistical formula**: σ = (πλ²) * (2J+1)/[(2j₁+1)(2j₂+1)] * (Γ_in Γ_out)/(Γ_total) - λ = h/√(2μE_cm) ≈ 1.4 fm - J, j₁, j₂ are spins (all zero for this example) - Γ_in = Γ_out = 0.5 MeV, Γ_total = 1 MeV → σ ≈ π(1.4 fm)² * (1)/(1*1) * (0.5×0.5)/(1) ≈ 3.1 mb. 6. **Interpret the result**: A cross‑section of ~3 millibarns means the reaction is relatively unlikely but measurable with a particle detector. **Conclusion**: The alpha particle collides with carbon‑12, forming a temporary 16O* nucleus, which then decays either back to alpha + carbon or by emitting a gamma ray to become stable 16O. The probability of each outcome is given by the partial widths in the cross‑section formula.