Related Formula
The Q-value or energy released (Q$Q$) during a nuclear reaction sequence is determined from mass defect (Δ m$\Delta m$):
Q = Δ m × 931.5 ~MeV$$Q = \Delta m \times 931.5 \mathrm{~MeV}$$
Core Logic
Adding the two equations together to obtain the single combined net nuclear reaction:
₃Li⁶ + ₀n¹ + ₁H² + ₁H³ arrow 2(₂He⁴) + ₁H³ + ₀n¹$${}_{3}\mathrm{Li}^{6} + {}_{0}\mathrm{n}^{1} + {}_{1}\mathrm{H}^{2} + {}_{1}\mathrm{H}^{3} \rightarrow 2\left({}_{2}\mathrm{He}^{4}\right) + {}_{1}\mathrm{H}^{3} + {}_{0}\mathrm{n}^{1}$$
Cancelling intermediate species appearing on both sides yields:
₃Li⁶ + ₁H² arrow 2(₂He⁴)$${}_{3}\mathrm{Li}^{6} + {}_{1}\mathrm{H}^{2} \rightarrow 2\left({}_{2}\mathrm{He}^{4}\right)$$
Step 1: Calculate Mass Defect
The mass defect Δ m$\Delta m$ of this net process is:
Δ m = M(Li) + M(₁H²) - 2 M(₂He⁴)$$\Delta m = M(\mathrm{Li}) + M({}_{1}\mathrm{H}^{2}) - 2 M({}_{2}\mathrm{He}^{4})$$
Substituting the given mass profiles:
Δ m = 6.01690 + 2.01471 - 2(4.00388)$$\Delta m = 6.01690 + 2.01471 - 2(4.00388)$$
Δ m = 8.03161 - 8.00776 = 0.02385 ~amu$$\Delta m = 8.03161 - 8.00776 = 0.02385 \mathrm{~amu}$$
Step 2: Compute Energy Released
Converting mass defect into MeV value:
Q = 0.02385 × 931.5 ~MeV ≈ 22.216 ~MeV$$Q = 0.02385 \times 931.5 \mathrm{~MeV} \approx 22.216 \mathrm{~MeV}$$
Rounding off gives approximately 22.22 ~MeV$22.22 \mathrm{~MeV}$.
Pattern Recognition
When chain equations share intermediate steps (like neutron consumption/generation or tritium tracking), add the algebraic steps together to deduce the overall net target process. This cuts out unnecessary individual constituent mass balances.
Chapter Mix
Class 12 Physics: Nuclei
