Energy released when two deuterons (_1mathrmH^2)$({}_{1}\mathrm{H}^{2})$ fuse to form a helium nucleus (_2mathrmHe^4)$({}_{2}\mathrm{He}^{4})$ is: (Given: Binding energy per nucleon of _1mathrmH^2 = 1.1${}_1\mathrm{H}^2 = 1.1$ MeV and binding energy per nucleon of _2mathrmHe^4 = 7.0${}_2\mathrm{He}^4 = 7.0$ MeV)

### Related Formula 1. Fusion reaction: _1mathrmH^2 + _1mathrmH^2 longrightarrow _2mathrmHe^4$${}_1\mathrm{H}^2 + {}_1\mathrm{H}^2 \longrightarrow {}_2\mathrm{He}^4$$ 2. Q-value (Energy Released) of a nuclear reaction: Q = textTotal Binding Energy (Products) - textTotal Binding Energy (Reactants)$$Q = \text{Total Binding Energy (Products)} - \text{Total Binding Energy (Reactants)}$$ ### Core Logic Let's compute the total binding energies: - **Reactants:** Two deuterons (_1mathrmH^2$({}_1\mathrm{H}^2$). - Number of nucleons in each deuteron = 2 - Binding energy per nucleon = 1.1 \ mathrmMeV$1.1 \ \mathrm{MeV}$ - Total Binding Energy of reactants: textBE_textreactants = 2 times [2 times 1.1 \ mathrmMeV] = 4.4 \ mathrmMeV$$\text{BE}_{\text{reactants}} = 2 \times [2 \times 1.1 \ \mathrm{MeV}] = 4.4 \ \mathrm{MeV}$$ - **Products:** One helium nucleus (_2mathrmHe^4$({}_2\mathrm{He}^4$). - Number of nucleons = 4 - Binding energy per nucleon = 7.0 \ mathrmMeV$7.0 \ \mathrm{MeV}$ - Total Binding Energy of products: textBE_textproducts = 4 times 7.0 \ mathrmMeV = 28.0 \ mathrmMeV$$\text{BE}_{\text{products}} = 4 \times 7.0 \ \mathrm{MeV} = 28.0 \ \mathrm{MeV}$$ ### Step 1: Calculate energy released The energy released (Q$Q$) in the fusion process is: Q = textBE_textproducts - textBE_textreactants$$Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactants}}$$ Q = 28.0 \ mathrmMeV - 4.4 \ mathrmMeV = 23.6 \ mathrmMeV$$Q = 28.0 \ \mathrm{MeV} - 4.4 \ \mathrm{MeV} = 23.6 \ \mathrm{MeV}$$ Thus, the energy released is 23.6 \ mathrmMeV$23.6 \ \mathrm{MeV}$. ### Pattern Recognition Sees: Q-value of fusion from binding energy per nucleon. Trap: Confusing "Binding Energy per nucleon" with the total binding energy of the nucleus. Always multiply by the mass number A$A$ first! Shortcut: Q = (A_textfinal times textBE_textfinal) - (A_textinitial times textBE_textinitial) = (4 times 7.0) - (2 times 2 times 1.1) = 28 - 4.4 = 23.6 \ mathrmMeV$Q = (A_{\text{final}} \times \text{BE}_{\text{final}}) - (A_{\text{initial}} \times \text{BE}_{\text{initial}}) = (4 \times 7.0) - (2 \times 2 \times 1.1) = 28 - 4.4 = 23.6 \ \mathrm{MeV}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei