### 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

### Core Logic In basic beta^-$\beta^-$ emission processes, a neutron decays inside a nucleus to satisfy lepton numbers and conservation rules: mathrmn rightarrow mathrmp + mathrme^- + overlinenu$\mathrm{n} \rightarrow \mathrm{p} + \mathrm{e}^- + \overline{\nu}$ ### Step 1: Conservation Cross-Check Charge Balance: 0 rightarrow (+1) + (-1) + 0 = 0$0 \rightarrow (+1) + (-1) + 0 = 0$ (Conserved) Lepton Family Index: 0 rightarrow 0 + (+1) + (-1) = 0$0 \rightarrow 0 + (+1) + (-1) = 0$ (Conserved via antineutrino entry). This perfectly isolates option (1). ### Pattern Recognition Negative beta emission is always accompanied by an antineutrino, whereas positive positron transformation releases a regular neutrino molecule. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

### Related Formula - **Nuclear Fission**: Heavy nucleus splits into intermediate lighter fragments after absorbing a neutron. - **Nuclear Fusion**: Extremely light isotopes combine to form heavier nuclei. - **Q-value**: Positive for exothermic nuclear processes (releasing energy) and negative for endothermic nuclear processes (absorbing energy). ### Core Logic Let us check each reaction: - **Reaction A**: ^1mathrmn + ^235_92mathrmU rightarrow ^140_54mathrmXe + ^94_38mathrmSr + 2^1_0mathrmn${}^1\mathrm{n} + {}^{235}_{92}\mathrm{U} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + 2{}^1_0\mathrm{n}$ This is a heavy Uranium nucleus absorbing a neutron and splitting into smaller fragments. This is the definition of **Nuclear Fission** (III). - **Reaction B**: 2mathrmH_2 + mathrmO_2 rightarrow 2mathrmH_2mathrmO$2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$ This represents the combination of hydrogen and oxygen molecules to form water, which is a classic exothermic **Chemical reaction** (I). - **Reaction C**: ^2_1mathrmH + ^2_1mathrmH rightarrow ^3mathrmHe + ^1_0mathrmn${}^2_1\mathrm{H} + {}^2_1\mathrm{H} \rightarrow {}^3\mathrm{He} + {}^1_0\mathrm{n}$ Light Deuterium nuclei fuse together to form Helium-3, releasing considerable energy (Q > 0$Q > 0$). This is **Fusion with positive Q value** (II). - **Reaction D**: ^1_1mathrmH + ^3_1mathrmH rightarrow ^2_1mathrmH + ^2_1mathrmH${}^1_1\mathrm{H} + {}^3_1\mathrm{H} \rightarrow {}^2_1\mathrm{H} + {}^2_1\mathrm{H}$ Proton and Tritium reacting to form Deuteron products. Since this reaction has products with a lower binding energy than the reactants, it is an endothermic process. Hence, it is **Fusion with negative Q value** (IV). ### Step 1: Alignment Let's summarize the matches: - A rightarrow$\rightarrow$ III - B rightarrow$\rightarrow$ I - C rightarrow$\rightarrow$ II - D rightarrow$\rightarrow$ IV This perfectly corresponds to Option (2). ### Pattern Recognition Identifying chemical vs. nuclear reactions is trivial (chemical reactions involve molecular change like 2mathrmH_2 + mathrmO_2$2\mathrm{H}_2 + \mathrm{O}_2$, whereas nuclear reactions involve changes in nuclear isotopes). Always use chemical reactions to instantly lock in a match (B-I) and narrow down options! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei Class 11 Chemistry: Chemical Bonding and Molecular Structure

### Related Formula N = N_0 e^-lambda t$N = N_0 e^{-\lambda t}$ where lambda$\lambda$ is the decay constant. ### Core Logic Initially, only material P is present, so its mass decreases exponentially from a maximum value to zero. Material Q is formed from P and then decays into R, so its mass initially increases from zero, reaches a maximum, and then decreases to zero. Material R is stable and accumulated over time, so its mass increases continuously from zero and levels off at a maximum value equal to the initial mass of P. ### Step 1: Graphical Identification Looking at the options, option (2) correctly depicts the exponential decay of P, the transient rise and fall of Q, and the continuous growth of R to a stable value.

### Pattern Recognition For sequential decay P rightarrow Q rightarrow R$P \rightarrow Q \rightarrow R$, parent P$P$ always starts at max and drops to 0. Intermediate Q$Q$ starts at 0, peaks, and returns to 0. Final stable product R$R$ starts at 0 and grows asymptotically to max value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei