Which of the following pair of nuclei are isobars of the element?

### Related Formula textNumber of Moles = fractextGiven MasstextMolar Mass$$\text{Number of Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}}$$ N = textMoles times N_A$$N = \text{Moles} \times N_A$$ E_texttotal = N times E_textper\_fission$$E_{\text{total}} = N \times E_{\text{per\_fission}}$$ ### Core Logic Given mass of U-235 = 47 \, mathrmg$47 \, \mathrm{g}$. Molar mass of U-235 = 235 \, mathrmg/mol$235 \, \mathrm{g/mol}$. Number of moles = frac47235 = frac15 text moles$\frac{47}{235} = \frac{1}{5} \text{ moles}$. ### Step 1: Calculate Total Atoms Total number of U-235 atoms (N$N$): N = frac15 times 6 times 10^23 = 1.2 times 10^23 text atoms$$N = \frac{1}{5} \times 6 \times 10^{23} = 1.2 \times 10^{23} \text{ atoms}$$ ### Step 2: Calculate Total Energy E_texttotal = 1.2 times 10^23 times 190 \, mathrmMeV$$E_{\text{total}} = 1.2 \times 10^{23} \times 190 \, \mathrm{MeV}$$ E_texttotal = 228 times 10^23 \, mathrmMeV$$E_{\text{total}} = 228 \times 10^{23} \, \mathrm{MeV}$$ Comparing with alpha times 10^23 \, mathrmMeV$\alpha \times 10^{23} \, \mathrm{MeV}$, we get alpha = 228$\alpha = 228$. ### Pattern Recognition Direct stoichiometry-like energy yield. Mass / Molar Mass -> Atoms -> Total Energy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

### Core Logic Statement I: The binding energy per nucleon does not monotonically increase with mass. It reaches a maximum around iron (A approx 56$A \approx 56$) and then gradually decreases for heavier nuclei. Hence, Statement I is false. Statement II: Nuclei inherently tend toward states of greater stability. Thus, nuclei with lower binding energy per nucleon undergo reactions (fission or fusion) to form nuclei with higher binding energy per nucleon. Hence, Statement II is true. ### Step 1: Final Conclusion Statement I is false but Statement II is true. Therefore, option (4) is correct. ### Pattern Recognition The binding energy curve is a classic plateau curve peaking at Iron-56. Statements claiming monotonic trends across all masses in nuclear physics are generally false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

### 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