Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) The density of the copper (binom6429mathrmCu)$(\binom{64}{29}\mathrm{Cu})$ nucleus is greater than that of the carbon (binom126mathrmC)$(\binom{12}{6}\mathrm{C})$ nucleus. [cite: 20] Reason (R): The nucleus of mass number A has a radius proportional to mathrmA^1/3$\mathrm{A}^{1/3}$. [cite: 21] In the light of the above statements, choose the most appropriate answer from the options given below: [cite: 22]

### Related Formula Empirical relationship for nuclear radius vs. mass number: R = R_0 A^1/3 implies R^3 propto A$$R = R_0 A^{1/3} \implies R^3 \propto A$$ ### Core Logic Set up the scaling ratio between the two nuclei: left(fracR_1R_2right)^3 = fracA_1A_2$$\left(\frac{R_1}{R_2}\right)^3 = \frac{A_1}{A_2}$$ Given parameters: A_1 = 64$A_1 = 64$, R_1 = 4.8$R_1 = 4.8$, R_2 = 4$R_2 = 4$. left(frac4.84 ight)^3 = frac64A_2 implies (1.2)^3 = frac64A_2$$\left(\frac{4.8}{4} ight)^3 = \frac{64}{A_2} \implies (1.2)^3 = \frac{64}{A_2}$$ 1.728 = frac64A_2 implies A_2 = frac641.728 = 27$$1.728 = \frac{64}{A_2} \implies A_2 = \frac{64}{1.728} = 27$$ ### Step 1: Solve for Target Target Form We are given that A_2 = frac1000x$A_2 = \frac{1000}{x}$: 27 = frac1000x implies x = frac100027 approx 37.037$$27 = \frac{1000}{x} \implies x = \frac{1000}{27} \approx 37.037$$ *Note on official key calculation path step error check:* Let's check the solution text transcription matrix values: A = frac641.44 times 1.2 = frac1000x implies x = frac144 times 1264 = 27$$A = \frac{64}{1.44 \times 1.2} = \frac{1000}{x} \implies x = \frac{144 \times 12}{64} = 27$$ Following the exact PDF text calculation step configuration: x = 27$x = 27$. ### Pattern Recognition Mass number 64 corresponds to 4^3$4^3$, and mass number 27 corresponds to 3^3$3^3$. The radii ratio scales linearly as 4.8 : 4.0 = 1.2 = 4 : 3$4.8 : 4.0 = 1.2 = 4 : 3$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

### Related Formula Q = E_textreleased = B.E._textproducts - B.E._textreactants$$Q = E_{\text{released}} = B.E._{\text{products}} - B.E._{\text{reactants}}$$ ### Core Logic Calculate total binding energies: - Reactant (Initial High Mass Nuclide): 236 times 7.6text MeV$236 \times 7.6\text{ MeV}$ - Products (Two Middle Mass Nuclides): 2 times (118 times 8.6)text MeV = 236 times 8.6text MeV$2 \times (118 \times 8.6)\text{ MeV} = 236 \times 8.6\text{ MeV}$ ### Step 1: Subtract values to find net energy Q = (236 times 8.6) - (236 times 7.6)$$Q = (236 \times 8.6) - (236 \times 7.6)$$ Q = 236 times (8.6 - 7.6) = 236 times 1 = 236text MeV$$Q = 236 \times (8.6 - 7.6) = 236 \times 1 = 236\text{ MeV}$$ ### Pattern Recognition Factoring out the total common nucleon coefficient (A = 236$A = 236$) upfront reduces arithmetic step durations down to a basic difference calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

### Related Formula The Q-value or energy released (Q$Q$) during a nuclear reaction sequence is determined from mass defect (Delta m$\Delta m$): Q = Delta m times 931.5 mathrm~MeV$$Q = \Delta m \times 931.5 \mathrm{~MeV}$$ ### Core Logic Adding the two equations together to obtain the single combined net nuclear reaction: _3mathrmLi^6 + _0mathrmn^1 + _1mathrmH^2 + _1mathrmH^3 rightarrow 2left(_2mathrmHe^4right) + _1mathrmH^3 + _0mathrmn^1$${}_{3}\mathrm{Li}^{6} + {}_{0}\mathrm{n}^{1} + {}_{1}\mathrm{H}^{2} + {}_{1}\mathrm{H}^{3} \rightarrow 2\left({}_{2}\mathrm{He}^{4}\right) + {}_{1}\mathrm{H}^{3} + {}_{0}\mathrm{n}^{1}$$ Cancelling intermediate species appearing on both sides yields: _3mathrmLi^6 + _1mathrmH^2 rightarrow 2left(_2mathrmHe^4right)$${}_{3}\mathrm{Li}^{6} + {}_{1}\mathrm{H}^{2} \rightarrow 2\left({}_{2}\mathrm{He}^{4}\right)$$ ### Step 1: Calculate Mass Defect The mass defect Delta m$\Delta m$ of this net process is: Delta m = M(mathrmLi) + M(_1mathrmH^2) - 2 M(_2mathrmHe^4)$$\Delta m = M(\mathrm{Li}) + M({}_{1}\mathrm{H}^{2}) - 2 M({}_{2}\mathrm{He}^{4})$$ Substituting the given mass profiles: Delta m = 6.01690 + 2.01471 - 2(4.00388)$$\Delta m = 6.01690 + 2.01471 - 2(4.00388)$$ Delta m = 8.03161 - 8.00776 = 0.02385 mathrm~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 times 931.5 mathrm~MeV approx 22.216 mathrm~MeV$$Q = 0.02385 \times 931.5 \mathrm{~MeV} \approx 22.216 \mathrm{~MeV}$$ Rounding off gives approximately 22.22 mathrm~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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

### Related Formula Q = Delta M c^2$$Q = \Delta M c^2$$ Q = sum K.E._textproducts$$Q = \sum K.E._{\text{products}}$$ ### Core Logic The nuclear fission reaction can be written as: (mathrmX) rightarrow (mathrmY) + (mathrmZ) + (mathrmP)$$(\mathrm{X}) \rightarrow (\mathrm{Y}) + (\mathrm{Z}) + (\mathrm{P})$$ The parent mass is M$M$. Three similar daughter nuclei are formed, each with mass approx fracM3$\approx \frac{M}{3}$. The total energy released due to the mass defect Delta M$\Delta M$ is Delta M c^2$\Delta M c^2$. This energy is equally distributed among the three identical daughter nuclei as kinetic energy (assuming parent is at rest). ### Step 1: Equate Energy Delta M c^2 = frac12 left(fracM3right) V^2 + frac12 left(fracM3right) V^2 + frac12 left(fracM3right) V^2$$\Delta M c^2 = \frac{1}{2} \left(\frac{M}{3}\right) V^2 + \frac{1}{2} \left(\frac{M}{3}\right) V^2 + \frac{1}{2} \left(\frac{M}{3}\right) V^2$$ Delta M c^2 = 3 times frac12 left(fracM3right) V^2$$\Delta M c^2 = 3 \times \frac{1}{2} \left(\frac{M}{3}\right) V^2$$ Delta M c^2 = frac12 M V^2$$\Delta M c^2 = \frac{1}{2} M V^2$$ ### Step 2: Solve for V V^2 = frac2 Delta M c^2M$$V^2 = \frac{2 \Delta M c^2}{M}$$ V = c sqrtfrac2 Delta MM$$V = c \sqrt{\frac{2 \Delta M}{M}}$$ ### Pattern Recognition Since total mass of the products is M$M$ (ignoring the tiny mass defect for kinetic energy calculations), the total kinetic energy frac12 M V^2$\frac{1}{2} M V^2$ equals the released energy Delta M c^2$\Delta M c^2$. The number of identical fragments doesn't change the velocity expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei

### Related Formula Nuclear radius is empirically related to mass number by: R = R_0 A^1/3$R = R_0 A^{1/3}$ ### Core Logic Given R_1 = fracR_22$R_1 = \frac{R_2}{2}$ where A_2 = 192$A_2 = 192$. We need to find A_1$A_1$. ### Step 1: Forming the Ratio fracR_1R_2 = left(fracA_1A_2right)^1/3$$\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$$ frac12 = left(fracA_1192right)^1/3$$\frac{1}{2} = \left(\frac{A_1}{192}\right)^{1/3}$$ ### Step 2: Cubing Both Sides left(frac12right)^3 = fracA_1192$$\left(\frac{1}{2}\right)^3 = \frac{A_1}{192}$$ frac18 = fracA_1192$$\frac{1}{8} = \frac{A_1}{192}$$ A_1 = frac1928 = 24$$A_1 = \frac{192}{8} = 24$$ ### Pattern Recognition Since R propto A^1/3$R \propto A^{1/3}$, scaling R$R$ by k$k$ means scaling A$A$ by k^3$k^3$. Half the radius (k = 1/2$k = 1/2$) means 1/8$1/8$th the mass number. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclei