Choose the correct nuclear process from the below options [p: proton, n: neutron, e⁻$\mathbf{e}^{-}$ : electron, e⁺$\mathrm{e}^{+}$ : positron, v: neutrino, ν$\overline{\nu}$ : antineutrino]

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: ¹n + ²³⁵₉₂U arrow ¹⁴⁰₅₄Xe + ⁹⁴₃₈Sr + 2¹₀n${}^1\mathrm{n} + {}^{235}_{92}\mathrm{U} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + 2{}^1_0\mathrm{n}$

This is a heavy Uranium nucleus absorbing a neutron and splitting into smaller fragments. This is the definition of Nuclear Fission (III).

• Reaction B: 2H₂ + O₂ arrow 2H₂O$2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$

This represents the combination of hydrogen and oxygen molecules to form water, which is a classic exothermic Chemical reaction (I).

• Reaction C: ²₁H + ²₁H arrow ³He + ¹₀n${}^2_1\mathrm{H} + {}^2_1\mathrm{H} \rightarrow {}^3\mathrm{He} + {}^1_0\mathrm{n}$

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: ¹₁H + ³₁H arrow ²₁H + ²₁H${}^1_1\mathrm{H} + {}^3_1\mathrm{H} \rightarrow {}^2_1\mathrm{H} + {}^2_1\mathrm{H}$

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 arrow$\rightarrow$ III
• B arrow$\rightarrow$ I
• C arrow$\rightarrow$ II
• D arrow$\rightarrow$ IV

This perfectly corresponds to Option (2).

Pattern Recognition

Identifying chemical vs. nuclear reactions is trivial (chemical reactions involve molecular change like 2H₂ + O₂$2\mathrm{H}_2 + \mathrm{O}_2$, whereas nuclear reactions involve changes in nuclear isotopes). Always use chemical reactions to instantly lock in a match (B-I) and narrow down options!

Chapter Mix

Class 12 Physics: Nuclei Class 11 Chemistry: Chemical Bonding and Molecular Structure

Related Formula

where λ$\lambda$ 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 arrow Q arrow 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.

Chapter Mix

Class 12 Physics: Nuclei

Related Formula

R = R₀ A1/3$R = R_0 A^{1/3}$ [cite: 667]

ρ = MassVolume = (mₙ A)/((4)/(3)π R³)$$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m_n A}{\frac{4}{3}\pi R^3}$$ [cite: 664]

Core Logic

Substituting the expression for radius R$R$ into the density equation: [cite: 664]

ρ = mₙ A(4)/(3)π (R₀ A1/3)³ = (mₙ A)/((4)/(3)π R₀³ A) = (mₙ)/((4)/(3)π R₀³)$$\rho = \frac{m_n A}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \frac{m_n A}{\frac{4}{3}\pi R_0^3 A} = \frac{m_n}{\frac{4}{3}\pi R_0^3}$$ [cite: 664]

As observed, the mass number A$A$ cancels out perfectly, implying that the density of all nuclei is roughly identical and constant regardless of their mass numbers[cite: 664, 666]. Thus, the nuclear density of copper is equal to that of carbon, meaning Assertion (A) is incorrect[cite: 20, 663]. Reason (R) is correct since R ∝ A1/3$R \propto A^{1/3}$ is a foundational empirical law of nuclear physics[cite: 21, 668].

Pattern Recognition

Nuclear mass scales with A$A$, while volume scales with R³ ∝ (A1/3)³ = A$R^3 \propto (A^{1/3})^3 = A$[cite: 664]. Therefore, Density ∝ (A)/(A) = constant$\text{Density} \propto \frac{A}{A} = \text{constant}$[cite: 664, 666]. Always look out for options asserting varying nuclear densities across heavy vs light elements—it is a common trap.

Chapter Mix

Class 12 Physics: Nuclei

Related Formula

Empirical relationship for nuclear radius vs. mass number:

Core Logic

Set up the scaling ratio between the two nuclei:

Given parameters: A₁ = 64$A_1 = 64$, R₁ = 4.8$R_1 = 4.8$, R₂ = 4$R_2 = 4$.

Step 1: Solve for Target Target Form

We are given that A₂ = (1000)/(x)$A_2 = \frac{1000}{x}$:

Note on official key calculation path step error check: Let's check the solution text transcription matrix values:

Following the exact PDF text calculation step configuration: x = 27$x = 27$.

Pattern Recognition

Mass number 64 corresponds to 4³$4^3$, and mass number 27 corresponds to 3³$3^3$. The radii ratio scales linearly as 4.8 : 4.0 = 1.2 = 4 : 3$4.8 : 4.0 = 1.2 = 4 : 3$.

Chapter Mix

Class 12 Physics: Nuclei

Related Formula

Core Logic

Calculate total binding energies:

• Reactant (Initial High Mass Nuclide): 236 × 7.6 MeV$236 \times 7.6\text{ MeV}$
• Products (Two Middle Mass Nuclides): 2 × (118 × 8.6) MeV = 236 × 8.6 MeV$2 \times (118 \times 8.6)\text{ MeV} = 236 \times 8.6\text{ MeV}$

Step 1: Subtract values to find net energy

Pattern Recognition

Factoring out the total common nucleon coefficient (A = 236$A = 236$) upfront reduces arithmetic step durations down to a basic difference calculation.

Chapter Mix

Class 12 Physics: Nuclei