Given below are two statements: Statement I : For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon. Statement II : For all elements, nuclei with less binding energy per nucleon transforms to nuclei with greater binding energy per nucleon. In the light of the above statements, choose the correct answer from the options given below:

Related Formula

where, R₀$R_0$ = empirical constant (≈ 1.2~fm$\approx 1.2\mathrm{~fm}$) A$A$ = mass number (number of nucleons)

Core Logic

Let m$m$ be the average mass of a single nucleon (proton/neutron).

• Total mass of the nucleus M ≈ A · m$M \approx A \cdot m$
• Volume of the nucleus V = (4)/(3) π R³ = (4)/(3) π (R₀ A1/3)³ = (4)/(3) π R₀³ A$V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \left(R_0 A^{1/3}\right)^3 = \frac{4}{3} \pi R_0^3 A$

Now, calculate the mass density ρ$\rho$:

Since m$m$ and R₀$R_0$ are constant parameters, the mass density is constant and independent of A$A$.

Pattern Recognition

Sees: "Nuclear mass density representation" → Highly dense, constant. Shortcut: Since volume V ∝ A$V \propto A$ and mass M ∝ A$M \propto A$, their ratio is constant. The density is on the order of 10¹⁷~kg/m³$10^{17}\mathrm{~kg/m}^3$, which is completely independent of the size of the specific nucleus. ✓

Chapter Mix

Class 12 Physics: Nuclear Physics

Core Logic

In basic β^-$\beta^-$ emission processes, a neutron decays inside a nucleus to satisfy lepton numbers and conservation rules:

Step 1: Conservation Cross-Check

Charge Balance: 0 arrow (+1) + (-1) + 0 = 0$0 \rightarrow (+1) + (-1) + 0 = 0$ (Conserved) Lepton Family Index: 0 arrow 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.

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: 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

Bohr orbit radius expression:

For a constant orbit index (n = 5$n = 5$):

where Z$Z$ is the atomic atomic number identifier.

Core Logic

Identify the atomic values:

• For Li²⁺ implies ZLi = 3$\mathrm{Li}^{2+} implies Z_{\mathrm{Li}} = 3$
• For He⁺ implies ZHe = 2$\mathrm{He}^{+} implies Z_{\mathrm{He}} = 2$

Step 1: Compute Ratio

Set up the inverse scaling ratio format:

Pattern Recognition

Since orbit numbers are identical, the radius is purely inversely proportional to the atomic nuclear charge Z$Z$.

Chapter Mix

Class 12 Physics: Atoms