7.9 MeV α$\alpha$-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) \_\_\_\_ m. [ 14π ε₀ = 9 × 10⁹ Nm² / C² and electron charge = 1.6 × 10⁻¹⁹ C]$\left[\frac{1}{4\pi \epsilon_{0}} = 9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2} \text{ and electron charge } = 1.6 \times 10^{-19} \mathrm{C}\right]$

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