Atoms and Nuclei Previous Year Questions — JEE Main Physics
23 past-year Atoms and Nuclei questions from JEE Main (Physics).
Q12 (2025)
Energy released when two deuterons $({}_{1}\mathrm{H}^{2})$ fuse to form a helium nucleus $({}_{2}\mathrm{He}^{4})$ is:
(Given: Binding energy per nucleon of ${}_1\mathrm{H}^2 = 1.1$ MeV and binding energy per nucleon of ${}_2\mathrm{He}^4 = 7.0$ MeV)
- $8.1 \ \mathrm{MeV}$
- $5.9 \ \mathrm{MeV}$
- $23.6 \ \mathrm{MeV}$
- $26.8 \ \mathrm{MeV}$
### Related Formula
1. Fusion reaction:
$${}_1\mathrm{H}^2 + {}_1\mathrm{H}^2 \longrightarrow {}_2\mathrm{He}^4$$
2. Q-value (Energy Released) of a nuclear reaction:
$$Q = \text{Total Binding Energy (Products)} - \text{Total Binding Energy (Reactants)}$$
### Core Logic
Let's compute the total binding energies:
- **Reactants:** Two deuterons $({}_1\mathrm{H}^2$).
- Number of nucleons in each deuteron = 2
- Binding energy per nucleon = $1.1 \ \mathrm{MeV}$
- Total Binding Energy of reactants:
$$\text{BE}_{\text{reactants}} = 2 \times [2 \times 1.1 \ \mathrm{MeV}] = 4.4 \ \mathrm{MeV}$$
- **Products:** One helium nucleus $({}_2\mathrm{He}^4$).
- Number of nucleons = 4
- Binding energy per nucleon = $7.0 \ \mathrm{MeV}$
- Total Binding Energy of products:
$$\text{BE}_{\text{products}} = 4 \times 7.0 \ \mathrm{MeV} = 28.0 \ \mathrm{MeV}$$
### Step 1: Calculate energy released
The energy released ($Q$) in the fusion process is:
$$Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactants}}$$
$$Q = 28.0 \ \mathrm{MeV} - 4.4 \ \mathrm{MeV} = 23.6 \ \mathrm{MeV}$$
Thus, the energy released is $23.6 \ \mathrm{MeV}$.
### 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 $A$ first!
Shortcut: $Q = (A_{\text{final}} \times \text{BE}_{\text{final}}) - (A_{\text{initial}} \times \text{BE}_{\text{initial}}) = (4 \times 7.0) - (2 \times 2 \times 1.1) = 28 - 4.4 = 23.6 \ \mathrm{MeV}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q15 (2025)
Considering Bohr's atomic model for hydrogen atom:
(A) the energy of H atom in ground state is same as energy of $\mathrm{He}^+$ ion in its first excited state.
(B) the energy of H atom in ground state is same as that for $\mathrm{Li}^{++}$ ion in its second excited state.
(C) the energy of H atom in its ground state is same as that of $\mathrm{He}^+$ ion for its ground state.
(D) the energy of $\mathrm{He}^+$ ion in its first excited state is same as that for $\mathrm{Li}^{++}$ in its ground state
Choose the correct answer from the options given below:
- (B), (D) only
- (A), (B) only
- (A), (D) only
- (A), (C) only
### Related Formula
$$E_n = -13.6 \frac{Z^2}{n^2}\mathrm{~eV}$$
where $Z$ is the atomic number and $n$ is the principal quantum number.
### Core Logic
Let's check each statement by calculating energy values:
- **H atom ground state ($Z = 1, n = 1$):**
$$E_H = -13.6 \frac{1^2}{1^2} = -13.6\mathrm{~eV}$$
- **Statement (A):** $\mathrm{He}^+$ ion in its first excited state ($Z = 2, n = 2$):
$$E_{\mathrm{He}^+, n=2} = -13.6 \frac{2^2}{2^2} = -13.6\mathrm{~eV}$$
This is equal to $E_H$. Thus, **Statement (A) is correct**.
- **Statement (B):** $\mathrm{Li}^{++}$ ion in its second excited state ($Z = 3, n = 3$):
$$E_{\mathrm{Li}^{++}, n=3} = -13.6 \frac{3^2}{3^2} = -13.6\mathrm{~eV}$$
This is equal to $E_H$. Thus, **Statement (B) is correct**.
- **Statement (C):** $\mathrm{He}^+$ ion in its ground state ($Z = 2, n = 1$):
$$E_{\mathrm{He}^+, n=1} = -13.6 \frac{2^2}{1^2} = -54.4\mathrm{~eV} \neq E_H$$
Thus, **Statement (C) is incorrect**.
- **Statement (D):** $\mathrm{He}^+$ in first excited state ($Z=2, n=2$) versus $\mathrm{Li}^{++}$ in ground state ($Z=3, n=1$):
$$E_{\mathrm{Li}^{++}, n=1} = -13.6 \frac{3^2}{1^2} = -122.4\mathrm{~eV} \neq -13.6\mathrm{~eV}$$
Thus, **Statement (D) is incorrect**.
### Step 1: Final Conclusion
Statements (A) and (B) are the only correct statements.
### Pattern Recognition
Bohr energy is proportional to $\frac{Z^2}{n^2}$. If the ratio of $\frac{Z}{n}$ is the same for two systems, their quantum state energy levels will be identical.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q19 (2025)
For a nucleus of mass number $A$ and radius $R$, the mass density of nucleus can be represented as:
- $A^3$
- $A^{\frac{1}{3}}$
- $A^{\frac{2}{3}}$
- $\text{Independent of }A$
### Related Formula
$$R = R_0 A^{\frac{1}{3}}$$
$$\rho = \frac{\text{Mass of Nucleus}}{\text{Volume of Nucleus}}$$
where,
$R_0$ = empirical constant ($\approx 1.2\mathrm{~fm}$)
$A$ = mass number (number of nucleons)
### Core Logic
Let $m$ be the average mass of a single nucleon (proton/neutron).
- Total mass of the nucleus $M \approx A \cdot m$
- Volume of the nucleus $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \left(R_0 A^{1/3}\right)^3 = \frac{4}{3} \pi R_0^3 A$
Now, calculate the mass density $\rho$:
$$\rho = \frac{M}{V} = \frac{A \cdot m}{\frac{4}{3} \pi R_0^3 A} = \frac{3 m}{4\pi R_0^3}$$
Since $m$ and $R_0$ are constant parameters, the mass density is constant and **independent of $A$**.
### Pattern Recognition
Sees: "Nuclear mass density representation" → Highly dense, constant.
Shortcut: Since volume $V \propto A$ and mass $M \propto A$, their ratio is constant. The density is on the order of $10^{17}\mathrm{~kg/m}^3$, which is completely independent of the size of the specific nucleus. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclear Physics
Q12 (2025)
Choose the correct nuclear process from the below options
[p: proton, n: neutron, $\mathbf{e}^{-}$ : electron, $\mathrm{e}^{+}$ : positron, v: neutrino, $\overline{\nu}$ : antineutrino]
- $\mathrm{n}\rightarrow \mathrm{p} + \mathrm{e}^{-} + \overline{\mathrm{v}}$
- $\mathfrak{n}\to \mathfrak{p} + \mathfrak{e}^{-} + \mathfrak{v}$
- $\mathrm{n}\rightarrow \mathrm{p} + \mathrm{e}^{+} + \overline{\mathrm{v}}$
- $\mathrm{n}\rightarrow \mathrm{p} + \mathrm{e}^{+} + \mathrm{v}$
### Core Logic
In basic $\beta^-$ emission processes, a neutron decays inside a nucleus to satisfy lepton numbers and conservation rules:
$$\mathrm{n} \rightarrow \mathrm{p} + \mathrm{e}^- + \overline{\nu}$$
### Step 1: Conservation Cross-Check
Charge Balance: $0 \rightarrow (+1) + (-1) + 0 = 0$ (Conserved)
Lepton Family Index: $0 \rightarrow 0 + (+1) + (-1) = 0$ (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
Q13 (2025)
Match the LIST-I with LIST-II
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">LIST-I</th><th style="border: 1px solid #888; padding: 8px;">LIST-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. ${}^1\mathrm{n} + {}^{235}_{92}\mathrm{U} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + 2{}^1_0\mathrm{n}$</td><td style="border: 1px solid #888; padding: 8px;">I. Chemical reaction</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. $2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$</td><td style="border: 1px solid #888; padding: 8px;">II. Fusion with $+\mathrm{ve}$ Q value</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. ${}^2_1\mathrm{H} + {}^2_1\mathrm{H} \rightarrow {}^3\mathrm{He} + {}^1_0\mathrm{n}$</td><td style="border: 1px solid #888; padding: 8px;">III. Fission</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. ${}^1_1\mathrm{H} + {}^3_1\mathrm{H} \rightarrow {}^2_1\mathrm{H} + {}^2_1\mathrm{H}$</td><td style="border: 1px solid #888; padding: 8px;">IV. Fusion with $-\mathrm{ve}$ Q value</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- A-II, B-I, C-III, D-IV
- A-III, B-I, C-II, D-IV
- A-II, B-I, C-IV, D-III
- A-III, B-I, C-IV, D-II
### 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**: ${}^1\mathrm{n} + {}^{235}_{92}\mathrm{U} \rightarrow {}^{140}_{54}\mathrm{Xe} + {}^{94}_{38}\mathrm{Sr} + 2{}^1_0\mathrm{n}$
This is a heavy Uranium nucleus absorbing a neutron and splitting into smaller fragments. This is the definition of **Nuclear Fission** (III).
- **Reaction B**: $2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$
This represents the combination of hydrogen and oxygen molecules to form water, which is a classic exothermic **Chemical reaction** (I).
- **Reaction C**: ${}^2_1\mathrm{H} + {}^2_1\mathrm{H} \rightarrow {}^3\mathrm{He} + {}^1_0\mathrm{n}$
Light Deuterium nuclei fuse together to form Helium-3, releasing considerable energy ($Q > 0$). This is **Fusion with positive Q value** (II).
- **Reaction D**: ${}^1_1\mathrm{H} + {}^3_1\mathrm{H} \rightarrow {}^2_1\mathrm{H} + {}^2_1\mathrm{H}$
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 $\rightarrow$ III
- B $\rightarrow$ I
- C $\rightarrow$ II
- D $\rightarrow$ IV
This perfectly corresponds to Option (2).
### Pattern Recognition
Identifying chemical vs. nuclear reactions is trivial (chemical reactions involve molecular change like $2\mathrm{H}_2 + \mathrm{O}_2$, whereas nuclear reactions involve changes in nuclear isotopes). Always use chemical reactions to instantly lock in a match (B-I) and narrow down options!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Class 11 Chemistry: Chemical Bonding and Molecular Structure
