JEE (Main) · Physics, Class 12
Atoms and Nuclei: JEE Main PYQ, 2020–2025
Atoms and Nuclei (Physics, Class 12) has been asked 167 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.5 questions a paper, 5% of all Physics questions. It is the 9th most-asked of 20 Physics chapters. The 2025 shifts had 18 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 14in 14 papers |
| 2021 | 40in 26 papers |
| 2022 | 36in 24 papers |
| 2023 | 23in 12 papers |
| 2024 | 36in 18 papers |
| 2025 | 18in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Bohr model45
- Radioactivity45
- Hydrogen spectrum33
- Nuclear binding energy29
- Nuclear fission and fusion10
- Rutherford model and alpha scattering5
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10 Atoms and Nuclei questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Atoms and Nuclei
In the fusion reaction deuterium + tritium gives helium-4 + neutron, the binding energies are: deuterium 2.22 MeV, tritium 8.48 MeV, helium-4 28.3 MeV. What is the energy released in this reaction?
Answer: (A)
The energy released equals the total binding energy of the products minus the total binding energy of the reactants. The reactants have 2.22 MeV (deuterium) and 8.48 MeV (tritium), totalling 10.70 MeV. The product helium-4 has 28.3 MeV. Subtracting gives 28.3 - 10.70 = 17.6 MeV, the well-known energy from D-T fusion.
Q2 · Physics · Atoms and Nuclei
A stationary heavy nucleus undergoes fission into two fragments whose masses are in the ratio 2 : 3. If the total kinetic energy released is 200 MeV, the kinetic energy carried by the lighter fragment is:
Answer: (A)
The nucleus is initially at rest, so the fragments have equal and opposite momenta, p₁ = p₂ = p. Kinetic energy is p²/2m, so K₁/K₂ = m₂/m₁. With m₁ : m₂ = 2 : 3 we get K₁ : K₂ = 3 : 2. The lighter fragment therefore takes 3/5 of 200 MeV = 120 MeV, and the heavier takes 80 MeV.
Q3 · Physics · Atoms and Nuclei
Consider Rutherford's planetary model of a hydrogen atom, with the electron in a circular orbit of radius r about the proton under Coulomb attraction. According to classical electromagnetism, the accelerating electron radiates energy and spirals inward. Treating each stage as a nearly circular orbit, by what factor does the orbital frequency of the electron increase when the orbit radius shrinks from r to r/4?
Answer: (A)
For a circular orbit, ke²/r² = mv²/r, so v² = ke²/(mr) and v ∝ r^(-1/2). The orbital frequency is f = v/(2πr) ∝ r^(-3/2). When r becomes r/4, f increases by 4^(3/2) = 8. Because the frequency changes continuously as the electron spirals in, the classical model predicts a continuous spectrum and an unstable atom, which does not match the observed line spectrum.
Q4 · Physics · Atoms and Nuclei
Why do the electrons in an atom cause negligible deflection of an alpha particle in Rutherford's experiment?
Answer: (A)
Momentum transfer in a collision is limited by the mass of the lighter body. Electrons weigh about 1/7300 of an alpha particle, so even a full collision with an electron would change the alpha particle's direction only slightly. Rutherford's large-angle scattering therefore came from the massive, positive nucleus alone.
Q5 · Physics · Atoms and Nuclei
Two ⁶Li nuclei fuse to form one ¹²C nucleus. The binding energy per nucleon is 5.33 MeV for ⁶Li and 7.68 MeV for ¹²C. What is the energy released in this fusion?
Answer: (A)
Energy released = BE(products) − BE(reactants). BE(¹²C) = 12 × 7.68 = 92.16 MeV. BE(2 × ⁶Li) = 12 × 5.33 = 63.96 MeV. Energy released = 92.16 − 63.96 ≈ 28.2 MeV. The product is more tightly bound, so energy is released.
Q6 · Physics · Atoms and Nuclei
A beam of electrons, each with kinetic energy 12.5 eV, is passed through hydrogen gas whose atoms are all initially in the ground state (E₁ = −13.6 eV). Assuming each atom can absorb energy from at most one electron collision and that excited atoms then de-excite by emitting photons, how many different spectral lines can the gas emit?
Answer: (A)
Excitation energies from n = 1 are 10.2 eV (n = 2), 12.09 eV (n = 3) and 12.75 eV (n = 4). A bombarding electron can transfer any amount of its energy, so it can excite an atom to any level whose excitation energy is at most 12.5 eV. The highest level reached is therefore n = 3. Atoms in n = 3 can emit 3→2, 3→1 and 2→1; atoms in n = 2 emit 2→1. The distinct lines number 3 (= 3C₂). A photon of 12.5 eV would be absorbed by none of these atoms, because photon absorption requires an exact energy match.
Q7 · Physics · Atoms and Nuclei
In Rutherford's alpha-scattering experiment, an alpha particle with a large impact parameter (its path passes far from the nucleus) is deflected through:
Answer: (A)
The repulsive Coulomb force between the alpha particle and the nucleus falls off with distance. A larger impact parameter means the particle stays farther away, feels a weaker and shorter-lived push, and is deflected by a smaller angle. Small impact parameter gives the large-angle scattering.
Q8 · Physics · Atoms and Nuclei
A singly ionised helium ion (He⁺, Z = 2) emits a photon when its electron falls from n = 4 to n = 2. Using En = −13.6 Z²/n² eV, this photon has the same wavelength as the photon from which transition in a hydrogen atom?
Answer: (A)
For He⁺, E4 = −54.4/16 = −3.4 eV and E2 = −54.4/4 = −13.6 eV, so ΔE = 10.2 eV. Hydrogen n = 2 to n = 1 gives −3.4 − (−13.6) = 10.2 eV. Equal photon energy means equal wavelength.
Q9 · Physics · Atoms and Nuclei
The Sun radiates power of 3.8 × 10²⁶ W. Suppose hydrogen fusion converts 0.7% of the mass of the hydrogen fused into energy. Assume that 10% of the Sun's mass (2 × 10³⁰ kg) is hydrogen available for fusion, and that the Sun keeps radiating at the present rate. For about how many years would this fuel last? (c = 3 × 10⁸ m s⁻¹, 1 year = 3.15 × 10⁷ s)
Answer: (A)
Mass converted to energy per second: Δm = P/c² = 3.8 × 10²⁶ / 9 × 10¹⁶ ≈ 4.2 × 10⁹ kg/s. Since only 0.7% of the hydrogen fused becomes energy, hydrogen fused per second = 4.2 × 10⁹ / 0.007 ≈ 6.0 × 10¹¹ kg/s. The available fuel is 0.10 × 2 × 10³⁰ = 2 × 10²⁹ kg, so the time is 2 × 10²⁹ / 6.0 × 10¹¹ ≈ 3.3 × 10¹⁷ s. Dividing by 3.15 × 10⁷ s per year gives about 1.05 × 10¹⁰ years.
Q10 · Physics · Atoms and Nuclei
The radius of the first Bohr orbit of hydrogen is 0.53 Å. What is the radius of the third orbit?
Answer: (B)
In the Bohr model the orbit radius grows as the square of the quantum number: rₙ = n² r₁, so r₃ = 9 × 0.53 = 4.77 Å.
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All 167 real JEE Main questions from Atoms and Nuclei.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Atoms and Nuclei in JEE Main.
How many questions come from Atoms and Nuclei in JEE Main?
Atoms and Nuclei (Physics, Class 12) has been asked 167 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.5 questions a paper, 5% of all Physics questions. It is the 9th most-asked of 20 Physics chapters. The 2025 shifts had 18 from it in all.
Which topics of Atoms and Nuclei are asked most in JEE Main?
Bohr model (45), Radioactivity (45), Hydrogen spectrum (33), Nuclear binding energy (29), counted over the papers from 2020–2025.
Is Atoms and Nuclei important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 9 of 20 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Atoms and Nuclei JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from Atoms and Nuclei with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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