Atoms and Nuclei — NEET Physics MCQs with answers

Atoms and Nuclei is a Class 12 Physics chapter in the NEET (UG) syllabus. NEET720 has 662 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Bohr model of hydrogen atom · Alpha scattering and Rutherford model · Rutherford's nuclear model · Bohr model of the hydrogen atom · Bohr model · Hydrogen spectral series · Hydrogen spectra · Nuclear size, mass and binding energy · Fission and fusion · de Broglie wavelength in atomic context · Rutherford's model · Nuclear notation · Nuclear Binding Energy · Radioactivity · Atomic Spectra · Rutherford Scattering · Nuclear Energy · Nuclear Force · Nuclear Fission · Nuclear Composition · Nuclear Reactions · Hydrogen Spectrum · Nuclear Structure · Atomic Models · Binding Energy · Nuclear Stability · Size of the nucleus · Nuclear fusion · Nuclear size · Nucleus · Bohr model of the atom · de Broglie hypothesis and Bohr orbits · X-ray spectra · Nuclear size and density · Mass defect and binding energy · Binding energy per nucleon curve · Radioactive decay law · Hydrogen-like atoms · Mass-energy equivalence · Nuclear fission and fusion

8 free Atoms and Nuclei practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Fission and fusion

   Nuclear fission is best described as which of the following processes?

   • A.A heavy nucleus splitting into two lighter nuclei of comparable mass, releasing energy
   • B.Two light nuclei combining into a heavier nucleus, releasing energy
   • C.A nucleus spontaneously emitting an alpha particle to become a slightly lighter nucleus
   • D.A nucleus absorbing a neutron and becoming a heavier isotope without splitting
   Show answer and explanation

   Answer: A. A heavy nucleus splitting into two lighter nuclei of comparable mass, releasing energy

   Fission is the splitting of a heavy nucleus (like U-235) into two medium-mass fragments plus a few neutrons, releasing energy because the fragments have higher binding energy per nucleon than the original heavy nucleus.

   In nuclear fission, a heavy, neutron-rich nucleus such as uranium-235 absorbs a neutron and becomes unstable, splitting into two lighter 'daughter' nuclei of roughly comparable mass (rather than a small fragment breaking off), along with a few free neutrons. Energy is released because, from the binding-energy-per-nucleon curve, mid-mass fragments are more tightly bound per nucleon than the original heavy nucleus — the difference in binding energy appears as kinetic energy of the fragments and neutrons.

   Common mistake: Confusing fission with fusion or with simple radioactive decay processes

2. Question 2 · easy · Fission and fusion

   Why is nuclear fission of a very heavy nucleus (such as uranium) an energy-releasing process, while splitting a light, already tightly-bound nucleus such as helium-4 into two deuterons would require an input of energy?

   • A.Uranium's nucleons sit on the binding-energy-per-nucleon curve well below the peak (near A ≈ 56), so splitting it moves nucleons toward higher binding energy; helium-4 already sits near a local high point, so splitting it moves nucleons toward lower binding energy
   • B.Because uranium contains more neutrons than protons, and neutron-rich nuclei always release energy when they split
   • C.Because uranium is radioactive while helium-4 is a stable nucleus, and only radioactive nuclei can release fission energy
   • D.Because the strong nuclear force is completely absent in heavy nuclei, so no energy is needed to pull them apart
   Show answer and explanation

   Answer: A. Uranium's nucleons sit on the binding-energy-per-nucleon curve well below the peak (near A ≈ 56), so splitting it moves nucleons toward higher binding energy; helium-4 already sits near a local high point, so splitting it moves nucleons toward lower binding energy

   Energy release in splitting a nucleus depends on whether the fragments move to a higher binding energy per nucleon than the original — true for heavy nuclei (below the curve's peak) but false for already well-bound light nuclei like helium-4 (near a local high point).

   The binding-energy-per-nucleon curve rises steeply for light nuclei, peaks near A ≈ 56 (iron/nickel region), and falls slowly for heavy nuclei. Uranium (A ≈ 238) sits on the falling side, well below the peak; when it splits into medium-mass fragments (A ≈ 90-140), those fragments sit closer to the peak, i.e. more tightly bound per nucleon, so mass decreases and energy is released. Helium-4, by contrast, already has an unusually high binding energy per nucleon (about 7.1 MeV) for such a light nucleus — splitting it into two deuterons (each about 1.1 MeV/nucleon) would move to a lower binding energy per nucleon, requiring an energy input rather than releasing energy. The single governing criterion is always the change in binding energy per nucleon, not the neutron-proton ratio or radioactivity.

   Common mistake: Assuming any nuclear splitting releases energy, regardless of where the parent nucleus sits on the BE/nucleon curve

3. Question 3 · medium · Nuclear size, mass and binding energy

   On the binding energy per nucleon curve, nucleus P (A = 60) has a value of 8.7 MeV/nucleon and nucleus Q (A = 230) has a value of 7.2 MeV/nucleon. Which nucleus is more stable, and why?

   • A.Q, because it has a larger mass number
   • B.Q, because it has more total binding energy (higher A × BE/nucleon)
   • C.P, because it has a higher binding energy per nucleon
   • D.Both are equally stable since both lie on the same curve
   Show answer and explanation

   Answer: C. P, because it has a higher binding energy per nucleon

   A higher binding energy per nucleon means each nucleon is, on average, more strongly bound, requiring more energy to remove — this is the standard measure of nuclear stability, not total binding energy or mass number.

   Stability of a nucleus is judged by binding energy per nucleon, not by total binding energy or mass number, because it measures the average energy needed to remove one nucleon. Nucleus P has 8.7 MeV/nucleon while nucleus Q has only 7.2 MeV/nucleon, so P is more tightly bound per nucleon and therefore more stable — this is consistent with real nuclei: mid-mass nuclei near iron are more stable than very heavy nuclei like those near A = 230, which is why heavy nuclei tend to undergo fission toward more stable, mid-mass products.

   Common mistake: Judging stability by mass number or total binding energy instead of the per-nucleon value

4. Question 4 · easy · Nuclear size, mass and binding energy

   Protons within a nucleus repel each other electrostatically, yet the nucleus remains stable and does not fly apart. Which force is responsible for holding the nucleus together against this repulsion?

   • A.Gravitational force between nucleons
   • B.The strong nuclear force
   • C.The weak nuclear force
   • D.Magnetic force between the spins of nucleons
   Show answer and explanation

   Answer: B. The strong nuclear force

   The strong nuclear force is a short-range, powerful attractive force acting between all nucleons (protons and neutrons alike), overwhelming the comparatively weaker electrostatic repulsion between protons at nuclear distances.

   At distances of about 1-2 fm, the strong nuclear force is roughly 100 times stronger than the electrostatic (Coulomb) repulsion between protons, and it acts equally between proton-proton, proton-neutron, and neutron-neutron pairs (charge-independent). This overwhelming short-range attraction is what binds nucleons into a stable nucleus despite the mutual repulsion of the protons. Gravity and magnetic interactions between nucleons are far too weak by comparison to play any significant role in nuclear binding.

   Common mistake: Confusing the strong nuclear force with the weak force or assuming gravity is significant at nuclear scales

5. Question 5 · easy · Alpha scattering and Rutherford model

   In the gold foil experiment, the vast majority of alpha particles were observed to pass through the foil with little or no deflection. This observation shows that

   • A.most alpha particles collide elastically with electrons and bounce back
   • B.the nucleus repels every alpha particle equally, producing a uniform small deflection for all of them
   • C.most of the space inside an atom is empty, so alpha particles travel through without encountering strong forces
   • D.alpha particles lose all their kinetic energy while crossing the foil, so they cannot be deflected
   Show answer and explanation

   Answer: C. most of the space inside an atom is empty, so alpha particles travel through without encountering strong forces

   Undeflected passage of most alpha particles means they encountered no significant force, i.e. the atom is mostly empty space with the nucleus occupying a tiny fraction of the atomic volume.

   Since Coulomb force falls off rapidly with distance, an alpha particle passing far from the tiny nucleus (which happens for the vast majority of particles, given how small the nucleus is compared to the atom) experiences negligible repulsion and travels almost straight through. This is direct evidence that the atom's volume is overwhelmingly empty space, with the nucleus occupying roughly 1 part in 10^12 of the atomic volume. Option A is wrong because electrons are far too light to deflect a much heavier alpha particle. Option B is wrong because the deflection angle actually varies enormously with impact parameter — it is not uniform. Option D confuses energy loss with deflection.

   Common mistake: Attributing the lack of deflection to electron collisions or assuming a uniform small deflection for every particle.

6. Question 6 · medium · Alpha scattering and Rutherford model

   Rutherford's scattering experiment used a very thin gold foil rather than a thick block of gold mainly because

   • A.gold has a low melting point, so only thin sheets could be prepared safely
   • B.a thin foil ensures alpha particles interact with essentially a single layer of nuclei, avoiding multiple scattering that would blur the angular distribution
   • C.gold is cheaper in thin sheet form, so cost was the deciding factor
   • D.a thick block would completely stop the alpha particles, so no scattering could be observed at all
   Show answer and explanation

   Answer: B. a thin foil ensures alpha particles interact with essentially a single layer of nuclei, avoiding multiple scattering that would blur the angular distribution

   Gold's malleability allows it to be hammered into an extremely thin foil (a few hundred atoms thick), so each alpha particle typically encounters only one nucleus, giving a clean single-scattering angular distribution.

   Gold was chosen because it is highly malleable and can be beaten into foils only a few hundred atoms thick. This thinness is essential: if the foil were thick, an alpha particle could undergo multiple small-angle scatterings from several nuclei in succession, and the resulting angular distribution would no longer cleanly reflect single Coulomb scattering off one nucleus, making the data far harder to interpret. The high atomic number of gold (Z = 79) also gives a strong Coulomb field, producing a measurable rate of large-angle scattering with a reasonably intense alpha source. Melting point and cost are not the physical reasons behind the choice.

   Common mistake: Attributing the choice of thin gold foil to unrelated material properties like cost or melting point.

7. Question 7 · easy · Alpha scattering and Rutherford model

   Rutherford's alpha-scattering apparatus was enclosed in an evacuated chamber. The main reason for evacuating the chamber was to

   • A.prevent the alpha particles from being scattered or absorbed by air molecules before reaching the detector, which would obscure the true scattering pattern from the foil
   • B.keep the gold foil cool so that thermal expansion does not change its thickness during the experiment
   • C.stop the alpha particles from losing their positive charge to oxygen in the air
   • D.increase the speed of the alpha particles by removing air resistance before they strike the foil
   Show answer and explanation

   Answer: A. prevent the alpha particles from being scattered or absorbed by air molecules before reaching the detector, which would obscure the true scattering pattern from the foil

   Alpha particles are easily scattered or stopped by ordinary air; a vacuum ensures that the only source of deflection is the interaction with nuclei in the gold foil, so the measured angular distribution is not distorted by collisions with air molecules.

   Alpha particles, though energetic, have a very short range in matter because they interact strongly with electrons and nuclei of any material they pass through, including air. If Rutherford's apparatus were not evacuated, alpha particles would suffer additional random scattering and absorption from nitrogen and oxygen molecules in the air along their path from source to foil and from foil to detector. This would add spurious deflections unrelated to the gold nuclei, corrupting the angular distribution being measured. Evacuating the chamber removes this confounding source of scattering, ensuring that any observed deflection can be confidently attributed to the interaction with the gold foil's nuclei.

   Common mistake: Attributing the vacuum requirement to thermal or charge effects rather than to eliminating unwanted scattering by air molecules.

8. Question 8 · medium · Alpha scattering and Rutherford model

   A student sketches the path of a scattered alpha particle as a straight line moving toward the nucleus that suddenly bends at one sharp point (like a ball bouncing off a wall) and then continues as another straight line. The actual trajectory of an alpha particle being Coulomb-scattered by a nucleus is instead

   • A.a smoothly curving path (a hyperbola, with the nucleus at one focus) that bends gradually and most sharply near the point of closest approach, since the repulsive Coulomb force acts continuously and grows stronger as the particle gets nearer
   • B.a straight line all the way to the nucleus, followed by an instantaneous reversal, exactly as the student sketched, because the force only acts at the single instant of 'collision'
   • C.a circular arc centred on the nucleus, similar to a planet's circular orbit around a star
   • D.a series of small straight-line segments, since the alpha particle only interacts with the nucleus at discrete moments in time
   Show answer and explanation

   Answer: A. a smoothly curving path (a hyperbola, with the nucleus at one focus) that bends gradually and most sharply near the point of closest approach, since the repulsive Coulomb force acts continuously and grows stronger as the particle gets nearer

   The Coulomb repulsive force acts continuously on the alpha particle throughout its approach and departure, so the trajectory is a smooth hyperbolic curve (with the nucleus at the outer focus), curving most sharply near the distance of closest approach — not a straight line with a sudden kink.

   Because the alpha particle and nucleus repel each other via the Coulomb force at every separation (not just at one 'contact' instant), the alpha particle's path continuously bends as it approaches and recedes from the nucleus. Far from the nucleus, the force is weak and the path is nearly straight; as the particle gets closer, the repulsion strengthens and the curvature increases, reaching its sharpest near the point of closest approach; then, as the particle moves away again, the path straightens out once more. Solving the equations of motion for this inverse-square repulsive force shows the trajectory is exactly a hyperbola with the nucleus at the far focus (for a non-zero impact parameter) — a smoothly and continuously curving open path, not a straight-line-then-sudden-bend or a closed circular/elliptical orbit.

   Common mistake: Picturing the scattering event as an instantaneous billiard-ball-like collision at a single point, rather than a continuous curving path under a distance-dependent force.

Questions about Atoms and Nuclei for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (662 active practice questions in this chapter today).