Q.Describe the experimental setup of the Geiger-Marsden (alpha-particle scattering) experiment on a thin gold foil, and state its three key observations.

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set DS1 markMCQ
   Q.The path of scattered $\alpha$-particle is:

   i) circular

   ii) parabolic

   iii) elliptical

   iv) hyperbolic

   ›Reveal solutionSolution

   The scattered $\alpha$-particle travels along a hyperbola with the nucleus at the focus.

   Concept. In Rutherford's gold-foil experiment, a positively charged $\alpha$-particle approaching the positively charged nucleus experiences a repulsive inverse-square Coulomb force. An unbound particle moving under an inverse-square repulsive force traces one branch of a hyperbola, with the nucleus located at the focus. The impact parameter decides how sharply it is deflected, but the trajectory is always hyperbolic (not circular, parabolic or elliptical — those correspond to bound or special-energy orbits).

   ✓Final answer

   (iv) hyperbolic.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Radioactive source used in the Geiger-Marsden scattering experiment is

   (a) 235/92 U

   (b) 214/83 Bi

   (c) 141/56 Ba

   (d) 89/36 Kr

   ›Reveal solutionSolution

   Geiger and Marsden bombarded a thin gold foil with alpha particles emitted by the naturally radioactive isotope Bismuth-214 (214/83 Bi, historically called 'radium C').

   In Rutherford's classic alpha-scattering experiment, a beam of fast alpha particles from a naturally radioactive source was collimated and directed onto a thin gold foil, and the scattering angles of the alpha particles were observed using a rotatable detector. The radioactive source used to supply these energetic alpha particles was Bismuth-214 (214/83 Bi).

   ✓Final answer

   (b) 214/83 Bi.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The force which causes scattering of alpha particles in Rutherford's alpha particle scattering experiment is

   (a) Gravitational force

   (b) Coulomb force

   (c) Magnetic force

   (d) Nuclear force

   ›Reveal solutionSolution

   Rutherford's alpha-scattering results are explained by electrostatic repulsion between the positively charged alpha particle and a tiny, dense, positively charged nucleus.

   In Rutherford's gold-foil experiment, most alpha particles passed through nearly undeflected, but a small fraction were scattered through large angles, and a very few even bounced almost straight back. Rutherford explained this by proposing that almost all of the atom's positive charge and mass is concentrated in a tiny central nucleus. As an alpha particle (also positively charged) approaches this nucleus, the Coulomb (electrostatic) repulsive force between the two positive charges grows very large at close range, deflecting - and occasionally nearly reversing - the alpha particle's path. Gravitational force is far too weak, magnetic force does not apply to charges at rest/moving without a magnetic field being invoked here, and nuclear force acts only at nuclear-contact distances, not over the range where this scattering occurs.

   ✓Final answer

   (b) Coulomb force (electrostatic repulsion between the alpha particle and the nucleus).

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): The force of repulsion between atomic nucleus and alpha particle varies with distance according to inverse square law. Reason (R): Rutherford did alpha scattering experiment.

   (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

   (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

   (c) Assertion (A) is true, but Reason (R) is false.

   (d) Assertion (A) and Reason (R) both are false.

   ›Reveal solutionSolution

   The inverse-square nature of the nuclear repulsion force comes from Coulomb's law, not from the mere fact that Rutherford performed the scattering experiment.

   Assertion (A) is true: the Coulomb repulsion between the positively charged nucleus and an alpha particle is $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{r^2} \propto \dfrac{1}{r^2}$, an inverse-square law. Reason (R) is also true — Rutherford did perform the alpha-particle scattering experiment — but this is a historical/experimental fact, not the physical reason WHY the force obeys an inverse-square law (that reason is Coulomb's law of electrostatics). So R does not correctly explain A.

   ✓Final answer

   Both A and R are true, but R is not the correct explanation of A (option b).

5. CBSE 2025Set 55/5/11 markMCQ
   Q.Assertion (A): The presence of only a few alpha particles at a scattering angle of $180^{\circ}$ led Rutherford to the discovery of the nucleus. Reason (R): The size of the nucleus is approximately $10^{-5}$ times the size of an atom, and therefore only a few alpha particles are rebounded. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, but R is false. (D) Both A and R are false.

   ›Reveal solutionSolution

   In Rutherford's gold-foil experiment only a very small fraction of alpha particles bounced back at large angles (up to $180^{\circ}$), which told him the atom's positive charge and mass are concentrated in a tiny nucleus — so A is true. The nucleus is indeed only about $10^{-5}$ times the size of the atom, and it is exactly this minute size that makes rebounding encounters so rare — so R is true and it correctly explains A. The answer is (A).

   1. Understanding the assertion (A).

      In Rutherford's experiment, most alpha particles passed through the thin gold foil undeflected, a few were deflected through small angles, and an extremely small number — about 1 in 8000 — were turned through very large angles, some rebounding almost along their original path ($\approx 180^{\circ}$). A fast, comparatively heavy, positively charged alpha particle can be turned straight back only by an intense repulsive force from something very massive and positively charged concentrated in a tiny region. From the rarity of these rebounds together with the fact that they happened at all, Rutherford concluded that the atom's positive charge and nearly all its mass sit in a minute central core — the nucleus. So A is true.

   2. Understanding the reason (R).

      The radius of a nucleus is of the order of $10^{-15}\,\text{m}$, while the radius of an atom is of the order of $10^{-10}\,\text{m}$. Their ratio is

   $\frac{10^{-15}\,\text{m}}{10^{-10}\,\text{m}} = 10^{-5}$

   so the nucleus is about $10^{-5}$ times the size of the atom. So R is also true.

   3. Does R explain A? Yes — this is the crucial link. Because the nucleus occupies such a minute fraction of the atom, the atom is almost entirely empty space. An alpha particle is thrown back through a large angle only when it approaches the nucleus almost head-on, and since the "target" is $10^{-5}$ times the atom's size, such near head-on encounters are extremely rare. Most alpha particles never come close to any nucleus and sail straight through. The tiny size of the nucleus is therefore precisely why only a few alpha particles rebound — which is the observation the assertion describes. R is the correct explanation of A.

   Watch out

   A common mistake is to reason that since only 1 in 8000 particles rebounded, the experiment "failed" or the assertion is doubtful. It is the other way round: the fewness of the rebounds proved the positive charge is NOT spread over the whole atom (as Thomson's model claimed — in that model no alpha particle could ever bounce back), while the existence of rebounds proved a concentrated, massive nucleus exists. Both facts together led to the nuclear model.

   ✓Final answer

   The correct option is (A) — both A and R are true, and R is the correct explanation of A.

6. CBSE 2025Set X11 markMCQ
   Q.In Rutherford's $\alpha$-ray scattering experiment, $\alpha$-particles of specific energy are projected towards a thin gold foil. If the impact parameter for the $\alpha$-particles is zero, the angle of scattering is

   (a) $\theta = 0^\circ$

   (b) $\theta = 90^\circ$

   (c) $\theta = 180^\circ$

   (d) $\theta = 45^\circ$

   ›Reveal solutionSolution

   (c) $\theta = 180^\circ$ Impact parameter $b = 0$ means the $\alpha$-particle is aimed directly at the nucleus (head-on approach). It decelerates, stops at the distance of closest approach, and is

   ✓Final answer

   (c) $\theta = 180^\circ$

   Impact parameter $b = 0$ means the $\alpha$-particle is aimed directly at the nucleus (head-on approach). It decelerates, stops at the distance of closest approach, and is repelled straight back along its incoming path, so it is scattered through $180^\circ$. (In general $b \propto \cot(\theta/2)$; $b=0 \Rightarrow \cot(\theta/2)=0 \Rightarrow \theta/2 = 90^\circ \Rightarrow \theta = 180^\circ$.)

7. CBSE 2025Set IMPROVEMENT1 markMCQ
   Q.[FIGURE: Three diagrams labelled (I), (II) and (III), each showing an alpha particle (α) approaching, and passing near, a nucleus N (marked as a dot). Path (I) shows the α-particle travelling in an almost straight line directly over N. Path (II) shows the α-particle's path curving/bending as it passes near N. Path (III) shows the α-particle travelling in a straight line, offset from N.] In the following pictures three paths of an α-particle passing near a nucleus (N) are shown. Which of these paths will be correct?

   (a) All three paths

   (b) Only path I

   (c) Only path II

   (d) Only path III

   ›Reveal solutionSolution

   Since the nucleus and the α-particle are both positively charged, the Coulomb force between them is repulsive, so the α-particle's trajectory must be a hyperbola that curves away from the nucleus (never crossing through it or continuing undeviated).

   In Rutherford's scattering experiment, an α-particle approaching a nucleus experiences a repulsive Coulomb force (both are positively charged) that increases as the particle gets closer to the nucleus. This causes the α-particle's path to bend continuously away from the nucleus, tracing a smooth hyperbolic trajectory with the nucleus at the outer focus — the particle is deflected but the deflection increases smoothly as it approaches, and the path never touches or crosses the nucleus, nor does it remain a straight line near the nucleus. Of the three sketched paths, only the smoothly curving path (II) is consistent with this repulsive, continuously-deflecting interaction; the other two either show no bending near the nucleus or an unphysical bend.

   ✓Final answer

   (c) Only path II is the correct trajectory.

8. CBSE 2025Set ANNUAL1 mark
   Q.Define impact parameter.

   ›Reveal solutionSolution

   Impact parameter (b) measures how far off-centre an incoming particle's original straight-line path is aimed relative to the nucleus, which determines its scattering angle.

   In scattering experiments (e.g. Rutherford's alpha-particle scattering), the impact parameter b is defined as the perpendicular distance of the initial velocity vector of the incident particle from the centre of the nucleus (target), measured while the particle is still far away and moving in a straight line, unaffected by the nucleus.

   A smaller impact parameter corresponds to a closer approach to the nucleus and hence a larger scattering angle; b = 0 corresponds to a head-on collision (180° scattering), while a large b gives negligible deflection.

   ✓Final answer

   Impact parameter is the perpendicular distance between the initial velocity direction of the incident particle and the centre of the target nucleus.

9. CBSE 2024Set ANNUAL1 markMCQ
   Q.Choose the correct option from the given options and fill in the blank - At the suggestion of Rutherford, Geiger and Marsden performed a scattering experiment. In this experiment they directed a beam of ________ at a thin gold foil.

   (a) $\gamma$-rays

   (b) $\alpha$-particles

   (c) $\beta$-particles

   (d) Neutrons

   ›Reveal solutionSolution

   The Geiger-Marsden experiment used alpha particles to probe the atom's structure and led to the nuclear model.

   At the suggestion of Rutherford, Geiger and Marsden performed the alpha-particle scattering experiment: a beam of $\alpha$-particles (from a radioactive source) was directed at a thin gold foil, and the scattering pattern observed (most passing straight through, a few deflected at large angles) led Rutherford to propose the nuclear model of the atom.

   ✓Final answer

   (b) $\alpha$-particles

10. CBSE 2024Set FS1 markMCQ
    Q.The path of scattered $\alpha$-particle is:

    (i) circular

    (ii) parabolic

    (iii) elliptical

    (iv) hyperbolic

    ›Reveal solutionSolution

    An $\alpha$-particle deflected by the repulsive Coulomb field of the nucleus follows a hyperbolic path — option (iv).

    Concept. In Rutherford's $\alpha$-scattering experiment the positively charged $\alpha$-particle is repelled by the positive nucleus through the inverse-square Coulomb force $F\propto 1/r^2$.

    Why hyperbola. A repulsive inverse-square force gives an open (unbound) orbit. Just as an attractive $1/r^2$ force can give an ellipse, a repulsive $1/r^2$ force gives a hyperbola, with the nucleus at the outer focus. The particle comes in, is deflected, and recedes to infinity.

    ✓Final answer

    (iv) hyperbolic

11. CBSE 2024Set ANNUAL1 markMCQ
    Q.The value of scattering angle of alpha particle for maximum value of impact parameter is -

    (a) 90°

    (b) 60°

    (c) 45°

    (d) 0°

    ›Reveal solutionSolution

    Impact parameter b and scattering angle θ are inversely related in Rutherford scattering — the largest b (particle passing far from the nucleus) gives almost no deflection.

    In Rutherford's alpha-scattering theory, the relation between impact parameter b and scattering angle θ is:

    $b = \dfrac{Ze^2}{4\pi\varepsilon_0\, E}\cot\left(\dfrac{\theta}{2}\right)$

    This shows b and θ are inversely related: as b increases (the alpha particle's path passes farther from the nucleus, feeling weaker repulsion), θ decreases.

    For the maximum possible value of b (a particle travelling essentially undisturbed, far from the nucleus), $\cot(\theta/2) \to \infty$, which requires $\theta \to 0°$ — the particle goes almost straight through with negligible deflection.

    ✓Final answer

    (d) 0°.

12. CBSE 2024Set ANNUAL1 markMCQ
    Q.The force which causes scattering of alpha particles in the Rutherford alpha particle scattering experiment is

    (a) Gravitational force

    (b) Coulomb force

    (c) Magnetic force

    (d) Nuclear force

    ›Reveal solutionSolution

    Alpha particles are scattered by the concentrated positive charge of the nucleus purely through electrostatic (Coulomb) repulsion; nuclear and gravitational forces play no role at the distances involved.

    In Rutherford's gold-foil experiment, fast alpha particles ($+2e$) approaching a heavy nucleus (charge $+Ze$) are repelled because like charges repel. At the impact parameters and energies used, the alpha particles never get close enough to feel the (extremely short-range) strong nuclear force, and the gravitational force between such tiny masses is utterly negligible. The entire large-angle scattering, including rare near-180-degree deflections, is explained purely by the inverse-square Coulomb repulsion between the alpha particle and the small, massive, positively charged nucleus - this is precisely what led Rutherford to propose the nuclear model of the atom.

    ✓Final answer

    (b) Coulomb force.