Briefly explain the Geiger-Marsden experiment. Show the variation of the number of particles scattered (N) with scattering angle ($\theta$) in this experiment. What is the main conclusion that can be inferred from this plot ?

Briefly explain the Geiger-Marsden experiment. Show the variation of the number of particles scattered (N) wit

Question

\theta

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The Geiger-Marsden experiment fired alpha particles at thin gold foil and measured scattering angles. The number of scattered particles varies as $N \propto \frac{1}{\sin^4(\theta/2)}$ , revealing that atoms contain a tiny, dense, positively charged nucleus.

The Experiment

In 1909, under Rutherford's direction, Hans Geiger and Ernest Marsden bombarded a thin gold foil (about 400 atoms thick) with a beam of alpha particles from a radioactive source. The experimental setup was elegantly simple: alpha particles passed through a collimating slit, struck the foil, and scattered particles were detected by observing scintillations on a zinc sulfide screen that could be moved to different angles.

The expectation, based on J.J. Thomson's "plum pudding" model (where positive charge was spread uniformly throughout the atom with electrons embedded within), was that alpha particles would pass through with only minor deflections—perhaps a degree or two at most. The diffuse positive charge couldn't exert a strong enough force to cause large-angle scattering.

What they observed was shocking: while most alpha particles did pass straight through or deflected by small angles, a tiny fraction (about 1 in 8000) scattered through angles greater than 90°, some even bouncing almost straight back. Rutherford famously remarked it was "as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

The Scattering Law

Rutherford's theoretical analysis, treating the scattering as a Coulomb interaction between the alpha particle (charge $+2e$ ) and a concentrated positive charge (the nucleus), led to a precise prediction. The number of alpha particles $N$ detected at a scattering angle $\theta$ follows:

N(\theta) \propto \frac{1}{\sin^4(\theta/2)}

This relationship can be understood through the impact parameter: alpha particles passing closer to the nucleus experience stronger Coulomb repulsion and scatter through larger angles. The $\sin^4(\theta/2)$ dependence emerges from the geometry of hyperbolic trajectories in an inverse-square force field.

The Variation with Scattering Angle

The plot of $N$ versus $\theta$ shows a dramatic variation:

Scattering Angle $\theta$	Relative Number $N$	Behavior
Small ( $\theta \to 0°$ )	Very large ( $N \to \infty$ )	Most particles undeflected
Moderate ( $\theta \approx 30°-60°$ )	Moderate	Noticeable scattering
Large ( $\theta \approx 90°-180°$ )	Very small	Rare backscattering events

The curve is steeply hyperbolic: at small angles, $\sin(\theta/2) \approx \theta/2$ is small, so $N$ is enormous. As $\theta$ increases toward 180° (backscattering), $\sin(\theta/2) \to 1$ , but the fourth power in the denominator means even at large angles the function drops precipitously.

A typical plot shows:

A near-vertical rise as $\theta \to 0°$
Rapid decay for increasing $\theta$
A long tail extending to 180°, with non-zero (though tiny) counts even at backscattering angles

Tip

The experimental verification was quantitative: Geiger and Marsden measured $N$ at various angles and confirmed the $\sin^{-4}(\theta/2)$ dependence held over several orders of magnitude, validating Rutherford's nuclear model.

The Main Conclusion …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set DS1 markMCQ
$\alpha$
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 …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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. Gravitati …

Start your 14-day free trial to see the full solution →
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 physica …

Start your 14-day free trial to see the full solution →
CBSE 2025Set 55/5/11 markMCQ
$180^{\circ}$
›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).

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.

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.

Does R explain A? …
Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$\alpha$
(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 …

Start your 14-day free trial to see the full solution →
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, no …

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
CBSE 2024Set FS1 markMCQ
$\alpha$
(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 $F\propto 1/r^2$ .

…

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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-1 …

Start your 14-day free trial to see the full solution →

Q.Briefly explain the Geiger-Marsden experiment. Show the variation of the number of particles scattered (N) with scattering angle ( $\theta$ ) in this experiment. What is the main conclusion that can be inferred from this plot ?

Concept understanding — Geiger-Marsden Alpha-Scattering Experiment

The Experiment

The Scattering Law

The Variation with Scattering Angle

The Main Conclusion …

Q.Briefly explain the Geiger-Marsden experiment. Show the variation of the number of particles scattered (N) with scattering angle (θ\thetaθ) in this experiment. What is the main conclusion that can be inferred from this plot ?

The Experiment

The Scattering Law

The Variation with Scattering Angle

The Main Conclusion …

🎓Unlock everything free for 14 days

Q.Briefly explain the Geiger-Marsden experiment. Show the variation of the number of particles scattered (N) with scattering angle ( $\theta$ ) in this experiment. What is the main conclusion that can be inferred from this plot ?

Unlock everything free for 14 days