PhysicsCBSE Class 12 · Atoms

Alpha-Particle Scattering and Rutherford's Model of the Atom Class 12

Updated 15 October 2026 · Fact-checked

In the Geiger-Marsden experiment, alpha particles fired at thin gold foil mostly passed straight through, but a few were deflected through large angles and some bounced back. Rutherford concluded that an atom's positive charge and almost all its mass sit in a tiny nucleus, with electrons around it. The model explains scattering but cannot explain atomic stability or line spectra.

Understand Alpha-Particle Scattering and Rutherford's Model

Before 1911, Thomson's model pictured the atom as a sphere of positive charge with electrons embedded in it, like seeds in a watermelon. Rutherford's students tested this by firing fast alpha particles at a thin gold foil, and the result overturned the picture.

The experiment. Alpha particles (helium nuclei, charge +2e) of about 5.5 MeV from a radioactive source were collimated into a narrow beam and aimed at a very thin gold foil (about $2.1 \times 1 0^{- 7}$ m thick). A movable zinc sulphide screen with a microscope counted the tiny flashes made by scattered particles at different angles θ. The foil was kept thin so that most alpha particles would be scattered at most once.

Observations.

Most alpha particles passed through the foil undeflected or deflected by very small angles.
A small fraction were deflected through large angles; about 1 in 8000 was deflected by more than 90°, and a very few bounced almost straight back.
The number scattered fell steeply as the scattering angle increased.

Conclusions. A smeared-out positive charge could never turn back a fast, heavy alpha particle. So:

Most of the atom is empty space (most particles pass through).
All the positive charge and almost all the mass are concentrated in a tiny central nucleus (a strong repulsion from a small, massive charge turns a few particles through large angles).
The nucleus is about $1 0^{- 15}$ to $1 0^{- 14}$ m across, roughly $1 0^{4}$ to $1 0^{5}$ times smaller than the atom (about $1 0^{- 10}$ m).
Electrons move around the nucleus, as planets move round the Sun, held by electrostatic attraction. This is Rutherford's nuclear model.

Distance of closest approach. For a head-on collision, an alpha particle of kinetic energy K slows down, stops momentarily at distance d from the nucleus (charge Ze) and turns back. At that point all its kinetic energy has become electrostatic potential energy: $K = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{d}$ , so $d = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{K}$ . This gives an upper limit to the nuclear size; for 5 MeV alpha particles on gold it is about $4.5 \times 1 0^{- 14}$ m. Doubling K halves d.

Impact parameter. The impact parameter b is the perpendicular distance of the initial velocity of the alpha particle from the centre of the nucleus. Small b (aimed nearly at the nucleus) gives large deflection; b = 0 gives head-on collision and θ = 180°; large b gives small deflection. Since the nucleus is tiny, very few particles have small b, which is why so few are scattered through large angles.

Limitations of Rutherford's model. (1) Stability: an electron moving in a circle is accelerating, and classical electromagnetism says an accelerated charge radiates energy; the electron should lose energy and spiral into the nucleus in a tiny fraction of a second, so atoms could not be stable. (2) Spectra: as the electron spiralled in, its frequency of revolution would change continuously, so the atom should emit a continuous spectrum, but atoms emit sharp line spectra. Bohr's model fixed both problems.

Alpha-Particle Scattering and Rutherford's Model: formulas and key facts

Distance of closest approach: $d = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{K}$; Head-on collision of an alpha particle (charge 2e) of kinetic energy K.
Energy at closest approach: $K = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Z e )}{d}$; All KE turned into electrostatic PE.
Projectile of charge ze: $d = \frac{1}{4 π ε _{0}} \frac{z Z e ^{2}}{K}$; z = 1 for a proton, 2 for an alpha particle.
Size comparison: $R_{a t o m} \sim 1 0^{- 10} m, R_{n u c l e u s} \sim 1 0^{- 15} to 1 0^{- 14} m$; The atom is mostly empty space.
Impact parameter and deflection: $b ↓\Rightarrow θ ↑; b = 0 \Rightarrow θ = 180°$; Qualitative relation.

How to solve Alpha-Particle Scattering and Rutherford's Model questions

Use this method for scattering questions.

1For closest approach, write K = k(2e)(Ze)/d and convert K from MeV to joule (1 MeV = 1.6 × 10⁻¹³ J).
2Solve for d, or for K if d is given.
3For comparisons, note d ∝ (charge of projectile × Z)/K.
4For impact parameter questions, relate small b to large deflection and b = 0 to 180°.
5For observations and conclusions, pair each observation with what it shows.
6For limitations, give the two classical failures: stability and line spectra.

Alpha-Particle Scattering and Rutherford's Model in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

How empty an atom is

Suppose you scale up an atom so that its nucleus (about $1 0^{- 14}$ m across) becomes a 1 cm marble. Take the atom to be about $1 0^{- 10}$ m across.

(a) By what factor is the atom larger than its nucleus? (b) How large would the scaled-up atom be? (c) Use this to explain why most alpha particles go straight through a gold foil.

Try it first, then show the solutionSolution

Ratio = $\frac{1 0 ^{- 10}}{1 0 ^{- 14}} = 1 0^{4}$ .
Scaled atom: 1 cm × 10⁴ = 100 m across, about the size of a sports stadium with a marble at its centre.
An alpha particle crossing such an atom is very unlikely to pass near the tiny nucleus; with only light electrons in its way, it continues almost undeflected.

Answer: (a) about 10⁴; (b) about 100 m; (c) the nucleus is so small that most alpha particles never come close to it.

Memory hook: A marble in a stadium: that is a nucleus in its atom.

Energy during a head-on approach

Imagine an 8 MeV alpha particle heading straight for a gold nucleus (Z = 79), which stays fixed.

(a) Find the distance of closest approach. (b) What is the alpha particle's kinetic energy when it is twice that distance from the nucleus? (c) What happens after closest approach?

Try it first, then show the solutionSolution

$d = \frac{9 \times 1 0 ^{9} \times 2 \times 79 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{8 \times 1.6 \times 1 0 ^{- 13}} \approx 2.84 \times 1 0^{- 14}$ m.
At 2d the potential energy is half of 8 MeV, i.e. 4 MeV, so the kinetic energy is 8 − 4 = 4 MeV.
It is pushed back along its path, regaining all 8 MeV of kinetic energy as it moves away (180° scattering).

Answer: (a) about 2.84 × 10⁻¹⁴ m; (b) 4 MeV; (c) it is repelled straight back with its original energy.

Memory hook: Head-on alpha: kinetic energy climbs a Coulomb hill, stops at the top and rolls back.

Tricks to remember Alpha-Particle Scattering and Rutherford's Model

Short memory aids. Say each one aloud once, then use it on a question the same day.

Most through, few back
Most pass: empty space. Few deflect strongly: tiny, heavy, positive nucleus.
Use it when: Observations and conclusions.
Small b, big bend
The closer the aim, the larger the deflection; b = 0 means a 180° bounce.
Use it when: Impact parameter questions.
KE becomes PE at d
K = k(2e)(Ze)/d at the turning point.
Use it when: Closest approach numericals.
Two failures: spiral and spectrum
Rutherford's atom should spiral in and give a continuous spectrum; it does neither.
Use it when: Limitations of the model.

Alpha-Particle Scattering and Rutherford's Model: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Distance of closest approach · 2 marks · Very short answerBoard level

Find the distance of closest approach of a 6 MeV alpha particle to a gold nucleus (Z = 79) in a head-on collision.

Try it first, then show the solutionSolution

$K = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{d}$ , with K = 6 MeV = 9.6 × 10⁻¹³ J. (1)
$d = \frac{9 \times 1 0 ^{9} \times 2 \times 79 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{9.6 \times 1 0 ^{- 13}} \approx 3.79 \times 1 0^{- 14}$ m. (1)

Answer: About 3.79 × 10⁻¹⁴ m.

Example 2: Observations and conclusions · 3 marks · Short answerBoard level

Describe the main observations of the Geiger-Marsden alpha-scattering experiment and the conclusions Rutherford drew from them.

Try it first, then show the solutionSolution

Most alpha particles passed through the thin gold foil undeflected: most of the atom is empty space. (1)
A few were deflected through large angles and about 1 in 8000 by more than 90°: the positive charge and most of the mass are concentrated in a tiny region, the nucleus, which can repel an alpha particle strongly. (1)
The fraction scattered falls rapidly with angle: only particles passing very close to the tiny nucleus (small impact parameter) are deflected strongly. Electrons revolve round the nucleus. (1)

Answer: Mostly undeflected (empty space); rare large deflections (tiny, massive, positive nucleus); electrons orbit the nucleus.

Example 3: Changing energy and target · 2 marks · Very short answerHarder board level

Using the result of the first example, find the distance of closest approach for (a) a 12 MeV alpha particle on gold and (b) a 6 MeV alpha particle on silver (Z = 47).

Try it first, then show the solutionSolution

(a) d ∝ 1/K: 3.79 × 10⁻¹⁴/2 ≈ 1.90 × 10⁻¹⁴ m. (1)
(b) d ∝ Z: 3.79 × 10⁻¹⁴ × 47/79 ≈ 2.26 × 10⁻¹⁴ m. (1)

Answer: (a) about 1.90 × 10⁻¹⁴ m; (b) about 2.26 × 10⁻¹⁴ m.

Example 4: Impact parameter · 3 marks · Short answerHarder board level

Define impact parameter. Sketch (describe) the paths of alpha particles with large, small and zero impact parameters near a nucleus, and explain why so few alpha particles are scattered backwards.

Try it first, then show the solutionSolution

Impact parameter b: the perpendicular distance of the alpha particle's initial velocity line from the centre of the nucleus. (1)
Large b: path bends only slightly. Small b: path bends sharply round the nucleus. b = 0: the particle moves straight in, stops at the closest-approach distance and returns along the same line (θ = 180°). (1)
Backward scattering needs very small b; since the nucleus is tiny compared with the atom, very few particles in a broad beam happen to be aimed so close. (1)

Answer: b is the perpendicular distance of the initial path from the nucleus; smaller b gives larger deflection; few particles have tiny b, so backward scattering is rare.

Example 5: Comparing projectiles and reaching the nucleus · 5 marks · Long answerHOTS / competency

(a) A proton and an alpha particle approach the same nucleus head-on with the same speed. Find the ratio of their distances of closest approach. (2)
(b) The radius of a gold nucleus is about $1.2 \times 1 0^{- 15} \times 19 7^{1/3}$ m. Estimate the minimum energy (in MeV) an alpha particle needs to just touch it. (2)
(c) Why would such an alpha particle no longer follow the simple Coulomb result? (1)

Try it first, then show the solutionSolution

(a) d ∝ (charge of projectile)/K. Same speed: K ∝ m, so K_α = 4K_p. (1)
$\frac{d _{p}}{d _{α}} = \frac{e / K _{p}}{2 e / K _{α}} = \frac{K _{α}}{2 K _{p}} = 2$ . (1)
(b) R ≈ 1.2 × 10⁻¹⁵ × 5.82 ≈ 7.0 × 10⁻¹⁵ m. (1)
$K = \frac{9 \times 1 0 ^{9} \times 2 \times 79 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{7.0 \times 1 0 ^{- 15}} \approx 5.2 \times 1 0^{- 12}$ J ≈ 33 MeV. (1)
(c) At such distances the short-range nuclear force acts as well as the Coulomb force, so pure Coulomb scattering no longer describes the motion. (1)

Answer: (a) d_p : d_α = 2 : 1; (b) about 33 MeV; (c) the nuclear force becomes important.

Example 6: Limits of the nuclear model · 4 marks · Case-basedHarder board level

Read the passage and answer.

Rutherford's model pictures electrons orbiting a tiny nucleus like planets round the Sun, held by electrostatic attraction.
(a) What does classical electromagnetism predict for an electron moving in a circular orbit? (1)
(b) What would happen to the electron's orbit as a result? (1)
(c) What kind of spectrum would such atoms emit, and what do atoms actually emit? (1)
(d) Name the model that overcame these problems. (1)

Try it first, then show the solutionSolution

(a) It is accelerating (centripetal acceleration), so it should continuously radiate electromagnetic energy. (1)
(b) Losing energy, it should spiral into the nucleus very quickly, so atoms would be unstable. (1)
(c) Its revolution frequency would change continuously, so it would emit a continuous spectrum; real atoms emit sharp line spectra. (1)
(d) Bohr's model of the hydrogen atom. (1)

Answer: (a) continuous radiation; (b) spiral into the nucleus; (c) continuous versus observed line spectra; (d) Bohr's model.

Where students lose marks in Alpha-Particle Scattering and Rutherford's Model

The slips examiners see most often, and what to write instead.

Mistake: Using charge e instead of 2e for the alpha particle.
Forgetting the alpha particle is a helium nucleus.
Do this: Alpha particle charge = +2e.
Usually costs: The answer mark.
Mistake: Leaving K in MeV in the SI formula.
Unit slip.
Do this: 1 MeV = 1.6 × 10⁻¹³ J.
Usually costs: The answer mark.
Mistake: Stating that the nucleus is negatively charged or that electrons cause the large deflections.
Confusing roles of electrons and nucleus.
Do this: Electrons are far too light to deflect alpha particles appreciably; the large deflections come from the positive, massive nucleus.
Usually costs: 1 mark.
Mistake: Listing observations without conclusions.
Half answers.
Do this: Pair each observation with its conclusion.
Usually costs: 1 to 2 marks.

Exam tips

Alpha scattering and Rutherford's model were set in 2023 (55/5/1 Q30(a)), 2024 (55/5/1 Q16) and 2026 (55/1/1 Q8).
Draw the scattering set-up and a sketch of paths with different impact parameters; diagrams carry marks.
In closest-approach numericals, write K = k(2e)(Ze)/d before substituting.
Always add the two limitations when asked about Rutherford's model.
Use the 1 in 8000 (more than 90°) figure only as approximate; it is quoted in NCERT.

Practice questions on Alpha-Particle Scattering and Rutherford's Model

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In the alpha-scattering experiment, most alpha particles pass through the foil undeflected. This shows that
(a) the nucleus is negatively charged (b) most of the atom is empty space (c) electrons are heavy (d) gold is a good conductor
Hint
Nothing in the way.
Answer
(b) most of the atom is empty space.
Q2 · 1 mark · Very short answerBoard level
Define impact parameter. What is its value for an alpha particle scattered through 180°?
Hint
Head-on.
Answer
The perpendicular distance of the initial velocity of the alpha particle from the centre of the nucleus; for 180° scattering it is zero.
Q3 · 2 marks · Very short answerHarder board level
Find the distance of closest approach of a 4 MeV alpha particle to a nucleus of atomic number 50.
Hint
K = k(2e)(Ze)/d.
Answer
d = 9 × 10⁹ × 2 × 50 × (1.6 × 10⁻¹⁹)²/(4 × 1.6 × 10⁻¹³) = 3.6 × 10⁻¹⁴ m.
Q4 · 2 marks · Very short answerHarder board level
What kinetic energy (in MeV) must an alpha particle have to come within 10⁻¹⁴ m of a gold nucleus (Z = 79) head-on?
Hint
K = k(2e)(Ze)/d.
Answer
K = 9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)²/10⁻¹⁴ ≈ 3.64 × 10⁻¹² J ≈ 22.8 MeV.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Only a very small fraction of alpha particles are scattered backwards by a gold foil.
Reason (R): The nucleus occupies a very small part of the atom, so very few alpha particles have small impact parameters.
(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) A is false but R is true.
Hint
Link b and θ.
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
Why was a very thin gold foil used in the scattering experiment, and why gold?
Hint
Multiple scattering; Z and malleability.
Answer
A thin foil ensures that each alpha particle is scattered (mostly) only once, so the angle measured reflects a single encounter with a nucleus. Gold can be beaten into very thin foils and has a large Z, giving strong, easily observed deflections.
Q7 · 2 marks · Very short answerHOTS / competency
Give two reasons why Rutherford's model of the atom was not acceptable.
Hint
Accelerating charges radiate.
Answer
(1) An orbiting electron accelerates and should radiate energy, spiralling into the nucleus, so atoms would not be stable. (2) Such an atom would emit a continuous spectrum, whereas atoms emit sharp line spectra.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Charge of an alpha particle?Show answer
+2e.
2.Distance of closest approach formula?Show answer
d = k(2e)(Ze)/K.
3.Impact parameter for head-on collision?Show answer
Zero.
4.Order of nuclear size?Show answer
10⁻¹⁵ to 10⁻¹⁴ m.
5.Main conclusion of alpha scattering?Show answer
Positive charge and most mass are in a tiny nucleus.
6.Why is Rutherford's atom unstable classically?Show answer
An accelerating electron radiates energy and would spiral into the nucleus.

Alpha-Particle Scattering and Rutherford's Model: frequently asked questions

Why did Rutherford conclude that the nucleus is very small?

Only a tiny fraction of alpha particles were deflected through large angles; they must have passed very close to a small, concentrated charge. If the charge were spread out, none would be turned back.

Why don't electrons deflect alpha particles much?

An electron is about 7300 times lighter than an alpha particle, so collisions with electrons barely change the alpha particle's direction, just as a bowling ball is hardly deflected by a ping-pong ball.

What is the distance of closest approach?

The smallest distance an alpha particle reaches from the nucleus in a head-on collision, where all its kinetic energy has become electrostatic potential energy. It gives an upper limit to the size of the nucleus.

What are the limitations of Rutherford's model?

It cannot explain the stability of atoms (orbiting electrons should radiate and spiral in) or the line spectra of atoms (it predicts continuous spectra).

What is the impact parameter?

The perpendicular distance between the initial direction of motion of the alpha particle and the centre of the nucleus. The smaller it is, the larger the scattering angle.