ScienceCBSE Class 9 · Journey Inside the Atom

Thomson, Rutherford and Bohr Atomic Models and the Gold Foil Experiment (Class 9)

Updated 17 October 2026 · Fact-checked

J. J. Thomson (1897) discovered the electron and pictured the atom as a positive sphere with electrons embedded in it. In the 1911 gold foil experiment most alpha particles passed straight through, some were deflected through large angles and very few bounced back, so Rutherford concluded that the atom is mostly empty space with a tiny, dense, positively charged nucleus holding nearly all the mass, with electrons revolving round it. His model could not explain stability, which Bohr (1913) explained by allowing electrons only in fixed shells in which they do not lose energy.

Notice how the atom changes: electrons buried in a positive sphere, then a tiny central nucleus with empty space, then electrons in fixed orbits.

Illustration drawn for QuestionBuddy.

Understand Thomson, Rutherford and Bohr Atomic Models

Early ideas. Acharya Kanada spoke of parmanu, indivisible particles (in the Vaisesika Sutras), and Leucippus and Democritus of atomos, but these were ideas from thought. John Dalton's atomic theory (1808) was the first scientific one, based on experiments. Late in the nineteenth century radioactivity showed that atoms are made of smaller particles. Atoms are about 10⁻¹⁰ m across; a sheet of paper 0.1 mm thick is about a million atoms deep.

Thomson and the electron. In 1897 J. J. Thomson passed a high voltage through a gas at very low pressure in a glass tube. Rays went from the cathode (−) to the anode (+): cathode rays. Electric and magnetic fields showed they are streams of negatively charged particles much lighter than atoms: electrons. The rays were the same whatever the cathode metal or the gas, so electrons are in all atoms. The charge of an electron is −1.602 × 10⁻¹⁹ C (relative charge −1). Because atoms are neutral, Thomson's plum pudding (watermelon) model had a sphere of positive charge with electrons spread through it, the total positive charge equal to the total negative charge.

The gold foil experiment (1911). Geiger and Marsden, working with Rutherford, fired a narrow beam of alpha (α) particles (positively charged; later known as helium nuclei) at a very thin gold foil. Thomson's model predicted only small deflections. Instead: most α-particles passed straight through, so most of the atom is empty space; some were deflected through large angles, so the positive charge is concentrated in a tiny region that repels them; very few (about 1 in 20 000 in later counts) bounced straight back, so that region is very small, dense and holds nearly all the mass: the nucleus.

Rutherford's nuclear model. A tiny, dense, positively charged nucleus at the centre holds almost all the mass; electrons revolve round the nucleus like planets round the Sun. The nucleus (about 10⁻¹⁵ m) is about 10⁵ times smaller than the atom (about 10⁻¹⁰ m); if the atom were a cricket ground 100 m across, the nucleus would be a peppercorn at the centre. The nuclear charge is carried by protons (charge +1, much heavier than electrons); in a neutral atom the number of protons equals the number of electrons.

Limitation of Rutherford's model. An electron moving in a circle keeps changing direction, so it is accelerating. A charged particle that accelerates should lose energy, spiral inwards and fall into the nucleus, so the atom would collapse almost instantly. But atoms are stable.

Bohr's model (1913). Electrons move only in certain fixed circular orbits called shells, energy levels or stationary states, labelled K, L, M, N (n = 1, 2, 3, 4) outward from the nucleus. While in a shell an electron does not lose energy. Each shell has a definite energy, increasing outwards. An electron can jump between shells only by absorbing or giving out exactly the energy difference between them, and cannot stay between shells. Each shell holds a limited number of electrons. Bohr's model was later improved by the quantum mechanical model, which describes electron clouds.

Thomson, Rutherford and Bohr Atomic Models: formulas and key facts

Timeline: Dalton 1808
Thomson (electron) 1897
Geiger, Marsden and Rutherford (gold foil) 1911
Bohr 1913
Chadwick (neutron) 1932
Gold foil observations and conclusions: Most pass straight empty space
some deflected positive charge concentrated
few bounce back dense, small nucleus
Sizes: Atom about 10⁻¹⁰ m
nucleus about 10⁻¹⁵ m
ratio about 10⁵
Charges: $E l ec t r o n - 1.602 \times 1 0^{- 19} C (r e l a t i v e - 1) ∣ p r o t o n + 1$
Bohr's jumps: $E n er g y ab sor b e d or e mi tt e d = e n er g y d i f f er e n ce b e tw ee n t h e tw os h e l l s$

Most pass straight	empty space
some deflected	positive charge concentrated
few bounce back	dense, small nucleus

How to solve Thomson, Rutherford and Bohr Atomic Models questions

Use this method for model, experiment and scale questions.

1For an experiment, write each observation in one column and its conclusion in the next.
2Compare models with the same three headings: where the positive charge is, where the electrons are, and what the model cannot explain.
3For scale questions, find the ratio first (10⁻¹⁰ ÷ 10⁻¹⁵ = 10⁵) and then scale the model.
4For percentages in a scattering table, divide each count by the total and multiply by 100.
5For Bohr's jumps, subtract the energies of the shells; the jump is allowed only for exactly that amount.
6State which model fails and why when asked for a limitation.

Thomson, Rutherford and Bohr Atomic Models 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.

Scaling the atom up to a cricket ground

Suppose the atom is blown up to the size of a 100 m wide ground, and a hypothetical mapmaker wants to place its nucleus at the centre.

Using atom ≈ 10⁻¹⁰ m and nucleus ≈ 10⁻¹⁵ m, find the size of the nucleus on this scale.

Try it first, then show the solutionSolution

Ratio of atom to nucleus = 10⁻¹⁰ ÷ 10⁻¹⁵ = 10⁵.
Nucleus on the scale = 100 m ÷ 10⁵ = 10⁻³ m = 1 mm.
So the nucleus is about the size of a peppercorn, and the rest is empty space with a few electrons.

Answer: About 1 mm.

Memory hook: A peppercorn in a stadium.

Reading the gold foil result

Suppose a student repeats the idea of the gold foil experiment with a coin-shaped target hidden behind a curtain and rolls many marbles at it. Most marbles pass straight, a few swerve and one or two bounce back.

Explain what the student can say about the hidden target, using Rutherford's reasoning.

Try it first, then show the solutionSolution

Most marbles pass straight, so the target is mostly empty.
A few swerve, so some small part pushes them aside.
One or two bounce back, so there is a small, hard, massive part in the middle.

Answer: A small, dense, massive centre in a mostly empty region, like the nucleus.

Memory hook: Pass, swerve, bounce: empty, charged, dense.

Tricks to remember Thomson, Rutherford and Bohr Atomic Models

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

Pass, Deflect, Rebound
Most pass: empty space. Some deflect: positive charge concentrated. Very few rebound: tiny dense nucleus.
Use it when: Gold foil table in a 3-mark answer.
Plum pudding
Thomson: plums (electrons) in a pudding (positive sphere). Rutherford: sun (nucleus) and planets (electrons).
Use it when: Remembering the first two models.
Bohr: stairs not a ramp
Electrons stand on fixed steps (shells); they can hop from step to step but cannot stand between steps.
Use it when: Bohr's postulates.
1897, 1911, 1913
Electron (Thomson), nucleus (Rutherford), shells (Bohr) in that order.
Use it when: Dates and order of models.

Thomson, Rutherford and Bohr Atomic Models: solved examples

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

Example 1: Gold foil observations and conclusions · 3 marks · Short answerBoard level

Alpha particles were fired at a thin gold foil. (a) Give three observations. (1)
(b) Give the conclusion drawn from each. (1)
(c) Why was the result a surprise for Thomson's model? (1)

Try it first, then show the solutionSolution

(a) Most α-particles passed straight through; some were deflected through large angles; very few bounced straight back. (1)
(b) Most of the atom is empty space; the positive charge is concentrated in a small region; that region is very small, dense and holds almost all the mass (the nucleus). (1)
(c) Thomson's model, with positive charge spread through the whole atom, predicted that all particles should pass with little deflection. (1)

Answer: (a)-(b) pass: empty space; deflected: concentrated positive charge; rebound: dense nucleus; (c) the model predicted no large deflections.

Example 2: Rutherford's model: features and limitation · 3 marks · Short answerBoard level

(a) State two features of Rutherford's nuclear model. (1)
(b) State the limitation of the model. (1)
(c) How did Bohr resolve it? (1)

Try it first, then show the solutionSolution

(a) A tiny, dense, positively charged nucleus at the centre holds almost all the mass; electrons revolve round the nucleus. (1)
(b) A revolving electron is accelerating, so it should lose energy and spiral into the nucleus; the model cannot explain why atoms are stable. (1)
(c) Bohr proposed that electrons move only in fixed shells, and while in a shell an electron does not lose energy. (1)

Answer: (a) dense nucleus with revolving electrons; (b) cannot explain stability; (c) fixed shells with no energy loss.

Example 3: Scales and numbers in the atom · 5 marks · Long answerHarder board level

(Take atom ≈ 10⁻¹⁰ m, nucleus ≈ 10⁻¹⁵ m, electron charge 1.602 × 10⁻¹⁹ C.)
(a) How many times larger is the atom than its nucleus? (1)
(b) If an atom were 60 m across, what would be the size of its nucleus in mm? (1)
(c) A gold foil is 0.5 mm thick. About how many atoms deep is it? (1)
(d) Out of 4 000 000 α-particles, about how many would bounce back if 1 in 20 000 does? (1)
(e) Find the total charge of 5 × 10⁶ electrons. (1)

Try it first, then show the solutionSolution

(a) 10⁻¹⁰ ÷ 10⁻¹⁵ = 10⁵ times. (1)
(b) 60 m ÷ 10⁵ = 6 × 10⁻⁴ m = 0.6 mm. (1)
(c) 0.5 mm = 5 × 10⁻⁴ m; number of atoms = 5 × 10⁻⁴ ÷ 10⁻¹⁰ = 5 × 10⁶ atoms. (1)
(d) 4 000 000 ÷ 20 000 = 200. (1)
(e) Charge = 5 × 10⁶ × 1.602 × 10⁻¹⁹ = 8.01 × 10⁻¹³ C (negative). (1)

Answer: (a) 10⁵; (b) 0.6 mm; (c) 5 × 10⁶ atoms; (d) 200; (e) 8.01 × 10⁻¹³ C.

Example 4: Electron jumps in Bohr's model · 5 marks · Long answerHarder board level

In a hypothetical atom the energies of the first four shells are, in arbitrary units: K = 0, L = 10, M = 12, N = 13.
(a) How much energy must be absorbed for an electron to jump from K to L, and from L to M? (1)
(b) How much energy is given out when an electron falls from N to K? (1)
(c) Can an electron in K absorb 5 units? Give a reason. (1)
(d) Can an electron stay between L and M? (1)
(e) Why does Bohr's model not predict collapse of the atom, unlike Rutherford's? (1)

Try it first, then show the solutionSolution

(a) K to L: 10 − 0 = 10 units absorbed. L to M: 12 − 10 = 2 units absorbed. (1)
(b) N to K: 13 − 0 = 13 units given out. (1)
(c) No. The only allowed jumps from K absorb exactly 10, 12 or 13 units, the differences between shells; 5 does not match any. (1)
(d) No. An electron can only be in a shell, never between shells. (1)
(e) An electron in a shell does not lose energy, so it does not spiral into the nucleus. (1)

Answer: (a) 10 and 2 units; (b) 13 units emitted; (c) no; (d) no; (e) no energy loss in a shell.

Example 5: How empty is an atom, and how dense is its nucleus? · 4 marks · Case-basedHOTS / competency

Use the rounded sizes atom radius = 10⁻¹⁰ m and nucleus radius = 10⁻¹⁵ m, and treat both as spheres, so volume is proportional to (radius)³. Assume the nucleus holds practically all the mass of the atom.
(a) By what factor is the atom's volume larger than the nucleus's volume? (1)
(b) What fraction of the atom's volume does the nucleus fill? (1)
(c) How many times denser is the nucleus than the atom taken as a whole? (1)
(d) Use your answers to explain why most α-particles passed through the gold foil, and why the few that rebounded were turned back so sharply. (1)

Try it first, then show the solutionSolution

(a) Radius ratio = 10⁻¹⁰ ÷ 10⁻¹⁵ = 10⁵, so volume ratio = (10⁵)³ = 10¹⁵. (1)
(b) Fraction = 1 ÷ 10¹⁵ = 10⁻¹⁵. (1)
(c) Density = mass ÷ volume. The same mass sits in a volume 10¹⁵ times smaller, so the nucleus is about 10¹⁵ times denser than the atom as a whole. (1)
(d) The atom is almost all empty space (the nucleus fills 10⁻¹⁵ of the volume), so most α-particles meet nothing and pass straight through; the rare one that comes close to the dense, positive, massive nucleus is repelled strongly and can be turned back. (1)

Answer: (a) 10¹⁵; (b) 10⁻¹⁵; (c) about 10¹⁵ times; (d) mostly empty space, with a rare close approach to a massive positive nucleus.

Where students lose marks in Thomson, Rutherford and Bohr Atomic Models

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

Mistake: Stating the conclusions of the gold foil experiment without the observations.
Students remember 'nucleus' but not the three observations.
Wrong working
Rutherford found the nucleus.
Right working
Most α-particles passed straight (empty space); some deflected (concentrated positive charge); very few rebounded (small, dense nucleus).
Do this: Write observation and conclusion in pairs.
Usually costs: 1 to 2 marks.
Mistake: Saying electrons are inside the nucleus in Rutherford's model.
Confusion about the structure.
Wrong working
The nucleus contains protons and electrons.
Right working
The nucleus contains protons (and later-discovered neutrons); electrons revolve outside it.
Do this: Electrons are in shells outside the nucleus.
Usually costs: 1 mark.
Mistake: Writing that Rutherford's electrons do not accelerate.
Circular motion at constant speed is thought not to be accelerated.
Wrong working
The electron moves at steady speed so there is no problem.
Right working
The direction changes, so the electron accelerates and should lose energy.
Do this: Velocity includes direction.
Usually costs: 1 mark.
Mistake: Mixing up the order of discoveries.
Dates are not learned.
Wrong working
Rutherford discovered the electron.
Right working
Thomson: electron 1897; Rutherford: nucleus 1911; Bohr: shells 1913; Chadwick: neutron 1932.
Do this: Learn the timeline as a chain.
Usually costs: 1 mark.
Mistake: Allowing an electron to exist between shells.
The idea of continuous energy is natural.
Wrong working
An electron can absorb any amount of energy and settle anywhere.
Right working
It can only absorb or emit the exact energy difference between two shells.
Do this: Remember the stairs.
Usually costs: 1 mark.

Exam tips

Describe the gold foil experiment with the set-up (α-particles, thin gold foil), three observations and three conclusions.
Draw Thomson's and Rutherford's models side by side and label charges.
Use the exact words 'does not lose energy while in a shell' for Bohr's key postulate.
For scale questions, write the ratio as a power of ten first.
Link every model to the problem that the next model solved.

Practice questions on Thomson, Rutherford and Bohr Atomic Models

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

Q1 · 2 marks · Very short answerBoard level
Who discovered the electron, in which year, and how did the experiment show that electrons are present in all atoms?
Hint
Think of cathode rays and the metal and gas used.
Answer
J. J. Thomson discovered the electron in 1897 using cathode rays (1 mark). The rays were the same whatever the cathode metal or gas used, so electrons must be present in all atoms (1 mark).
Q2 · 3 marks · Short answerHarder board level
An atom is about 10⁻¹⁰ m across and its nucleus about 10⁻¹⁵ m. (a) How many times is the atom larger than the nucleus? (b) If an atom were 60 m wide, what would the nucleus measure in mm? (c) How many atoms deep is a foil 0.5 mm thick?
Hint
Ratio of the two sizes, then scale.
Answer
(a) 10⁻¹⁰ ÷ 10⁻¹⁵ = 10⁵ (1 mark). (b) 60 m ÷ 10⁵ = 6 × 10⁻⁴ m = 0.6 mm (1 mark). (c) 5 × 10⁻⁴ m ÷ 10⁻¹⁰ m = 5 × 10⁶ atoms deep (1 mark).
Q3 · 2 marks · Assertion-reasonBoard level
Assertion (A): According to Rutherford's model, atoms should not be stable. Reason (R): An electron moving in a circle is accelerating, so it should lose energy and spiral into the nucleus. Choose: (a) A and R both true, R explains A; (b) A and R both true, R does not explain A; (c) A true, R false; (d) A false, R true.
Hint
Recall why a circular path means acceleration.
Answer
(a). The revolving electron's direction keeps changing, so it accelerates and should radiate energy, causing collapse, which is why the model fails to explain stability.
Q4 · 3 marks · Short answerHarder board level
Compare Thomson's and Rutherford's models of the atom with respect to (a) where the positive charge is, (b) where the electrons are, and (c) which experiment contradicted Thomson's model.
Hint
Think of the plum pudding and the planetary picture.
Answer
(a) Thomson: positive charge spread through the whole atom as a sphere; Rutherford: concentrated in a tiny dense nucleus (1 mark). (b) Thomson: electrons embedded in the positive sphere; Rutherford: electrons revolve round the nucleus (1 mark). (c) The gold foil experiment (alpha scattering) contradicted Thomson's model (1 mark).
Q5 · 4 marks · Case-basedHOTS / competency
In a hypothetical atom the shell energies are, in arbitrary units, K = 0, L = 6, M = 8, N = 9. (a) Find the energy emitted when an electron falls from N to L and from M to K. (b) Can an electron in K absorb 7 units? Give a reason. (c) Which absorbed energies would lift it from K to M? (d) Explain how this idea answers the stability question that Rutherford's model could not.
Hint
Allowed jumps correspond to exact differences.
Answer
(a) N to L: 9 − 6 = 3 units; M to K: 8 − 0 = 8 units (1 mark). (b) No; 7 matches no energy difference from K (the allowed values are 6, 8 and 9) (1 mark). (c) Exactly 8 units (1 mark). (d) While an electron stays in a shell it does not lose energy, so it does not spiral inwards and the atom is stable (1 mark).

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.Who discovered the electron?Show answer
J. J. Thomson, in 1897.
2.What particles were fired at the gold foil?Show answer
Alpha particles.
3.What did the few rebounding particles show?Show answer
That the atom has a very small, dense, positively charged nucleus.
4.Who proposed fixed shells for electrons?Show answer
Niels Bohr, in 1913.
5.About how many times larger than the nucleus is the atom?Show answer
About 10⁵ times.
6.Who discovered the neutron?Show answer
James Chadwick, in 1932.

Thomson, Rutherford and Bohr Atomic Models: frequently asked questions

Why was a thin gold foil used?

Gold can be hammered into extremely thin sheets, so the α-particles meet only a few layers of atoms and most pass through.

Why did most alpha particles pass through the foil?

Most of an atom is empty space, because the nucleus is only about 10⁻⁵ times the atom's size.

What is the nucleus made of?

Protons and neutrons. Rutherford showed the protons; Chadwick found the neutron in 1932.

Did Bohr's model replace Rutherford's?

It improved it by adding fixed shells; the nucleus idea remains. The quantum model later improved Bohr's model.

Are electrons now thought to move in circles?

In the quantum mechanical model they form clouds in regions where they are likely to be found, not fixed orbits.