Chemistry · Structure of Atom

Thomson & Rutherford Atomic Models

Once the electron and proton were known to exist, the question became how they are arranged inside the atom. NCERT Unit 2 (§2.2) answers this through two competing pictures: J. J. Thomson's plum-pudding model of 1898 and the nuclear model that Ernest Rutherford built from his α-particle scattering experiment. The same section fixes the meaning of atomic number, mass number, isotopes, and isobars. This subtopic is a recurring NEET source of statement-based and species-counting questions.

Thomson's plum-pudding model

J. J. Thomson, who had identified the electron in 1897, proposed in 1898 the first concrete model of how charge is arranged within an atom. He pictured the atom as a sphere of radius approximately $10^{-10}\ \text{m}$ in which the positive charge is distributed uniformly, with the electrons embedded into it in the arrangement that gives the most stable electrostatic configuration. Because the negative electrons are evenly seeded through a continuous positive cloud, the atom as a whole is electrically neutral.

The model is known by several food-based names — the plum-pudding model, the raisin-pudding model, or the watermelon model. The image is of a pudding or watermelon of positive charge with the electrons sitting in it like plums or seeds. A point that NEET frequently tests is that, in this model, the mass of the atom is assumed to be spread uniformly over the whole atom rather than concentrated anywhere.

Figure 1 uniform (+) charge cloud electrons embedded

Figure 1. Thomson's atom: electrons (the "plums") set into a continuous, uniformly distributed sphere of positive charge. Mass and positive charge are spread evenly, so the model could explain neutrality but not later scattering data.

Thomson's model successfully explained the overall electrical neutrality of the atom and was an important conceptual advance, earning him the 1906 Nobel Prize in Physics for his work on the conduction of electricity by gases. However, it could not survive the experimental tests that followed. The decisive evidence came from a study of how positively charged α-particles behave when fired at thin metal foil.

It is worth being precise about what the model did and did not commit to. By insisting that both the positive charge and the mass were smeared out over the whole sphere, Thomson left the atom with no dense interior anywhere — there was no region capable of exerting a large, sudden force on a fast-moving projectile. This is the feature that the scattering experiment was about to test, because a diffuse charge distribution and a concentrated one make sharply different predictions for how an incoming particle is deflected.

Rutherford's α-scattering experiment

By the early twentieth century, radioactivity had been discovered by Henri Becquerel and developed by Marie Curie, Pierre Curie, Rutherford, and Frederick Soddy. Three kinds of emission — $\alpha$, $\beta$, and $\gamma$ rays — were known. Rutherford found that α-rays consist of high-energy particles carrying two units of positive charge and four units of atomic mass, and concluded they are helium nuclei: combining $\alpha$-particles with two electrons yields helium gas, $\ce{He^2+ + 2e^- -> He}$.

Rutherford and his students Hans Geiger and Ernest Marsden bombarded a very thin gold foil (thickness about $100\ \text{nm}$) with a stream of high-energy α-particles from a radioactive source. A circular fluorescent zinc sulphide screen surrounded the foil, so that wherever an α-particle struck the screen, a tiny flash of light was produced and could be counted. The geometry let the experimenters record both how many particles passed straight through and how many were deflected to large angles.

Figure 2 α source gold foil most: undeflected few: small angle ~1 in 20,000: ~180°

Figure 2. Schematic of the gold-foil experiment. The overwhelming majority of α-particles pass through undeflected (teal); a few are deflected through small angles (purple); about one in twenty thousand rebound through nearly $180^\circ$ (coral).

Observations and conclusions

On Thomson's model the gold atom's mass and positive charge would be spread evenly across each atom, so the energetic α-particles should pass through this diffuse distribution with at most small, gentle changes of direction. The results, however, were quite unexpected and are summarised below alongside the inference Rutherford drew from each.

ObservationConclusion drawn
Most α-particles passed through the foil undeflected. Most of the space inside the atom is empty.
A small fraction of α-particles were deflected through small angles. The positive charge is not spread throughout the atom (as Thomson presumed) but is concentrated in a very small volume that repels the positive α-particles.
A very few α-particles (about 1 in 20,000) bounced back, deflected by nearly $180^\circ$. An enormous repulsive force acted on these particles, so the dense positive core (the nucleus) occupies a negligibly small volume compared with the whole atom.

The size contrast that follows from the third conclusion is dramatic. The radius of the atom is about $10^{-10}\ \text{m}$, while the radius of the nucleus is about $10^{-15}\ \text{m}$. NCERT offers a memorable scaling: if a cricket ball represented the nucleus, the atom would have a radius of about 5 km. Almost all of the atom is empty space, with its mass and positive charge packed into a core five orders of magnitude smaller in radius.

NEET Trap

Do not say "most particles bounced back"

The headline result is that most α-particles passed straight through — that is what proved the atom is mostly empty. Only a very few (about 1 in 20,000) bounced back, and it was these rare large-angle deflections that revealed the tiny dense nucleus. Confusing the frequency of each outcome inverts the entire argument.

Frequent + undeflected → empty space. Rare + back-scattered → concentrated positive nucleus.

Rutherford's nuclear model

From these observations and conclusions Rutherford proposed the nuclear model of the atom. Three features define it. First, the positive charge and most of the mass of the atom are densely concentrated in an extremely small region that Rutherford named the nucleus. Second, the nucleus is surrounded by electrons that revolve around it at very high speed in circular paths called orbits. Third, the electrons and the nucleus are held together by electrostatic forces of attraction.

The picture deliberately resembles the solar system: the nucleus plays the role of the massive Sun while the electrons behave like the lighter, orbiting planets. This planetary analogy is intuitive, but as the closing section shows, it is precisely the orbital motion of the charged electron that the classical model could not reconcile with atomic stability.

Figure 3 + nucleus electrons in orbits

Figure 3. Rutherford's nuclear atom. Almost all the mass and the entire positive charge sit in a minute central nucleus (radius $\sim 10^{-15}\,\text{m}$); electrons revolve in circular orbits within a region $\sim 10^{-10}\,\text{m}$ across. Drawn far larger than scale.

Thomson vs Rutherford compared

The two models differ on every structural question that matters: where the positive charge sits, where the mass sits, and whether the electrons are static or moving. The table contrasts them directly.

FeatureThomson (plum-pudding)Rutherford (nuclear)
Distribution of positive chargeSpread uniformly through the whole sphereConcentrated in a tiny central nucleus
Distribution of massSpread uniformly over the atomAlmost entirely in the nucleus
ElectronsEmbedded (static) in the positive cloudRevolving in circular orbits around the nucleus
Empty spaceEffectively none; atom is filled chargeMost of the atomic volume is empty
Explains scattering data?No — predicts only small deflectionsYes — explains large-angle back-scattering
AnalogyPlum pudding / watermelonSolar system (Sun and planets)
Build on this

The instability flaw of Rutherford's model and the spectral evidence that follows feed straight into developments leading to the Bohr model.

Atomic number and mass number

With a nucleus established, the identity of an atom can be defined by counting its nucleons. The positive charge of the nucleus is due to the protons, whose charge is equal and opposite to that of the electron. The atomic number $Z$ is the number of protons in the nucleus, and for electrical neutrality it also equals the number of electrons in a neutral atom.

$$Z = \text{number of protons} = \text{number of electrons (in a neutral atom)}$$

For example, hydrogen has $Z = 1$ and sodium has $Z = 11$. The mass of the nucleus is due to both protons and neutrons, which are collectively called nucleons. The total number of nucleons is the mass number $A$.

$$A = \text{number of protons }(Z) + \text{number of neutrons }(n)$$

The composition of any atom is written using the element symbol with the mass number as a left superscript and the atomic number as a left subscript, $\ce{^{A}_{Z}X}$. The number of neutrons is always $A - Z$.

Worked Example

Calculate the number of protons, neutrons, and electrons in $\ce{^{80}_{35}Br}$, and assign the symbol to a species with 18 electrons, 16 protons, and 16 neutrons.

For $\ce{^{80}_{35}Br}$: $Z = 35$, $A = 80$, neutral species, so protons $=$ electrons $= 35$ and neutrons $= 80 - 35 = 45$.

For the second species: protons $= 16$, so the element is sulphur (S). Mass number $A = 16 + 16 = 32$. Electrons (18) exceed protons (16) by 2, so it is an anion of charge $2-$: $\ce{^{32}_{16}S^{2-}}$.

NEET Trap

Charge changes electrons, not protons or neutrons

When a species is an ion, decide whether protons outnumber electrons (cation) or electrons outnumber protons (anion) by the magnitude of the charge. The number of neutrons is always $A - Z$, and the number of protons is always $Z$ — neither depends on the charge.

Neutrons $= A - Z$ always · electrons $= Z -$ (charge) for a cation, $Z +$ (charge) for an anion.

Isotopes and isobars

Two classification terms follow naturally from the $\ce{^{A}_{Z}X}$ notation. Isotopes are atoms with the same atomic number but different mass numbers; they differ only in the number of neutrons in the nucleus. Isobars are atoms with the same mass number but different atomic numbers, so they belong to different elements.

ClassSameDifferentExamples
Isotopes Atomic number $Z$ Mass number $A$ (i.e. neutron count) $\ce{^{1}_{1}H}$, $\ce{^{2}_{1}H}$, $\ce{^{3}_{1}H}$ · $\ce{^{12}_{6}C}$, $\ce{^{13}_{6}C}$, $\ce{^{14}_{6}C}$ · $\ce{^{35}_{17}Cl}$, $\ce{^{37}_{17}Cl}$
Isobars Mass number $A$ Atomic number $Z$ (different elements) $\ce{^{14}_{6}C}$ and $\ce{^{14}_{7}N}$

Hydrogen is the textbook illustration of isotopy: 99.985% of hydrogen atoms are protium $\ce{^{1}_{1}H}$ (one proton, no neutron); a small fraction are deuterium $\ce{^{2}_{1}H}$ or D (one proton, one neutron, 0.015%); and trace amounts are tritium $\ce{^{3}_{1}H}$ or T (one proton, two neutrons). Chlorine occurs as $\ce{^{35}_{17}Cl}$ and $\ce{^{37}_{17}Cl}$, and carbon as the isotopes with 6, 7, and 8 neutrons.

A central consequence is that all isotopes of a given element show the same chemical behaviour. Chemical properties are controlled by the number of electrons, which is fixed by the number of protons; the number of neutrons in the nucleus has very little effect. This is exactly why NEET statements asserting that "all isotopes of an element show the same chemical properties" are marked correct.

Drawbacks of Rutherford's model

Rutherford's model resembles a miniature solar system, with the nucleus as the massive Sun and the electrons as orbiting planets. The Coulomb force between an electron and the nucleus, $F = k\,q_1 q_2 / r^2$, is mathematically of the same inverse-square form as the gravitational force that governs the planets, so it is tempting to expect equally stable orbits. The analogy, however, breaks down on one decisive point.

A body moving in a circular orbit is constantly accelerated, even at constant speed, because its direction of motion keeps changing. According to Maxwell's electromagnetic theory, an accelerated charged particle must continuously emit electromagnetic radiation — a feature absent for the uncharged planets. An orbiting electron would therefore radiate away energy drawn from its motion, its orbit would steadily shrink, and calculations show it should spiral into the nucleus in only about $10^{-8}\ \text{s}$.

Since this collapse does not happen and atoms are in fact stable, the Rutherford model cannot explain the stability of an atom. This single failure is the launch point for the quantum picture that followed.

The model is also silent about how electrons are distributed around the nucleus and about the discrete line spectra emitted by atoms. A planetary atom radiating continuously as it collapsed would be expected to emit a continuous spread of frequencies, yet experiments showed that excited atoms emit light only at sharp, characteristic wavelengths. Rutherford's model offered no way to account for this discreteness, and no rule to fix the size of the electron's orbit. These shortcomings, especially the stability problem, motivated Niels Bohr to introduce quantised stationary orbits in which an electron does not radiate — the subject of the next subtopic.

Quick Recap

Five lines to retain

  • Thomson (1898): uniform sphere of positive charge with electrons embedded; mass spread evenly; explains neutrality only.
  • Rutherford's α-scattering: most pass undeflected (empty atom), few deflected (concentrated +), ~1 in 20,000 rebound (tiny dense nucleus).
  • Nuclear model: mass and positive charge in a tiny nucleus ($\sim10^{-15}$ m); electrons orbit; held by electrostatic attraction.
  • Counting: $Z =$ protons $=$ electrons (neutral); $A = Z + n$; neutrons $= A - Z$ always; isotopes share $Z$, isobars share $A$.
  • Fatal flaw: an orbiting electron is accelerated, must radiate (Maxwell), and would spiral in within $\sim10^{-8}$ s — cannot explain atomic stability.

NEET PYQ Snapshot — Thomson & Rutherford Atomic Models

Statement-based and species-counting questions drawn from this section, with real NEET years.

NEET 2025

Dalton's atomic theory could not explain which of the following?

  1. Law of gaseous volume
  2. Law of conservation of mass
  3. Law of constant proportion
  4. Law of multiple proportion
Answer: (1) Law of gaseous volume

Dalton's theory accounted for the laws of chemical combination (conservation of mass, constant and multiple proportions) but not the law of gaseous volumes.

NEET 2023

Select the correct statements: A. Atoms of all elements are composed of two fundamental particles. B. The mass of the electron is $9.10939\times10^{-31}$ kg. C. All isotopes of a given element show the same chemical properties. D. Protons and electrons are collectively known as nucleons. E. Dalton's atomic theory regarded the atom as the ultimate particle of matter.

  1. B, C and E only
  2. A, B and C only
  3. C, D and E only
  4. A and E only
Answer: (1) B, C and E only

A is wrong (atoms have three fundamental particles); D is wrong (nucleons are protons and neutrons, not electrons). B, C, and E are correct.

NEET 2021

Tritium, a radioactive isotope of hydrogen, emits which of the following particles?

  1. Neutron (n)
  2. Beta ($\beta^-$)
  3. Alpha ($\alpha$)
  4. Gamma ($\gamma$)
Answer: (2) Beta ($\beta^-$)

Of hydrogen's three isotopes — protium $\ce{^{1}_{1}H}$, deuterium $\ce{^{2}_{1}H}$, tritium $\ce{^{3}_{1}H}$ — only tritium is radioactive and emits low-energy $\beta^-$ particles.

NEET 2020

The numbers of protons, neutrons and electrons in $\ce{^{175}_{71}Lu}$, respectively, are:

  1. 104, 71 and 71
  2. 71, 71 and 104
  3. 175, 104 and 71
  4. 71, 104 and 71
Answer: (4) 71, 104 and 71

$Z = 71$ gives protons $= 71$ and (neutral atom) electrons $= 71$; neutrons $= A - Z = 175 - 71 = 104$.

FAQs — Thomson & Rutherford Atomic Models

Common doubts on the two models, the scattering experiment, and atomic bookkeeping.

Why is Thomson's model called the plum-pudding model?

In Thomson's 1898 model the atom is a sphere of radius about 10^-10 m in which the positive charge is spread uniformly, and the electrons are embedded inside it to give the most stable electrostatic arrangement. The picture resembles a pudding or watermelon of positive charge with the electrons sitting in it like plums or seeds, which is why it is called the plum-pudding, raisin-pudding, or watermelon model. An important feature is that the mass of the atom is assumed to be distributed uniformly over the whole atom.

What were the three key observations of Rutherford's alpha-scattering experiment?

When a beam of alpha particles was directed at a thin gold foil: (i) most of the alpha particles passed straight through the foil undeflected; (ii) a small fraction were deflected through small angles; and (iii) a very few, about 1 in 20,000, bounced back, being deflected by nearly 180 degrees. These results were unexpected on the basis of Thomson's model, which predicted only small, gentle deflections.

What conclusions did Rutherford draw about the structure of the atom?

Rutherford concluded that most of the space in the atom is empty, since most alpha particles passed through undeflected; that the positive charge is not spread throughout the atom but is concentrated in a very small, dense region that repels and deflects the positively charged alpha particles; and that this region, the nucleus, occupies a negligibly small volume compared with the atom. The radius of the atom is about 10^-10 m while that of the nucleus is about 10^-15 m.

What is the difference between isotopes and isobars?

Isotopes are atoms with the same atomic number (Z) but different mass numbers (A), differing only in the number of neutrons; for example protium, deuterium, and tritium are the three isotopes of hydrogen. Isobars are atoms of different elements with the same mass number but different atomic numbers, for example carbon-14 and nitrogen-14. Isotopes of an element show the same chemical behaviour because chemical properties are governed by the number of electrons, which equals the number of protons.

Why does Rutherford's model fail to explain the stability of the atom?

In Rutherford's model the electron revolves around the nucleus in a circular orbit, so it is constantly accelerated because its direction of motion keeps changing. According to Maxwell's electromagnetic theory, an accelerated charged particle must continuously emit electromagnetic radiation. The electron would therefore lose energy, its orbit would steadily shrink, and calculations show it should spiral into the nucleus in only about 10^-8 s. Since atoms are in fact stable, the model cannot account for atomic stability.

How do you find the number of protons, neutrons, and electrons from the symbol AZX?

The atomic number Z (subscript) equals the number of protons, and the mass number A (superscript) equals protons plus neutrons, so the number of neutrons is A - Z. The number of neutrons is always A - Z whether the species is neutral or an ion. For a neutral atom the number of electrons equals Z; for a cation the electrons are fewer than Z by the charge, and for an anion they are greater than Z by the charge.