Subatomic Particles: Electron, Proton, Neutron

Atoms are not indivisible. Three subatomic particles build every atom in the periodic table. These are the electron, proton, and neutron. Dalton's Atomic Theory treated atoms as tiny solid spheres. But late-19th and early-20th century experiments cracked that sphere open. They revealed structure inside. This article follows the experimental trail. Cathode-ray tubes revealed the electron. Oil-drop measurements pinned its charge. Gold-foil scattering revealed a dense positive nucleus. Neutron bombardment revealed the last missing piece. Each discovery is a quantitative story. It gives a charge-to-mass ratio, a charge, a mass. Each is not just a name. By the end you will be able to state the charge and relative mass of each particle. You will explain how each was measured. You will see how these three numbers set up The Nuclear Atom. You will also see how they set up the definitions of Atomic Number and Mass Number.

Key takeaways

The electron was discovered by J. J. Thomson (1897) using cathode-ray tubes; it carries charge $-1$ and negligible mass.
Robert Millikan's oil-drop experiment (1909–1913) measured the electron's charge directly: $1.602 \times 10^{-19}\ \text{C}$ .
Ernest Rutherford's gold-foil experiment (1911) proved atoms have a tiny, dense, positively charged nucleus containing the protons.
The proton has charge $+1$ and a mass of about 1 atomic mass unit (amu); the neutron, discovered by James Chadwick (1932), has no charge and a mass just slightly larger than the proton's.
Relative masses: electron $\approx 1/1836$ amu, proton $\approx 1$ amu, neutron $\approx 1$ amu — this near-equality of proton and neutron mass is what makes mass number a useful whole-number label.
Key idea: every atom's identity and behavior trace back to just three numbers per particle — charge, mass, and count — and those numbers were pried out of nature one careful experiment at a time.

The core insight

An atom is mostly empty space organized around a minuscule, massive, positively charged nucleus (protons + neutrons), surrounded by a cloud of much lighter, negatively charged electrons. Charge balance and mass accounting for these three particles explain the entire structure of the periodic table.

The physical picture

Nineteenth-century chemists had no way to look inside an atom, so Dalton's model — atoms as featureless, indestructible balls — worked well enough to explain mass ratios in reactions. But by the 1890s, physicists were passing electricity through evacuated glass tubes and noticing something atoms were not supposed to have: pieces that could be pulled off. A cathode-ray tube is a sealed glass tube with most of the air removed and two metal electrodes connected to a high-voltage source. When voltage is applied, a glowing beam ("cathode ray") streams from the negative electrode (cathode) toward the positive electrode (anode). J. J. Thomson showed this beam:

travels in straight lines and casts sharp shadows,
is deflected by both electric and magnetic fields, always toward the positive plate or in the direction consistent with negative charge,
behaves identically no matter what metal the cathode is made of, or what trace gas remains in the tube.

That last point was the key. If the beam's properties never depend on what material produced it, the beam cannot be a property of that material — it must be a universal constituent of all matter. Thomson called these particles "corpuscles"; we call them electrons. By measuring how much the beam deflects under a known electric field and a known magnetic field, Thomson could solve for the ratio of charge to mass, $e/m$ , without needing to know $e$ or $m$ separately. He found $e/m$ to be more than a thousand times larger than the equivalent ratio for hydrogen ions in electrolysis — meaning the electron is either very highly charged or, far more plausibly given other evidence, extremely light. The missing number was $e$ itself. Robert Millikan supplied it with the oil-drop experiment. Tiny oil droplets, sprayed into a chamber and given a static charge by friction or by ionizing radiation, fall under gravity between two horizontal plates. Millikan adjusted the voltage on the plates until the electric force on a drop exactly balanced gravity, suspending the drop in midair. From the balancing voltage, the drop's radius (found from its terminal falling speed), and the known density of oil, he calculated the charge on each drop. Every drop's charge came out to be a whole-number multiple of a single smallest value:

e = 1.602 \times 10^{-19}\ \text{C}

That smallest value is the fundamental unit of electric charge — the magnitude of the charge on one electron. Combining this with Thomson's $e/m$ ratio gives the electron's mass directly, and it is astonishingly small — about 1/1836 the mass of a hydrogen atom. If atoms contain negative electrons but are overall electrically neutral, they must also contain positive charge — and since electrons are so light, that positive charge must carry essentially all the atom's mass. Thomson's own "plum pudding" model imagined the positive charge as a diffuse cloud with electrons studded through it, like fruit in a pudding. Ernest Rutherford tested this picture in 1911 by firing a beam of alpha particles (positively charged, relatively heavy particles from a radioactive source) at an extremely thin sheet of gold foil, with a fluorescent screen surrounding the foil to detect where the particles ended up. If Thomson's diffuse-pudding model were correct, the positive charge would be too spread out to deflect a heavy alpha particle by much — nearly all particles should sail through with only slight deflection. That is not what Rutherford saw. Most alpha particles did pass straight through, but a small fraction bounced back at large angles — some almost straight back toward the source. Rutherford's own description captured the shock: it was "almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you." The only way to explain occasional sharp deflections is if the positive charge (and nearly all the mass) is concentrated in an extremely small volume — a nucleus — with mostly empty space around it, through which most alpha particles pass undisturbed. This nucleus contains the proton, a particle with charge $+1$ (equal in magnitude, opposite in sign, to the electron's charge) and a mass close to that of a hydrogen atom. But proton mass alone could not account for the total mass of most atoms. A helium nucleus, for instance, carries charge $+2$ (two protons) yet weighs about four times a hydrogen atom, not two. Something electrically neutral had to be adding mass without adding charge. James Chadwick found it in 1932: bombarding beryllium with alpha particles released a penetrating, uncharged radiation, which he showed consisted of particles with almost exactly the proton's mass but zero charge — the neutron.

Figure

Cathode-ray tube: the beam bends toward the positive plate, revealing negative charge

Figure

Rutherford's gold-foil result: most alpha particles pass through; a few bounce back off a dense nucleus

The core relations

q_{\text{electron}} = -1.602 \times 10^{-19}\ \text{C} = -1 \ (\text{relative units})

q_{\text{proton}} = +1.602 \times 10^{-19}\ \text{C} = +1 \ (\text{relative units})

q_{\text{neutron}} = 0

where: charges are usually quoted in relative units (proton $= +1$ , electron $= -1$ ) because chemists care about charge balance far more often than the absolute coulomb value.

m_{\text{proton}} \approx m_{\text{neutron}} \approx 1\ \text{amu}, \qquad m_{\text{electron}} \approx \frac{1}{1836}\ \text{amu}

where: 1 amu (atomic mass unit) $= 1.6605 \times 10^{-27}\ \text{kg}$ , defined so that a carbon-12 atom has a mass of exactly 12 amu.

\frac{e}{m_e} = 1.759 \times 10^{11}\ \text{C/kg}

where: this is Thomson's measured charge-to-mass ratio for the electron, obtained from electric- and magnetic-field deflection of the cathode-ray beam, independent of any knowledge of $e$ or $m_e$ alone.

\text{atomic mass (amu)} \approx (\text{number of protons}) + (\text{number of neutrons})

where: the electron's mass is small enough (about 0.05% of a proton's) that it is dropped from this approximation entirely — the topic taken up properly in Atomic Number and Mass Number.

Derivation AFrom Thomson's e/m ratio and Millikan's charge to the electron's mass

Goal. Combine the two independently measured quantities — Thomson's charge-to-mass ratio $e/m_e$ and Millikan's elementary charge $e$ — to obtain the electron's mass $m_e$ , and check that it is indeed a tiny fraction of a proton's mass.

Step 1. Thomson's cathode-ray experiment measures the ratio $e/m_e$ by balancing electric and magnetic deflection forces on the beam; no assumption about $e$ or $m_e$ individually is needed, only the geometry of the tube and the applied field strengths. His result, refined by later measurements, is

\frac{e}{m_e} = 1.759 \times 10^{11}\ \text{C/kg}

Step 2. Millikan's oil-drop experiment measures $e$ directly, by finding the smallest common charge increment among many suspended droplets:

e = 1.602 \times 10^{-19}\ \text{C}

Step 3. Divide $e$ by the ratio $e/m_e$ to isolate $m_e$ :

m_e = \frac{e}{(e/m_e)} = \frac{1.602 \times 10^{-19}\ \text{C}}{1.759 \times 10^{11}\ \text{C/kg}}

m_e = 9.11 \times 10^{-31}\ \text{kg}

Step 4. Convert to atomic mass units using $1\ \text{amu} = 1.6605 \times 10^{-27}\ \text{kg}$ :

m_e = \frac{9.11 \times 10^{-31}\ \text{kg}}{1.6605 \times 10^{-27}\ \text{kg/amu}} = 5.49 \times 10^{-4}\ \text{amu}

Step 5 (limiting-case check). Compare this to a proton's mass of essentially 1 amu:

\frac{m_{\text{proton}}}{m_e} = \frac{1}{5.49\times 10^{-4}} \approx 1822

This is close to the accepted ratio of about 1836 (the small gap here comes from using rounded constants); the key qualitative result — that the electron is roughly 1800 times lighter than a proton — is confirmed independent of exactly which decimal values are carried through. As the proton-to-electron mass ratio grows without bound in this comparison, electron mass becomes negligible in any total-mass bookkeeping, which is exactly why atomic mass is computed from protons and neutrons alone.

Worked examples

Example 1 — Finding total charge and approximate mass of a fluorine atom. A neutral fluorine atom contains 9 protons, 9 electrons, and 10 neutrons (this is fluorine-19, the only naturally significant isotope). Find (a) the net charge of the atom and (b) its approximate mass in amu. (a) Net charge. Each proton contributes $+1$ and each electron contributes $-1$ in relative units:

\text{net charge} = (9)(+1) + (9)(-1) = 0

The atom is neutral, as expected — a neutral atom always has equal numbers of protons and electrons. (b) Approximate mass. Neutrons and protons each contribute about 1 amu; electrons are dropped as negligible:

\text{mass} \approx (9\ \text{protons})(1\ \text{amu}) + (10\ \text{neutrons})(1\ \text{amu}) = 19\ \text{amu}

This matches the mass number of fluorine-19 directly — the whole-number "19" you see on an isotope label is exactly this proton-plus-neutron count.

Example 2 — Reconstructing an ion's composition from charge and mass number. An ion has a mass number of 24, a charge of $+2$ , and contains 12 neutrons. Find the number of protons and electrons. Step 1: protons from mass number and neutron count. Mass number is protons plus neutrons:

24 = (\text{protons}) + 12 \implies \text{protons} = 24 - 12 = 12

Step 2: electrons from charge. A $+2$ charge means the atom has lost 2 electrons relative to its proton count:

\text{net charge} = (\text{protons})(+1) + (\text{electrons})(-1)

+2 = (12)(+1) + (\text{electrons})(-1)

\text{electrons} = 12 - 2 = 10

So this ion has 12 protons, 12 neutrons, and 10 electrons — it is a magnesium ion, $\text{Mg}^{2+}$ , since an element with 12 protons is magnesium. Sanity check (asymmetric, non-round numbers by design): protons (12) $\neq$ neutrons would also work fine here — they happen to be equal for this particular isotope, but nothing in the method required that; rerunning the same two equations with, say, 13 neutrons and the same $+2$ charge would still give 12 protons and 10 electrons, showing the charge equation and the mass-number equation are independent checks, not accidental coincidences of round numbers.

Limiting cases & common traps

Situation	What actually happens	Common mistake
Counting mass of an ion	Electron mass is dropped; use protons + neutrons only	Trying to subtract/add electron mass "for accuracy" — the effect is under 0.03% per electron and not tracked in this tier
Neutral atom charge check	Protons = electrons always, net charge = 0	Forgetting that neutral atoms still contain charged particles internally — neutral means balanced, not chargeless
Identifying an element from a proton count	Proton count alone fixes identity (this is atomic number)	Using mass number or neutron count to guess the element — only proton count is diagnostic
Very light isotopes (e.g., hydrogen-1)	Zero neutrons is allowed; mass number equals proton count	Assuming every atom must have neutrons — ordinary hydrogen has none
Ions with large charge	Electron count differs sharply from proton count	Forgetting the sign convention — losing electrons gives a positive ion, not negative

The chemist behind it

Ernest Rutherford began as a physicist studying radioactivity, but his gold-foil scattering result is arguably the single most important piece of evidence in the entire history of the atomic model. He did not personally fire the alpha particles at the foil — that painstaking, months-long counting work was carried out by his students Hans Geiger and Ernest Marsden, using a hand-counted scintillation screen in a darkened room. Rutherford's contribution was recognizing what the rare backward-scattering events meant: a concentrated nucleus, not a diffuse pudding. Rutherford had already won the Nobel Prize in Chemistry in 1908 — for earlier work on radioactive decay — before he did the gold-foil experiment that overturned atomic theory. He famously joked that this made him a chemist by Nobel citation but a physicist by training, remarking that all science is "either physics or stamp collecting." His nuclear model, refined through the 1910s, set the stage directly for The Nuclear Atom and for defining what would become atomic number.

Origins · 1 minErnest Rutherford and how this was discovered

Ernest Rutherford

1871–1937 · New Zealand-born physicist who directed the gold-foil scattering experiments at the University of Manchester, discovered the atomic nucleus, and later became the first to split an atom deliberately via nuclear reaction, at the Cavendish Laboratory in Cambridge

Ask the tutor: If an atom loses two electrons to become a $2+$ ion, does its mass number change — and why or why not?

Why don't we just use coulombs and kilograms for everyday chemistry instead of relative charge and amu?

The actual values ( $1.602\times10^{-19}\ \text{C}$ , $1.673\times10^{-27}\ \text{kg}$ , and so on) are inconveniently small for routine bookkeeping. Relative units — proton $= +1$ , electron $= -1$ , mass $\approx 1$ amu per nucleon — let you track charge balance and approximate mass with small whole numbers, which is all that most chemical reasoning at this level requires. The absolute values matter only when you need real physical quantities, such as converting to grams via Avogadro's Number-scale calculations elsewhere.

If the neutron has almost the same mass as the proton, why did it take decades longer to discover?

The neutron carries no charge, so it does not respond to electric or magnetic fields the way cathode rays or alpha particles do. Nearly every early detection method — deflection in fields, ionization trails, electroscopes — relies on a particle being charged. Chadwick had to infer the neutron's existence indirectly, from the momentum and energy of particles it knocked loose from a target, rather than by observing it directly.

Does the "plum pudding" model deserve to be called wrong, given that it correctly predicted electrons exist?

It correctly introduced the electron and neutrality, but its central claim — that positive charge is spread diffusely through the whole atomic volume — was falsified quantitatively by Rutherford's large-angle scattering. A diffuse charge distribution cannot deliver the strong, localized electric field needed to bounce a fast alpha particle almost straight backward, so the model was replaced, not just refined.

Why is the proton's charge exactly opposite the electron's charge, given that they were discovered separately and are very different particles?

This is an experimentally verified fact (charge neutrality of ordinary matter to extraordinary precision) rather than something derived from first principles at this level. It is one of the deep symmetries of nature; explaining why it holds exactly belongs to particle physics, well beyond this tier, but chemists rely on it constantly when balancing charge in atoms and ions.

How do these three particles connect to the periodic table itself?

The number of protons in a nucleus is the single property that defines which element an atom is — this proton count is the atomic number, and arranging elements by atomic number is exactly what builds the modern periodic table. That precise connection, along with how neutrons produce isotopes of the same element, is developed in Atomic Number and Mass Number.