Discovery of subatomic particles

For most of the nineteenth century, Dalton's atomic theory held that atoms are the smallest, indivisible bits of matter — tiny, solid, featureless balls. Then physicists started passing electricity through gas-filled glass tubes at very low pressure, and the "indivisible" atom began to crack open. A glowing ray bent toward a positive plate, a heavier ray travelled the opposite way, and a mysterious neutral radiation knocked protons out of wax. Each discharge-tube experiment revealed a different piece hiding inside the atom: first the electron, then the proton, and — nearly fifty years later — the neutron. This article walks through the three classic experiments that built the modern picture of a divisible atom, the charge and mass each particle carries, and the numbers examiners expect you to reproduce from memory.

The core insight

Cathode ray, canal ray and Chadwick's beryllium experiments respectively revealed the electron, proton and neutron as universal, quantised, massive constituents of every atom.

The core idea

Electron — cathode rays (Thomson, 1897). A discharge tube with gas at very low pressure and a high voltage across it produces a glow that bends toward a positively charged plate, proving the ray is a stream of negatively charged particles emitted from the cathode. Thomson measured the ratio of charge to mass for these cathode rays:

em=1.7588×1011 C kg−1\frac{e}{m} = 1.7588 \times 10^{11}\ \text{C kg}^{-1}

where: ee is the magnitude of the particle's charge (coulombs) and mm is its mass (kilograms). This ratio came out identical no matter which gas filled the tube or what the electrodes were made of — proof that electrons are a universal ingredient of all atoms, not something specific to one element.

Charge of the electron — Millikan's oil drop (1909). Millikan balanced charged oil droplets between an electric field and gravity and found every droplet carried a whole-number multiple of one fixed charge:

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

Combined with Thomson's e/me/m, this fixes the electron's mass at me=9.1×10−31 kgm_e = 9.1\times10^{-31}\ \text{kg}, about 1/18371/1837 of a hydrogen atom's mass.

Proton — canal rays (Goldstein, 1886). Using a perforated cathode, Goldstein observed rays travelling opposite to cathode rays: positively charged, and — unlike cathode rays — with an e/me/m that changed depending on the gas in the tube, since different gases leave behind different positive ions. The lightest such particle, from hydrogen gas, was named the proton: charge +1.602×10−19 C+1.602\times10^{-19}\ \text{C} and mass 1.6726×10−27 kg1.6726\times10^{-27}\ \text{kg}, about 1836 times heavier than the electron.

Neutron — Chadwick (1932). Atomic masses could not be explained by protons and electrons alone, so a neutral particle was predicted decades before it was found. Chadwick bombarded beryllium with alpha particles and detected a highly penetrating, uncharged radiation that knocked protons out of paraffin wax — the neutron, mass 1.675×10−27 kg1.675\times10^{-27}\ \text{kg}, charge zero, and slightly heavier than the proton.

Figure
Three discharge tube experiments showing cathode rays bending toward a positive plate for the electron canal rays bending the opposite way for the proton and a neutral beam knocking protons from wax for the neutron
Three experiments, three particles Cathode ray tube cathode − anode + + plate electron (−) bends toward + Perforated-cathode tube cathode (holes) proton (+) through holes, opposite way Chadwick’s setup Be + alpha paraffin wax proton knocked out neutron (0) — no bending, but ejects protons
Derivation AFinding the electron's mass from e/m and e

The goal is to combine Thomson's charge-to-mass ratio for cathode rays with Millikan's independently measured value of the electron's charge to calculate the electron's actual mass in kilograms — a single division that historically pinned down, for the first time, exactly how light the electron is compared with an atom, and that let scientists show a proton is roughly 1836 times heavier than an electron.

Step 1. Write down the two independently measured quantities. Thomson gave the ratio e/m=1.7588×1011 C kg−1e/m = 1.7588\times10^{11}\ \text{C kg}^{-1} from how far cathode rays deflect in electric and magnetic fields. Millikan gave the charge itself, e=1.602×10−19 Ce = 1.602\times10^{-19}\ \text{C}, from balancing charged oil drops against gravity.

Step 2. Rearrange the ratio to isolate the mass. Since e/me/m is a ratio, dividing the known charge by the known ratio leaves the mass alone:

me=e(e/m)m_e = \frac{e}{(e/m)}

Step 3. Substitute the two numbers and compute:

me=1.602×10−19 C1.7588×1011 C kg−1=9.11×10−31 kgm_e = \frac{1.602\times10^{-19}\ \text{C}}{1.7588\times10^{11}\ \text{C kg}^{-1}} = 9.11\times10^{-31}\ \text{kg}

Step 4. Interpret the result. This mass is about 1/18371/1837 of a hydrogen atom's mass, confirming that electrons are far too light to account for any atom's mass — almost all of an atom's mass must therefore sit in the heavier, later-discovered proton and neutron.

Worked example

Question. In Millikan's oil drop experiment, charges of 3.2×10−19 C3.2\times10^{-19}\ \text{C}, 4.8×10−19 C4.8\times10^{-19}\ \text{C} and 6.4×10−19 C6.4\times10^{-19}\ \text{C} were measured on three different drops. Show these are consistent with a single fundamental unit of charge, identify it, and state which particle carries that unit of charge.

Step 1. Look for a common factor by dividing each value by a trial unit charge of 1.6×10−19 C1.6\times10^{-19}\ \text{C}:

3.2×10−19=2×(1.6×10−19),4.8×10−19=3×(1.6×10−19),6.4×10−19=4×(1.6×10−19)3.2\times10^{-19} = 2\times(1.6\times10^{-19}), \quad 4.8\times10^{-19} = 3\times(1.6\times10^{-19}), \quad 6.4\times10^{-19} = 4\times(1.6\times10^{-19})

Step 2. Each measured charge is an exact integer multiple (2×2\times, 3×3\times, 4×4\times) of 1.6×10−19 C1.6\times10^{-19}\ \text{C}, with no leftover fraction — a strong sign that charge does not vary continuously but comes packaged in fixed units.

Step 3. State the conclusion. This is Millikan's quantisation of charge: every droplet's charge is nene for a whole number nn, where e=1.6×10−19 Ce = 1.6\times10^{-19}\ \text{C} is the fundamental unit — the magnitude of the charge on one electron. A drop measured at 3.2×10−19 C3.2\times10^{-19}\ \text{C} has simply picked up 2 extra electrons; one at 6.4×10−19 C6.4\times10^{-19}\ \text{C} has picked up 4.

Answer: the fundamental unit of charge is e=1.6×10−19 Ce = 1.6\times10^{-19}\ \text{C}, the magnitude of the electron's charge, and every measured drop charge is a whole-number multiple of it.

Common traps

  • Mixing up cathode rays (electrons, same e/me/m whatever the gas) with anode/canal rays (positive ions, e/me/m depends on the gas) — these two experiments prove the existence of two different particles.
  • Crediting Millikan with discovering the electron — he only measured its charge precisely; Thomson discovered the electron itself via the e/me/m deflection experiment.
  • Forgetting the neutron came much later (1932) than the electron (1897) and proton (1886), because a neutral particle does not respond to electric or magnetic fields and is far harder to detect.
  • Treating the proton and neutron as having exactly equal mass — the neutron is very slightly heavier.

Ask the tutor: Want a step-by-step walk-through of why cathode rays bend toward the positive plate while canal rays bend the other way, or how Millikan balanced a falling oil drop against an electric field — just ask.

Why does the electron's charge-to-mass ratio stay the same for every gas, but the proton's does not

Cathode rays are electrons torn off gas atoms regardless of what the gas is — every element gives the same tiny, light, negatively charged particle, so e/me/m is identical every time. Canal rays, in contrast, are the leftover positive ions of whichever gas was in the tube, and different gases leave ions of very different mass, so their e/me/m changes from gas to gas. Only hydrogen, the lightest gas, gives the lightest possible positive ion — the proton — which is why hydrogen was used to identify it.

If Millikan did not discover the electron, why is his experiment so important

Thomson's cathode ray experiment proved electrons exist and gave only their charge-to-mass ratio, not their charge or mass separately. Millikan's oil drop experiment measured the electron's actual charge for the first time, and just as importantly showed every droplet's charge was a whole-number multiple of one fixed value — proving that electric charge is quantised, not continuous. Once you have both e/me/m and ee separately, dividing gives the electron's mass directly.

Why did scientists suspect a neutral particle existed before Chadwick actually found one

Chemists knew each element's atomic mass from experiments, but the mass of the protons alone (plus the negligible electron mass) could not account for it — most atoms weighed roughly twice what their proton count alone predicted. This gap meant some extra mass had to be present without adding extra positive charge, so a neutral, roughly proton-sized particle was predicted well before 1932. Chadwick's bombardment of beryllium with alpha particles finally produced and detected that particle, the neutron, confirming the prediction.

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