Discovery of Sub-atomic Particles: Electron, Proton & Neutron
The atom stopped being indivisible in the 1850s, when glass tubes and high voltages started glowing. Seventy-four years of discharge tubes, oil drops and beryllium targets produced the three particles this page catalogues — with the discovery credit questions NTA loves.
How Were the Particles Found? — Complete Theory
The electron — from glowing tubes. Pass high voltage through a gas at very low pressure in a discharge tube and the cathode emits rays that make the far wall glow. These cathode rays travel in straight lines (they cast shadows of objects in their path) and behave as negatively charged particles — they deflect toward the positive plate in electric fields. In 1897 J. J. Thomson measured their charge-to-mass ratio by balancing electric and magnetic deflections: e/m = 1.758820 × 10¹¹ C/kg. The stunning part: this value is the same for every gas and every cathode material — the particle is a universal constituent of all matter.
The proton — from the other end. In 1886 E. Goldstein observed that a perforated cathode lets through new rays travelling opposite to cathode rays — anode rays or canal rays, streams of positively charged particles. Unlike cathode rays, their behaviour depends on the gas — each gas gives its own positive ions. The lightest, simplest positive particle comes from hydrogen: the proton, carrying charge +1.6022 × 10⁻¹⁹ C and mass 1.672 × 10⁻²⁷ kg — 1837 times the electron’s mass.
The charge — from falling drops. Thomson had e/m but neither e nor m alone. In 1909 Robert Millikan suspended tiny charged oil droplets between plates, balancing gravity against the electric force. Every droplet’s charge was a whole-number multiple of 1.6022 × 10⁻¹⁹ C — the fundamental charge. Combined: m = e ÷ (e/m) = 9.109 × 10⁻³¹ kg.
The neutron — the invisible one. Atomic masses demanded a neutral, massive nuclear particle, but neutrality made it invisible: no field deflection, no ion trails. In 1932 James Chadwick bombarded beryllium with α-particles and measured the protons knocked out of paraffin wax:
| Particle | Discoverer · year · experiment | Charge (C) | Relative charge | Mass (kg) | Relative mass (u ≈) |
|---|---|---|---|---|---|
| Electron e⁻ | Thomson 1897 (e/m) · Millikan 1909 (charge) | −1.6022 × 10⁻¹⁹ | −1 | 9.109 × 10⁻³¹ | 0 (1/1837) |
| Proton p | E. Goldstein 1886 (canal rays) | +1.6022 × 10⁻¹⁹ | +1 | 1.672 × 10⁻²⁷ | 1.00727 (≈ 1) |
| Neutron n | J. Chadwick 1932 (⁹Be + α) | 0 | 0 | 1.675 × 10⁻²⁷ | 1.00867 (≈ 1) |
Visualising the Hunt & Comparing the Particles
Chaar decades, teen particles — the timeline first, then a comparator that puts the three side by side, bars included.
Particle Comparator
e/m of the electron is the largest of any known particle — tiny mass, full charge. Proton’s e/m is 1837 times smaller; the neutron’s is exactly zero, jo uski 1932 tak chhupne ki wajah hai.
Solved Examples (Step-by-Step)
Data → algebra → identity. Jo chain yahan chalti hai, wahi comparator me live chalti hai.
Two measurements, one mass
Thomson measured e/m = 1.758820 × 10¹¹ C/kg and Millikan measured e = 1.6022 × 10⁻¹⁹ C. Calculate the mass of the electron.
- Formula
m = e ÷ (e/m)— the only way to split the ratio into its parts. - Substitute
m = 1.6022×10⁻¹⁹ ÷ 1.758820×10¹¹ - Result
m = 9.109 × 10⁻³¹ kg— verify on the comparator. - NoteYe isliye possible hua kyunki dono measurements alag experiments se aaye — history ka relay, physics ka division.
me = 9.109 × 10⁻³¹ kg
Why the proton’s e/m is 1837 times smaller
Calculate the e/m ratio of the proton and compare it with the electron’s 1.758820 × 10¹¹ C/kg. (mp = 1.672 × 10⁻²⁷ kg)
- Substitute
e/m(p) = 1.6022×10⁻¹⁹ ÷ 1.672×10⁻²⁷ - Result
e/m(p) = 9.58 × 10⁷ C/kg - Ratio
1.758820×10¹¹ ÷ 9.58×10⁷ = 1836 ≈ 1837 - WhyCharges are equal and opposite — so the e/m ratio is decided purely by mass, and the proton is 1837× heavier. Same charge, smaller e/m.
e/m(p) = 9.58 × 10⁷ C/kg — exactly 1/1837 of the electron’s
Balancing Chadwick’s nuclear reaction
Complete and balance the nuclear reaction Chadwick used: ⁹Be + ⁴He → ? + ?, and explain why the emitted particle was so hard to detect.
- Balance mass9 + 4 = 13 → products carry 13 nucleons total.
- Balance charge4 + 2 = 6 → total nuclear charge 6, beryllium’s own electrons aside.
- Result
⁹Be + ⁴He → ¹²C + ¹n— carbon-12 plus one neutron. - Why hiddenZero charge → no electric/magnetic deflection, no ionisation trail; only its knocking power (ejecting protons from paraffin) exposed it.
⁹Be + ⁴He → ¹²C + ¹n — neutrality was its camouflage
Key Data & Takeaways
Eight lines that solve this topic
Constants: e = 1.6022 × 10⁻¹⁹ C · me = 9.109 × 10⁻³¹ kg · mp = 1.672 × 10⁻²⁷ kg · mn = 1.675 × 10⁻²⁷ kg · 1 u = 1.6605 × 10⁻²⁴ g
- The electron was measured in pieces — e/m by Thomson, charge by Millikan, mass by division; credit questions live on this relay.
- Universality is one-directional — cathode rays identical for all gases; canal rays differ per gas. The most-tested swap in this section.
- e/m is a mass meter in disguise — equal-magnitude charges mean the ratio ranks particles purely by mass.
- Neutrality delayed the neutron — invisible to every field detector until Chadwick’s knockout experiment in 1932.
FAQs
Who discovered the electron, proton and neutron?
The electron emerged from cathode ray discharge tube experiments, with J. J. Thomson measuring its charge-to-mass ratio in 1897 and Robert Millikan determining its charge in 1909. The proton was identified by E. Goldstein through anode (canal) rays in 1886, with hydrogen giving the lightest positive particle. The neutron was discovered by James Chadwick in 1932 from beryllium bombarded with alpha particles.
Why is the e/m ratio of cathode rays the same for all gases?
Because cathode ray particles come from the cathode itself and are identical whatever gas or electrode material is used — proving the electron is a universal constituent of all matter. In contrast, the e/m ratio of anode (canal) rays depends on the gas, because the positive particles are simply ionised gas atoms or molecules, which differ from gas to gas.
How was the charge of the electron measured?
Robert Millikan measured it in 1909 using his oil drop experiment: tiny charged oil droplets were suspended between plates by balancing gravity against the electric force. The charges on droplets were always whole-number multiples of 1.6022 × 10⁻¹⁹ C — the charge of one electron. Dividing by Thomson's e/m ratio gave the electron's mass, 9.109 × 10⁻³¹ kg.
How was the neutron discovered and why did it take so long?
In 1932 James Chadwick bombarded beryllium with alpha particles — ⁹Be + ⁴He → ¹²C + n — and found an uncharged radiation that knocked protons out of paraffin. It took until 1932 because the neutron carries no charge: it is undeflected by electric or magnetic fields and leaves no ion trails, so every charge-based detection method missed it. Its mass, 1.675 × 10⁻²⁷ kg, is slightly greater than the proton's.
Practice Questions (With Solutions)
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QCC Notes — Class 11 Chemistry
Strictly NCERT-aligned notes for JEE Main & NEET, prepared by QCC Notes (Padho Likho JEE). Content follows the latest NCERT edition and current NTA exam pattern.