Measuring the Electron: e/m and e
Crossed fields and floating drops measure the electron.
Lesson at a glanceShow the key points
- Undeflected beam: v = E/B; with ½mv² = eV, e/m = E²/2VB² (41.2) = 1.76 × 10¹¹ C/kg.
- Oil drop: qE = weight at balance; radius from Stokes terminal speed.
- Charge comes in multiples of e = 1.6 × 10⁻¹⁹ C; m_e ≈ 9.1 × 10⁻³¹ kg.
Two classic experiments measured the electron: Thomson found its charge-to-mass ratio in 1897, and Millikan its charge about fifteen years later. Together they gave its mass.
The discovery of the electron
Discharge-tube experiments in the late 1800s convinced physicists that cathode rays were tiny negative particles. J J Thomson is credited with discovering the electron, not because he was first to run such experiments, but because he measured their properties and proposed that electrons are a part of every atom.
Thomson’s measurement of e/m
Electrons from the cathode C are accelerated through a potential difference V to the anode A; a narrow beam passes through holes in A and A′, between deflecting plates D₁ (+) and D₂ (−), and hits a fluorescent screen. The plates’ field E pushes the beam up. Coils C₁, C₂ add a magnetic field B, perpendicular to both E and the beam, pushing it down.
Adjust the fields until the spot returns to its undeflected position. Then eE = evB, so v = E/B. The speed also follows from the accelerating voltage, ½mv² = eV:
Electrons accelerated through 2000 V pass undeflected through crossed fields of 3.0 × 10⁴ V/m and 1.13 mT. Find their speed and e/m.
Millikan’s oil-drop experiment
Two accurately parallel plates a few millimetres apart sit in a glass-walled chamber. Oil droplets sprayed in through a hole in the top plate pick up charge by friction (X-rays or a radioactive source can help); lit from the side, each looks like a star through a telescope. A drop of mass m and charge q feels its weight mg, the electric force qE, buoyancy and viscous drag.
Most drops are negative, so with E downward the electric force is upward. Adjust E until a chosen drop hangs still (no drag then):
Switch the field off: the drop falls and soon reaches a steady speed v, when Stokes drag 6πηrv plus buoyancy balance its weight. That gives the radius, and then the charge:
Over thousands of drops, every charge was a small whole-number multiple of one value: e = 1.6 × 10⁻¹⁹ C, the charge of the electron. Millikan won the 1923 Nobel Prize for this work.
Plates 6.0 mm apart hold an oil drop still at 300 V (upper plate positive). With the field off it falls 2.45 mm in 30 s at steady speed. Oil density 900 kg/m³, air viscosity 1.8 × 10⁻⁵ N s/m² (neglect air’s density). Find the charge and the number of extra electrons.
- 1Thomson: crossed E and B with zero deflection give v = E/B; with ½mv² = eV, e/m = E²/2VB² (41.2).
- 2Millikan: balance a drop (qE = weight), find its radius from terminal speed (Stokes' law), get q.
- 3All charges are whole multiples of e = 1.6 × 10⁻¹⁹ C; then m_e ≈ 9.1 × 10⁻³¹ kg.
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Keep exploring
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