Learn7408Paper 3B · Option: Turning points in physicsThe discovery of the electronCathode rays, specific charge and Millikan

The discovery of the electron · Subtopic 1 of 1

The Discovery of the Electron

Heated cathodes emit electrons (thermionic emission); accelerated through $V$ , each gains $\frac12mv^2=eV$ . Thomson used crossed electric and magnetic fields ( $v=\frac EB$ ) and magnetic deflection to find the specific charge $\frac em\approx1.76\times10^{11}$ C kg⁻¹, about 1800 times that of a hydrogen ion. Millikan balanced charged oil drops ( $\frac{QV}d=mg$ ) and found every charge was a whole multiple of $1.6\times10^{-19}$ C.

AQA 7408 Paper 3B≈ 45 minRevision 1 · updated 2026-10-27

Section 1 of 14

Learning objectives

Explain thermionic emission and use $\frac12mv^2=eV$ .
Find $\frac em$ using crossed fields and circular motion.
Describe Millikan's experiment and the quantisation of charge.

Pause and think

In the 1890s, nobody knew whether cathode rays were waves or particles. What experiment could decide?

Section 2 of 14

Core idea

Start hereWhat are cathode rays?

In the 1890s, glowing beams in evacuated tubes puzzled physicists: waves or particles? J. J. Thomson bent them with fields and measured how much they bent. His answer, a particle far smaller than any atom, began subatomic physics.

The big idea

Crossed fields select speed $v=\frac EB$ ; a magnetic field alone gives $r=\frac{mv}{eB}$ , so $\frac em=\frac v{Br}$ . Millikan's balanced drops showed charge comes in units of $e$ .

1. Specific charge

e/m from accelerating voltage and radius

$\tfrac12mv^2=eV\Rightarrow v^2=\frac{2eV}m$
Electron gun.
$r=\frac{mv}{eB}\Rightarrow v=\frac{eBr}m$
Circular path.
$\frac{e^2B^2r^2}{m^2}=\frac{2eV}m$
Square the second and equate.
$\frac em=\frac{2V}{B^2r^2}$
Rearrange.

Worked exampleThomson-style measurement

$v=2.0\times10^7$ m s⁻¹, $B=1.0\times10^{-3}$ T, $r=0.114$ m.

Goal: $\frac em$ .
Strategy: $\frac em=\frac v{Br}$ .

$\frac{2.0\times10^7}{1.0\times10^{-3}\times0.114}$
Substitute.
$=1.75\times10^{11}\text{ C kg}^{-1}$
Calculate.

Answer

1.75\times10^{11}

C kg⁻¹.

✓ CheckAccepted value

1.76\times10^{11}

✓.

2. Millikan

Worked exampleBalanced drop

$m=3.26\times10^{-15}$ kg, $d=10$ mm, $V=1000$ V.

Goal: Charge.
Strategy: $\frac{QV}d=mg$ .

$Q=\frac{mgd}V$
Rearrange.
$=\frac{3.26\times10^{-15}\times9.81\times0.010}{1000}=3.2\times10^{-19}\text{ C}$
Substitute.

Answer

3.2\times10^{-19}

=2e

✓ CheckA whole number of

e

✓.

Discovering the electron

Drawing diagram…

Two experiments, two conclusions.

Key takeawayFinding the drop’s mass With the field off, the drop falls at terminal velocity:

mg=6\pi\eta rv

and

m=\frac43\pi r^3\rho

give

r

and so

m

Common misconceptione/m vs e Thomson found the ratio; Millikan found

e

; together they give the electron's mass.

Pause and think

How do Thomson and Millikan together give $m_e$ ?

Summary

$\frac12mv^2=eV$ .
$v=\frac EB$ , $\frac em=\frac v{Br}$ .
$\frac{QV}d=mg$ .

Specific charge: Charge per unit mass.
Thermionic emission: Electron release from a hot surface.
Quantised: Coming in whole-number multiples.

HistoryThomson’s Nobel. J. J. Thomson won the 1906 Nobel Prize for the electron; his son G. P. Thomson won it in 1937 for showing electrons diffract like waves.

Fun factOld televisions. Every cathode-ray TV was an electron gun with magnetic deflection coils sweeping the beam across the screen.

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