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The Discovery of the Electron

In low-pressure discharge tubes, rays came from the cathode. These cathode rays travel in straight lines, carry negative charge and are bent by electric and magnetic fields. In an electron gun a hot filament releases electrons by thermionic emission and an anode voltage V accelerates them: eV = ½mv². J. J. Thomson balanced an electric field against a magnetic field so the beam went straight: v = E/B. Then e/m = v² ÷ 2V ≈ 1.76 × 10¹¹ C kg⁻¹, about 1800 times the value for a hydrogen ion, showing the particle was very light and the same in every metal. Millikan balanced charged oil drops between plates (QV/d = mg) and found every charge was a whole-number multiple of e = 1.6 × 10⁻¹⁹ C: charge is quantised.

🎬 Step-by-step story

  1. In a glass tube with little air, rays come out of the cathode. A cross in their way makes a sharp shadow.
  2. Heat the cathode: electrons boil off. The anode pulls them and they speed up.
  3. An electric field bends the beam up. A magnetic field bends it down. Balance them: the beam goes straight.
  4. From the speed and the voltage we get e/m. It is huge, so the particle is tiny: the electron.
  5. Millikan: tune the plate voltage until an oil drop stops. Its charge is always a whole number of e.
  6. Your turn: change the voltage and the charge. Make the drop rise, fall or hang still.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

If the rays are invisible, how did people see them?

They make the glass or a fluorescent screen glow where they hit, and objects in their path cast shadows.

Why do we need to heat the cathode?

Electrons are held inside the metal. Heating gives some of them enough energy to escape the surface.

Why does the beam go straight when both fields are on?

The electric force pushes one way and the magnetic force the other way. When eE = Bev, they cancel.

Why does a large e/m mean a small mass?

If the charge is similar to an ion's, a value 1800 times larger means the mass must be about 1800 times smaller.

How did Millikan know the drop's mass?

He let it fall with no field. From its steady (terminal) speed and Stokes' law he found the radius and so the mass.

Why does the drop stop at only one voltage?

Only one value of V makes QV/d exactly equal to mg. Above it the drop rises; below it, it falls.

Cathode rays

In the 1800s scientists passed a high voltage across a glass tube with most of the air pumped out (a discharge tube). The gas glowed, and at very low pressure a glow appeared on the glass opposite the cathode (negative electrode). Something was coming from the cathode: cathode rays.

In a discharge tube, positive ions hit the cathode and knock out electrons; the glow comes from atoms excited by collisions and then giving out light.

Thermionic emission of electrons

When a metal is heated, some free electrons gain enough energy to escape from the surface. This is thermionic emission. In an electron gun a filament heated by a low-voltage current releases electrons in a vacuum. A positive anode at a high voltage V pulls them across; a hole in the anode lets a narrow beam through.

Work done by the field = gain in kinetic energy:

eV = ½ m v², so v = √(2eV ÷ m).

A vacuum is needed so the electrons do not collide with gas atoms. One electronvolt (1 eV) = 1.6 × 10⁻¹⁹ J is the energy an electron gains through 1 V.

Specific charge of the electron (e/m)

Specific charge = charge ÷ mass (C kg⁻¹).

Method 1: crossed fields (Thomson)

The beam passes between plates (field E = V_p ÷ d) and through a magnetic field B at right angles. The electric force eE pushes one way and the magnetic force Bev the other. Adjust until the beam is undeflected: eE = Bev → v = E ÷ B. Put this v into eV = ½mv²:

e/m = v² ÷ 2V = E² ÷ (2VB²)

Method 2: circular path in a magnetic field

In a magnetic field alone the beam moves in a circle (fine-beam tube): Bev = mv² ÷ r, so r = mv ÷ Be. With v from eV = ½mv²: e/m = 2V ÷ (B² r²).

Why it mattered

Thomson found e/m ≈ 1.76 × 10¹¹ C kg⁻¹, about 1800 times bigger than for the hydrogen ion (the largest known at the time, 9.6 × 10⁷ C kg⁻¹). Either the charge was huge or the mass tiny. Later it was clear the mass is tiny: the electron is a particle inside every atom.

Millikan's oil drop: measuring e

Tiny oil drops are sprayed between two horizontal plates a distance d apart. Some pick up charge by friction (or from X-rays).

  1. Field off: the drop falls and quickly reaches terminal speed v, when weight = air drag. By Stokes' law, drag = 6πηrv. With mg = (4/3)πr³ρg: r = √(9ηv ÷ 2ρg). This gives the drop's radius and mass.
  2. Field on: adjust the voltage V until the drop hangs still. Then electric force = weight: QV ÷ d = mg, so Q = mgd ÷ V.

Millikan measured many drops. Every charge was a whole-number multiple of 1.6 × 10⁻¹⁹ C. So charge is quantised and the smallest charge, e, is the charge of one electron. With e/m known, the electron mass m = 9.11 × 10⁻³¹ kg.

Try it: make a balloon "cathode ray" bender

Rub a balloon on dry hair, then hold it near a thin, slow stream of water from a tap. The stream bends: charge feels an electric force, just as the beam does between the plates. In the 3D free play, set n = 2 and find the voltage where the drop stops; check it with V = mgd ÷ Q.

Key formulas and definitions

Worked examples

1. An electron is accelerated through 500 V. Find its kinetic energy in J and in eV.

E_k = eV = 1.6 × 10⁻¹⁹ × 500 = 8.0 × 10⁻¹⁷ J = 500 eV.

2. Find the speed of an electron accelerated through 2000 V (e/m = 1.76 × 10¹¹ C kg⁻¹).

v = √(2 × 1.76 × 10¹¹ × 2000) = √(7.04 × 10¹⁴) = 2.65 × 10⁷ m s⁻¹.

3. A beam is undeflected when E = 4.0 × 10⁴ V m⁻¹ and B = 1.5 × 10⁻³ T. Find v.

v = E ÷ B = 4.0 × 10⁴ ÷ 1.5 × 10⁻³ = 2.67 × 10⁷ m s⁻¹.

4. Continue: the anode voltage was 2000 V. Find e/m.

e/m = v² ÷ 2V = (2.67 × 10⁷)² ÷ 4000 = 7.11 × 10¹⁴ ÷ 4000 = 1.78 × 10¹¹ C kg⁻¹.

5. A drop of mass 4.9 × 10⁻¹⁵ kg is held still between plates 6.0 mm apart at 900 V. Find its charge and the number of extra electrons.

Q = mgd ÷ V = 4.9 × 10⁻¹⁵ × 9.81 × 6.0 × 10⁻³ ÷ 900 = 3.2 × 10⁻¹⁹ C = 2e. Two extra electrons.

6. With the field off, an oil drop (ρ = 900 kg m⁻³) falls at a terminal speed of 1.0 × 10⁻⁴ m s⁻¹ in air (η = 1.8 × 10⁻⁵ Pa s). Find its radius.

r = √(9ηv ÷ 2ρg) = √(9 × 1.8 × 10⁻⁵ × 1.0 × 10⁻⁴ ÷ (2 × 900 × 9.81)) = √(9.2 × 10⁻¹³) = 9.6 × 10⁻⁷ m ≈ 1 µm.

7. In a fine-beam tube, electrons accelerated through 250 V move in a circle of radius 5.0 cm in B = 1.07 × 10⁻³ T. Find e/m.

e/m = 2V ÷ (B² r²) = 500 ÷ ((1.07 × 10⁻³)² × 0.050²) = 500 ÷ (2.86 × 10⁻⁹) = 1.75 × 10¹¹ C kg⁻¹.

Common mistakes

Practice quiz

1. Cathode rays are:
2. Thermionic emission means electrons are released by:
3. For an undeflected beam in crossed fields:
4. Millikan found that charge is:
5. The specific charge of the electron is about:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What did J. J. Thomson discover?

He showed cathode rays are particles with a specific charge about 1800 times that of a hydrogen ion, the same for all metals: the electron (1897).

What is the formula in Millikan's oil drop experiment?

For a stationary drop QV/d = mg, so Q = mgd/V. The mass comes from the terminal speed using Stokes' law.

What is thermionic emission?

The release of electrons from a heated metal surface, used in electron guns, old TV tubes and X-ray tubes.

Where this is taught

England (GCSE, A level)Year 133.12 Turning points in physics

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