Electric Current through Gases
Glowing tubes, cathode rays, the electron, and vacuum valves.
Discharge Through Gases
How a gas at low pressure lights up and conducts.
Gases normally insulate. Give them a high enough voltage at low enough pressure, though, and they glow and conduct, in patterns that led straight to the discovery of the electron.
Making a gas conduct
A gas has few free charged particles. Cosmic rays create some ions, but they recombine about as fast. To pass a steady current we must keep producing charged particles: for example by applying a large voltage to a gas at low pressure, by heating a metal in a vacuum until it emits electrons, or by sending X-rays through the gas.
The discharge tube
A sealed glass tube about 30 cm long and 4 cm wide has a metal electrode at each end and a side tube for pumping out the gas. The electrodes connect to the secondary of an induction coil: the one at the negative terminal is the cathode, the other the anode.
As the voltage is raised, sparks eventually jump. The least voltage that causes sparking is the sparking potential. Paschen found it depends only on the product of pressure and electrode separation:
A cylindrical tube, closed at one end with a cathode fixed there, has a movable piston carrying the anode. The gas is kept at constant temperature. Does the sparking potential change as the piston slides in or out?
What happens as the pressure falls
Typical values for a 30 cm tube (the exact pressures depend on the tube, voltage and gas):
- ~10 cm of Hg: irregular streaks of light and a crackling noise.
- ~10 mm: the streaks broaden into a luminous positive column almost filling the tube, with a buzzing sound. Its colour depends on the gas: reddish for air, bright red for neon, bluish for CO₂.
- ~3–4 mm: the column shortens; a bluish cathode (negative) glow appears, separated from it by the Faraday dark space.
- ~1 mm: the glow detaches from the cathode, leaving the Crookes dark space next to it.
- ~0.1 mm: the column splits into bright and dark bands: striations.
- ~0.01 mm: striations and glow vanish, the Crookes dark space fills the tube, and the glass walls fluoresce (yellow-green for soda glass).
- Lower still: the current dwindles and the tube stops conducting.
Why the glow has this pattern
- The few ions present are accelerated by the field; with enough voltage they ionize molecules they strike, and ionization snowballs. At low pressure, freed electrons travel far before attaching to anything.
- Positive ions accelerated into the cathode knock out more electrons. Light electrons race away while heavy positive ions linger, so a positive charge builds up near the cathode, making an intense field there.
- Electrons leaving the cathode travel about one mean free path before they collide and ionize, so the glow starts a little way from the cathode: that gap is the Crookes dark space.
- Passing through the positive charge in the glow region slows the electrons and they lose their ionizing power: the Faraday dark space (longer, because the field there is weaker). Reaccelerated, they ionize again, starting the positive column.
- Repeated cycles of ionizing, slowing and reaccelerating give the striations.
- As pressure drops the mean free path grows, so the Crookes dark space lengthens; when it exceeds the tube, electrons hit the glass directly and make it fluoresce.
Electrons in a 20 cm discharge tube have a mean free path of 30 cm. How long is the Crookes dark space?
- 1Gases conduct only if charged particles are continually produced.
- 2Sparking potential depends on pd: Paschen's law V = f(pd) (41.1).
- 3Falling pressure: sparks → positive column → cathode glow + Faraday dark space → Crookes dark space → striations → fluorescence → no conduction.
- 4Crookes dark space ≈ one electron mean free path.
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Cathode Rays & Canal Rays
Invisible beams of electrons and ions.
When the Crookes dark space fills the tube, something invisible streams from the cathode and makes the far wall glow. Crookes, Thomson and others pinned down what it was.
Properties of cathode rays
- They leave the cathode at right angles to its surface, wherever the anode is. A concave cathode focuses them at its centre of curvature, where a fluorescent material glows brightest.
- They travel in straight lines: a metal cross in their path casts a sharp shadow on the far wall.
- They push on what they hit: a light mica paddle wheel in their path turns.
- They heat what they hit: a blackened platinum strip at the focus of a concave cathode becomes red-hot.
- They make many crystals, minerals and salts fluoresce.
- They produce X-rays when they strike a solid, especially a metal.
- Electric and magnetic fields deflect them exactly as they would a stream of negative particles, whatever gas, cathode metal or anode position is used. Thomson showed this with a tube that makes a small glowing dot, which moves when a magnet or charged rod is brought near.
- They ionize gas they pass through.
- They pass through thin metal foils.
- They blacken photographic plates.
All of this is explained at once if cathode rays are simply a stream of fast electrons.
Canal rays (positive rays)
Drill holes in the cathode and run the tube at about 1 mm of Hg: faint luminous streams emerge from the holes on the far side of the cathode. These canal rays are the slow positive ions, made near the cathode, that pass through the holes. Electric and magnetic fields deflect them like positive particles, and they too cause fluorescence.
- 1Cathode rays: leave the cathode normally, travel straight, exert force, heat, cause fluorescence, make X-rays, ionize, penetrate foils, fog plates.
- 2Deflected like negative charges, independent of gas or cathode: they are electrons.
- 3Canal rays: positive ions streaming through holes in the cathode.
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Measuring the Electron: e/m and e
Crossed fields and floating drops measure the electron.
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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Thermionic Emission & the Diode
Hot cathodes and one-way valves.
Heat a metal enough and electrons boil off its surface. Collect them on a second electrode and you have a one-way valve for current: the diode.
Thermionic emission
At high temperature electrons escape from a metal’s surface: thermionic emission, and the escaped electrons are thermions. An escaping electron leaves the metal positively charged, which pulls it back, so only electrons with enough spare kinetic energy get out.
At ordinary temperatures almost no free electron has that much energy; heating raises their energies until some can escape. The emission current follows
An emitter has work function 4.0 eV. By what factor does its thermionic current rise between 1800 K and 2200 K?
Raising a surface from 700 K to 900 K multiplies its emission rate by 50. Find the work function.
For tungsten, A = 60 × 10⁴ A m⁻² K⁻² and φ = 4.55 eV. A cathode of area 3.0 × 10⁻⁵ m² runs at 2200 K. Find its emission current.
The diode valve
A cathode (filament) that can be heated, and an anode (plate) to collect electrons, sealed in an evacuated bulb. The cathode may be a tungsten or thoriated-tungsten wire heated directly by a current, or a nickel tube coated with barium oxide, heated by a coil inside it. The anode is a nickel cylinder around the cathode. A battery makes the anode positive by the plate voltage Vp; the resulting plate current ip flows round the circuit.
Characteristics
- The electrons between cathode and anode form a negative space charge that repels newly emitted electrons back. Raising Vp overcomes it, and ip rises: far from saturation, ip = kVp3/2 (the Langmuir–Child law).
- At large Vp every emitted electron is collected and the current saturates. A hotter cathode gives a higher saturation current.
- With the anode negative, electrons are pushed back and no current flows: the diode conducts one way only.
Raising a diode’s plate voltage from 60 V to 64 V increases its current from 30 mA to 38 mA. Find its dynamic plate resistance there.
Rectification
Turning AC into one-way current is rectification.
- Half-wave: an AC source, a diode and a resistor in series. Current flows only in the half-cycles when the anode is positive.
- Full-wave: a transformer’s secondary has a centre tap C; its ends A and B go to the anodes of two diodes, whose cathodes join one end of the load, the other end returning to C. In each half-cycle one diode conducts, and in both the current crosses the load the same way.
- 1Thermionic emission: i = AST²e^(−φ/kT) (41.3); φ is the work function.
- 2Diode: space-charge-limited i_p = kV_p^(3/2), then saturation; one-way conduction.
- 3Dynamic plate resistance r_p = ΔV_p/Δi_p (41.4).
- 4Half-wave and full-wave rectification.
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The Triode & Amplification
A grid that controls a big current.
Put a wire mesh between a diode’s cathode and anode and a small voltage on it controls a big current. That made the triode the first amplifier.
The triode
A triode is a diode with a wire grid between cathode and anode: three terminals. Thermions pass through gaps in the grid to the anode. With the grid at zero and the anode positive it behaves like a diode. A negative grid voltage Vg repels the electrons and reduces the plate current; a positive one helps them along. Because the grid is close to the cathode, a change in Vg has much more effect than the same change in Vp:
Characteristics and constants
Plate current is measured as Vp and Vg are varied with rheostats. Plots of ip against Vp at constant Vg are the anode (plate) characteristics; ip against Vg at constant Vp, the mutual characteristics. A triode is normally operated on the straight parts of these curves.
When both voltages change, Δip = ΔVp/rp + gmΔVg. Keeping ip fixed (Δip = 0):
A triode has rp = 10 kΩ and gm = 2.5 mA/V. Find μ, and the change in plate voltage that offsets a 1.0 V drop in grid voltage.
The triode as an amplifier
A small varying signal vs (say from a microphone) is added to the grid voltage: ΔVg = vs. A load resistor RL is in the plate circuit. With the operating point on the linear part of the mutual characteristic, the plate current follows the signal, and the voltage across RL changes by v₀ = RLΔip, a much larger copy of the signal.
The gain
With a battery V in the plate circuit, Vp = V − RLip, so ΔVp = −v₀. Then
The triode above (μ = 25, rp = 10 kΩ) drives a 30 kΩ load. Find the voltage gain and the output for a 0.40 V signal.
- 1Triode: a grid between cathode and anode; i_p = k(V_p + μV_g)^(3/2).
- 2Constants r_p, g_m, μ with μ = r_p g_m (41.9).
- 3Amplifier gain A = μ/(1 + r_p/R_L) (41.10).