Physics › Turning points › Cathode rays and the electron
Cathode rays and the electron
In the 1890s nobody could say what streamed from the negative electrode of an evacuated discharge tube. Thomson measured the specific charge of those rays, showed it was the same whatever the electrode was made of, and identified the first subatomic particle.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on Current, charge and the direction problem and Force on a moving charge.
IN THIS TOPIC
- Describe how cathode rays are produced in a discharge tube.
- Explain thermionic emission, and how an electron gun turns it into a beam.
- Use eV = ½mv² for an electron accelerated from rest.
- Outline one determination of e/m, and explain why Thomson's result mattered.
COMMON MISCONCEPTION
The atom is the smallest unit of matter; nothing can be pulled out of one.
Glows in a tube
Seal a gas at very low pressure into a glass tube, put several thousand volts across two electrodes, and the tube glows. Nineteenth-century physicists found that the glow traces something streaming from the negative electrode, the cathode, and named the something cathode rays before anyone knew what they were. The mechanism runs in stages. The pd first ionises some of the remaining gas. Positive ions then strike the cathode and knock particles out of it, and those particles accelerate away towards the anode, exciting the gas they pass through into glowing.
The rays behaved like nothing respectable. They travelled in straight lines and cast sharp shadows, yet electric and magnetic fields bent them, and the direction of the bending said they carried negative charge. Whatever they were, they were charged particles, and identical ones emerged whatever metal the cathode was made from and whatever gas filled the tube. That universality is the first crack in the atom's unbreakable reputation.
The electron gun
The discharge tube is a messy source. Every experiment that follows uses a clean one, thermionic emission. Heat a metal filament and its free electrons gain enough kinetic energy to escape the surface, much as molecules evaporate from a liquid. Place a positive anode nearby in a vacuum and the escaped electrons accelerate towards it. Drill a hole in the anode and a narrow, fast beam sails through. That arrangement is an electron gun.
The beam's speed comes straight from the work-energy idea. An electron of charge e falling through a pd V has work eV done on it, and starting from next to nothing, all of it becomes kinetic energy:
WORKED EXAMPLE
The speed out of the gun
An electron gun accelerates electrons from rest through 2500 V. Calculate the speed of the emerging beam (e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg).
eV = ½mv2, so v = = √(2 × 1.60 × 10−19 × 2500 / (9.11 × 10−31)).
v = 3.0 × 107 m s−1. A tenth of the speed of light, from a bench-top supply.
That startling fraction is worth remembering; it returns in the relativity lessons, where this formula's validity at high speed is finally questioned.
Thomson and the specific charge
In 1897 J J Thomson measured the one number the beam would give up, its specific charge e/m, meaning the charge per kilogram. One classic route uses crossed fields. Send the beam between charged plates, which push it one way, and add a magnetic field at right angles, tuned to push it back exactly the other. When the beam runs straight, the electric force eE equals the magnetic force Bev, so the speed is simply v = E/B. No clock is needed anywhere in the measurement.
With v known, the gun's own accelerating pd finishes the job: eV = ½mv2 rearranges to e/m = v2/2V.
WORKED EXAMPLE
e/m from a balanced beam
A beam passes undeflected through crossed fields of E = 5.0 × 104 V m−1 and B = 2.0 × 10−3 T, having been accelerated through 1.78 kV. Find the beam speed and the specific charge of its particles.
v = E/B = 5.0 × 104 / (2.0 × 10−3) = 2.5 × 107 m s−1.
e/m = v2/2V = (2.5 × 107)2 / (2 × 1780) = 1.76 × 1011 C kg−1.
Two measured field strengths and one dial reading on the supply. Nothing more was needed, and the answer is the modern value to three figures.
The number's significance lies in a comparison. Before Thomson, the largest specific charge known belonged to the hydrogen ion, the lightest atom stripped of its electron. Thomson's particles beat it by a factor of about 1800, so either they carried absurdly more charge or they were absurdly lighter. Evidence pointed to lighter. That conclusion rewrote chemistry, because the cathode-ray particle, soon named the electron, is a constituent of atoms, torn from any metal and any gas alike. Atoms have parts.
GUIDED PRACTICE
The hydrogen benchmark
Calculate the specific charge of the hydrogen ion, a proton of mass 1.67 × 10−27 kg carrying e = 1.60 × 10−19 C, and compare it with the electron's 1.76 × 1011 C kg−1.
Show the working
e/mp = 1.60 × 10−19 / (1.67 × 10−27) = 9.6 × 107 C kg−1.
The electron's specific charge is about 1800 times larger. Same size of charge, so the electron must be about 1800 times lighter than the lightest atom: a particle smaller than atoms themselves.
INDEPENDENT PRACTICE
A faster gun
Using e/m = 1.76 × 1011 C kg−1, find the speed of electrons accelerated from rest through 5.0 kV.
Show the working
v = = √(2 × 1.76 × 1011 × 5000) = 4.2 × 107 m s−1.
Note the shortcut. With the specific charge in hand, neither e nor m is needed separately. Fourteen per cent of light speed, and the classical formula is already starting to creak.
ASSESSMENT FOCUS
- The discharge-tube story runs in order, and marks follow the order. Low-pressure gas, kilovolt pd, ionisation, positive ions striking the cathode, particles released and accelerated towards the anode, gas glowing along the path.
- Thermionic emission is one sentence. Electrons in a heated metal gain enough kinetic energy to escape its surface. The word "heated" carries the mark, so do not write "electrons are released from the filament" and stop there.
- eV = ½mv2 assumes the electron starts from rest in a vacuum. State that assumption, and quote the emerging speed to two significant figures.
- For the crossed-fields method the logic scores as much as the algebra. Undeflected means eE = Bev, so v = E/B, and the accelerating pd then gives e/m = v2/2V.
- Thomson's significance needs all three clauses. e/m about 1800 times the hydrogen ion's, so the particle is far lighter than the lightest atom, so atoms have smaller parts inside them.
CHECK YOURSELF
Show that the classical formula eV = ½mv² predicts electrons reaching the speed of light at an accelerating pd of about 260 kV, and comment on what this suggests.
Show a hint
Set v = c and solve for V using e/m = 1.76 × 1011 C kg−1.
Show the answer
V = v2/(2e/m) = (3.0 × 108)2 / (2 × 1.76 × 1011) = 2.6 × 105 V, about 260 kV.
Laboratory supplies exceed this easily, yet no electron has ever been observed at or beyond the speed of light.
So the classical formula must fail at high speeds. What actually happens near 260 kV, and why, is the business of this unit's relativity lessons.
Cathode rays are electrons, boiled off a hot filament and accelerated by eV = ½mv².
Thomson measured e/m, about 1800 times the hydrogen ion's value.
Same charge, far less mass, so atoms must have smaller parts inside them.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the cathode rays and the electron questions page.
CHECK YOUR PROGRESS
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- Describe how cathode rays are produced in a discharge tube.
- Explain thermionic emission, and how an electron gun turns it into a beam.
- Use eV = ½mv² for an electron accelerated from rest.
- Outline one determination of e/m, and explain why Thomson's result mattered.
Open the full revision checklist to track your progress across the whole unit.