Physics › Turning points › Cathode rays and the electron
Cathode rays and the electron
In the 1890s a glowing tube posed a question nobody could answer: what streams from a negative electrode through near-empty space? Measuring one ratio settled it, broke the atom's reputation as unbreakable, and handed physics its first fundamental particle.
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 use eV = ½mv² for an accelerated electron.
- Outline one determination of e/m and explain why Thomson's result mattered.
WHAT YOU PROBABLY THINK
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 pd first ionises some of the remaining gas; positive ions slamming into the cathode knock particles out of it, and those particles accelerate away toward 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; the clean one, used in every experiment that follows, is thermionic emission. Heat a metal filament and its free electrons gain enough kinetic energy to escape the surface, like evaporation from a liquid. Place a positive anode nearby in a vacuum and the escaped electrons accelerate toward it; drill a hole in the anode and a narrow, fast beam sails through. The 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 = √2eV/m = √(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 honesty 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, 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 required.
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) = 1.8 × 1011 C kg−1.
Two measured field strengths and one dial reading on the supply: that is the entire experiment, and the answer is the modern value.
The number's significance lies in a comparison. The largest specific charge known before Thomson belonged to the hydrogen ion, the lightest atom stripped of its electron. Thomson's particles beat it by a factor of about 1800: either they carried absurdly more charge, or they were absurdly lighter. Evidence pointed to lighter, and the conclusion rewrote chemistry: the cathode-ray particle, soon named the electron, is a constituent of atoms, torn from any metal and any gas alike. Atoms have parts.
YOUR TURN
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, before opening the working.
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.
TRY IT UNSEEN
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(e/m)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.
THE EXAM BIT
- The discharge-tube story runs in order: low-pressure gas, kilovolt pd, ionisation, positive ions striking the cathode, particles released and accelerated toward 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.
- eV = ½mv2 assumes the electron starts from rest in a vacuum; state it, 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 is the comparison: e/m about 1800 times the hydrogen ion's, so the particle is far lighter than the lightest atom, so atoms have smaller parts. All three clauses.
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 by heat, accelerated by eV = ½mv².
Thomson's e/m, 1800 times hydrogen's, proved atoms have smaller parts.
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.
CHECK YOUR PROGRESS
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- Describe how cathode rays are produced in a discharge tube.
- Explain thermionic emission and use eV = ½mv² for an accelerated electron.
- 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.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.