The discovery of the electron (3.12.1)
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Cathode rays are produced when a potential difference is applied across a discharge tube.
A discharge tube contains a negatively charged electrode – an anode – and a positively charged electrode – a cathode – separated in a chamber filled with inert gas.
Cathode rays were first observed in 1876. The term ‘cathode rays’ was used to describe the glow that appeared on the walls of a discharge tube.
In 1897, J.J. Thomson demonstrated that cathode rays:
- possess energy, momentum, and mass
- have a negative charge
- always have the same properties regardless of the type of gas in the discharge tube and the type of material the cathode is made from
- have a mass-to-charge ratio much greater than hydrogen ions.
The large mass-to-charge ratio of the cathode rays suggested that they were either very small or had a very large charge.
Thomson concluded that all atoms contain cathode ray particles, which are now known as electrons. The electron was the first subatomic particle to be discovered. It is important to note that cathode rays are beams of electrons.
When a metal is heated, the free (delocalised) electrons within the metal gain kinetic energy. If enough thermal energy is supplied to a metal, the electrons at the surface gain enough kinetic energy to escape. This is known as thermionic emission.
The work function of a metal is the minimum kinetic energy an electron needs to escape the surface.
Thermionic emission occurs when the kinetic energy of the electrons on a metal’s surface exceeds the work function. For many metals, thermionic emission becomes significant at temperatures above
Electrons emitted via thermionic emission can be accelerated by an electric field. This is the principle of operation in an electron gun.
The diagram below shows the inner structure of an electron gun.
An electron gun produces a narrow beam of electrons:
- A coil is heated electrically, which in turn heats a metal cathode.
- Electrons are emitted from the cathode via thermionic emission.
- An electric field applied by a potential difference between the cathode and anode accelerates the electrons towards a cylindrical anode.
- Only the electrons directed at the hole in the anode can pass through, creating a narrow beam.
- The electrons move at constant velocity beyond the anode due to the absence of an electric field.
When a charged particle is accelerated by a potential difference, the work done on the particle is:
Where:
- is the charge of the particle
- is the potential difference.
The work done on the particle is converted into kinetic energy. Since the charge of the electron is the work done on an electron by a potential difference is:
Therefore, the kinetic energy of the electron accelerated by a potential is:
The above equation is used to define a unit known as the electronvolt, . One electronvolt is the kinetic energy of an electron after it has been accelerated by a potential difference of Therefore, one electronvolt has a value of
The kinetic energy, measured in electronvolts of an electron accelerated by an electric field is equal to the accelerating voltage (in volts).
Question walkthrough
Finding accelerating voltage of an electron
Rearranges eV=½mv² to find the potential difference an electron is accelerated through, given its final velocity, mass, and charge.
When Thomson discovered the electron, there was no experimental method for measuring its charge or mass. However, he was able to measure its specific charge. The specific charge of a charged particle is defined as the charge per unit mass. For the electron, this is equal to:
Where:
- is the electron charge,
- is the electron mass,
Calculating the specific charge of the electron gives:
The specific charge for the proton is equal to:
The specific charge of the electron is approximately 1800 times larger than that of the proton, suggesting that the electron has either a very small mass or a very large charge.
When the electron was discovered, the proton had the largest specific charge measured. Thomson assumed that the electron and proton had the same size charge and that the electron had a much smaller mass. Experiments later proved this to be true.
An electron’s specific charge can be measured using its motion in a magnetic field. An apparatus known as a fine beam tube is used for this. The diagram below shows the components of a fine beam tube.
In a fine beam tube, electrons are accelerated by an electron gun and fired at a low-pressure gas. The gas atoms absorb energy from the electrons, causing the electrons in the atoms to be excited to higher energy levels. When these electrons return to the ground state, they emit light, creating a glowing electron beam.
External magnetic coils create a uniform magnetic field within the glass bulb. The electron beam is fired at right angles to the external magnetic field, causing it to curve due to the centripetal magnetic force applied to the electrons.
The radius of the beam’s circular path is related to the specific charge of the electron.
The specific charge of a charged particle can be measured using a fine beam tube. An electron in a fine beam tube follows a circular trajectory because it is fired at right angles to the external magnetic field. The electron experiences a centripetal magnetic force.
Equating the general equation for centripetal force to the magnetic force on the electron gives:
Where:
- is the velocity of the electron,
- is the radius of the circular path the electron follows,
- is the electron charge,
- is the mass of the electron, and
- is the magnetic field strength.
The above equation can be rearranged to obtain the velocity of the electron:
Substituting the above equation into gives:
Cancelling like terms and rearranging to make the subject gives the equation for the specific charge of an electron:
All quantities on the right side of the above equation can be measured using a fine beam tube.
Question walkthrough
Finding magnetic field via beam tube
Rearranges the specific charge equation e/m_e=2V/(B²r²) to find the magnetic field strength in a fine beam tube from the accelerating voltage and radius of the electron path.
Robert Millikan set out to determine the absolute charge of the electron. This was accomplished using the setup shown in the diagram below.
In Millikan’s experiments, an atomiser was used to create a fine mist of oil droplets. These droplets were charged by friction; positive droplets formed from the loss of electrons, and negative droplets from the gain of electrons.
The droplets fell through a hole in the top plate, where they could be viewed under a microscope equipped with a scale to measure distances.
A variable electric field was applied between the top and bottom plates. The field exerted a force on the charged oil droplets. The field strength was varied by changing the potential difference.
For Millikan’s experiment, if the electric field is turned off, then the only forces acting on the oil droplets are the:
- weight – acting downwards
- viscous force from the air – acting upwards.
The diagrams below show the forces acting on the oil droplet in the two stages of its fall.
Initially, as the oil droplet falls, the viscous drag force is less than the weight, and the oil droplet speeds up, as shown by the left diagram.
As the oil droplet speeds up, the viscous drag force increases until it equals the weight, and the oil droplet falls at terminal velocity, as shown in the diagram on the right.
In Millikan’s experiment (with the electric field turned off), when the oil droplets fall at terminal velocity, the weight and the viscous drag force are equal. Millikan knew the density of oil and the viscosity of air and, therefore, was able to calculate the radius using the following method.
Since the viscous drag force is described using Stokes’ law:
Then, at terminal velocity, this is equal to the weight:
Where:
- is the mass of the oil droplet
- is the viscosity of air
- is the radius of the droplet
- is the velocity of the droplet.
The mass of the oil droplet can be found from:
Where:
- is the density
- is the volume.
The volume of the oil droplet is equal to that of a sphere, i.e. Substituting this into the above equation leads to:
Rearranging for the radius gives:
In Millikan’s experiment, the oil droplets fall under gravity. When an electric field is applied, some droplets experience an upward force, causing them to decelerate. The electric field can be adjusted until the droplets become stationary, at which point the viscous force equals zero.
With the viscous force being zero, the two forces acting on the droplets are:
- the weight, acting downwards
- the force due to the electric field – acting upwards.
The force due to the electric field is given by:
Where:
- is the potential difference
- is the charge
- is the distance between the electrically charged metal plates.
Since the drop is stationary, the force due to the electric field must equal the weight:
The radius can be determined with the electric field turned off. Therefore, the only unknown in the above equation is
Millikan could find the charge of the oil droplets, and repeated this calculation over many droplets. He found that the charge on each droplet was always a whole number multiple of
Question walkthrough
Drawing force diagrams for oil drops
Draws free-body diagrams for a charged oil droplet in Millikan’s experiment under three conditions — accelerating under gravity, falling at terminal velocity, and held stationary by an electric field.
The results from Milikan’s oil drop experiment showed that charge exists as whole number multiples of and cannot exist in smaller quantities.
Millikan assumed the value of to be the charge of the electron. Later experiments confirmed Milikan’s conclusions. The value of is known as the fundamental unit of charge.
Finding the charge of an electron allowed the mass to be calculated using the value for the specific charge that J.J. Thomson had previously determined. This demonstrated that the electron was the lightest particle discovered at the time.
Question walkthrough
Finding charge in Millikan’s experiment
Balances the electric force (QV/d) against weight (mg) to find the charge on a stationary oil drop in Millikan’s experiment, expressing the result as a multiple of e.