This experiment shows that a cathode ray is made up of particles with a constant ratio of charge to mass (e/m). A magnetic field bends the cathode rays into a closed circle.
Apparatus and materials
Fine beam tube and stand
Power supply, 0-250 V (High Tension, HT), with special shrouded connecting leads
6.3 V supply for the heater filament (this is often included on the EHT supply)
Demonstration meters, 2
DC dial: 5 amp
DC dial: 300 volts
Power supply, low voltage, variable, 0 - 12 V
Battery pack, 6V, 2
Health & Safety and Technical notes
Make all connections with the power supply off.
Connecting leads used for the HT voltages must have adequate insulation.
The HT supply can deliver a fatal current. Use 4 mm leads with plugs having sprung shrouds for all high-voltage connections. The teacher supervising in a darkened room must be aware of the hazards and their control.
Do not connect or disconnect leads once the HT is switched on.
The simplest way to measure the magnetic field strength is to use a Hall probe. These are available from most equipment manufacturers. They come as a stand-alone unit which gives a digital read-out of magnetic field strength.
Hall probe sensors are also available with some computer interfaces – as one of the sensors that plugs into the interface. If you already have one of these interface kits, this might be a cheaper option than a stand-alone Hall probe.
For this experiment, students don’t necessarily need to know how the Hall probe works – you can introduce it simply as a device for measuring magnetic field strength.
A Teltron-type tube works best for this demonstration because you can use the electron gun that sends out a vertical beam. This can then be bent into a complete circle.
Follow the manufacturer’s instructions for setting up the fine beam tube.
Ensure that you can identify the following:
- The 6.3 V supply to the cathode heater (if you connect the wrong voltage to the heater you can easily damage the tube beyond repair).
- The HT supply to the anode. Set this to zero. The negative terminal of the HT goes to a socket, which is often near to the heater terminals.
A tube which has not been used for a while may not emit electrons. It may be possible to encourage it to do so by increasing the heater voltage by around 1 V or so. Monitor it carefully. Ensure that the normal heater current is only slightly exceeded.
a It is best to measure the magnetic field from the coils before setting up the tube. Set up the coils without the fine beam tube – one on each side of where the tube will be. Connect them in series with each other, with the ammeter and with the 12-volt power supply (or battery, rheostat and the switch).
b Set the current in the coils to about 2.5 amps.
c Use the Hall probe to measure the magnetic field.
d Switch the coils off for now. Do not adjust the rheostat or power supply from now on.
e Set up the fine beam tube in its special stand.
f Select the gun which gives a vertical electron beam. (There may be a selection switch.)
g Connect the 6.3 V supply to the filament. (Some HT supplies available for school use incorporate this output. Alternatively, it is possible to use a separate low voltage transformer.)
h Connect the negative terminal of the HT supply to the filament and the positive terminal to the anode.
i Connect a voltmeter between the anode and cathode in order to measure the accelerating voltage, which should be about 250 volts.
j Set the HT voltage to zero and switch on the 6.3 V supply to the heater filament.
k When the filament is glowing, carefully increase the anode voltage. At a voltage, which may be as low as 50 volts, it should be possible to see the fine beam. As the voltage is slowly increased, the beam will lengthen and strike the glass envelope of the tube.
l Switch on the current to the coils. The beam should be bent into a circular path.
m Adjust the accelerating voltage until the beam forms a complete circle inside the tube – coming back around on itself. (Do not change the current in the coils unless you really have to. If you do change it, you will need to measure the field again).
n Increasing and decreasing the accelerating voltage on the electron gun will show how the radius of the path of the electrons varies with the speed of the electrons for a constant magnetic field. The faster the electrons, the more difficult they are to bend, so the radius of the path increases.
o Each group of pupils should measure the path diameter D and the accelerating voltage. To measure D, simply hold a ruler outside the tube. In the darkened room, the ruler should be illuminated. (A Perspex ruler with a small electric lamp taped to one end - and covered with masking tape so that no direct light emerges – works well.)
p Increasing the current through the field coils for a constant voltage on the electron gun will show how the circular path of the electrons changes with the strength of the magnetic field. The stronger the magnetic field, the more effect it has on the electrons and the circular path reduces in radius. If you change the current in the field coils, be sure you return it to the value at which you measured the field strength.
1 This experiment is best demonstrated to the students in groups of four to five in a darkened room if full value is to be obtained.
2 Always reduce the anode voltage to zero when not actually observing the beam because the tube has a finite life time.
3 The electron beam is visible because there is a low-pressure gas in the tube. Electrons striking the gas molecules give them energy, which is then released as light. Hydrogen gas glows blue and helium gas glows green.
4 It is worth reminding students of the catapult field on a current carrying wire before showing the deflection of electrons. (See Related Experiments, below.) In the case of the magnetic force on a beam of electrons we must change the expression from the force of a magnetic field on a current carrying wire into the force of a magnetic field on a moving charge. Given that the beam is made of electrons, the direction of the current is towards the cathode.
The force F on a wire of length L carrying current I in a magnetic field B is F = ILB.
Here I = e / t and v = L / t, where e is the electron charge and v is its speed.
So F = evB
5 In the fine-beam tube the catapult force of the magnetic field is perpendicular to the stream of negatively charged electrons and so a uniform magnetic field will hold the stream in a circular orbit provided the electrons move at a constant speed. The magnetic field pulls the electrons into an orbit rather like a tether that holds a whirling ball.
6 If the tube is twisted slightly in its holder then the circular motion of the beam combines with a linear component of the beam to make a spiral.
7 It is helpful if students practise measuring the diameter of the beam by measuring the diameter of a wire loop without being allowed to bring the wire loop near to the rule. Practice with the loop will save time when they all measure the beam diameter.
8 The best modification so far produced forms a virtual image of an illuminated scale inside the tube, in the plane of the electron stream. To do this, place a vertical sheet of clean plate glass just in front of the tube. Place an illuminated scale in front of the sheet at such a distance that the image of the scale, behind the sheet, is in the middle of the tube. This does make measurements easier; but we do not recommend adding this complication except with a very able group.
9 The emphasis is to bring students into contact with real experiments on electrons and not to worry too much about precision, which may be beyond the design of the apparatus. Students should be encouraged to devise refinements, such as measuring the diameter of the electron beam, but too many refinements will make the experiments very complicated and the simple elegance of it will be lost.
10 When JJ Thomson measured the value of e/m he did not have the luxury of a heated cathode producing electrons all at the same speed, and so he did not know the speed of the electrons. He had to deflect his beam of electrons with a magnetic field and then return them to their undeflected position with an electric field.
11 It is worth discussing with students what they can infer from this experiment (and what they can’t).
- The fact that the cathode rays are deflected shows that they have charge; the direction of the deflection shows the charge is negative (which was already assumed because they emerged from a cathode).
- The fact that they are deflected in a curved path and not deflected through a right angle shows that they have inertia or mass. From this, we assume that the cathode ray is made up of particles.
- The fact that the whole beam is kept intact shows that the particles are all deflected by the same amount.
- It is likely that the beam is made up of particles that are all the same. But, at this stage, we cannot infer this for certain.
- We can infer that the ratio of the charge to mass of all the particles is the same. If there are some particles in the beam that have twice the charge, they must also have twice the mass. This is because they all follow the same orbit; the radius of the orbit depends on the acceleration. In turn, the acceleration depends on the force on the particles and their mass. The doubled charge will cause the force to be twice as big but the doubled mass will mean that the acceleration is the same.
12 Using the results to find e/m (the specific charge).
The electrons are accelerated in an electric field. Increasing the accelerating voltage increases the speed of the electrons. We can get an expression for the speed using the idea that the electrons lose electrical potential energy in the electric field and gain kinetic energy. So:
Kinetic energy gained = electrical potential energy lost
1/2 mv2 = eV (equation 1)
where m is the mass of the electron, e its charge, v its velocity and V the accelerating voltage
These moving electrons enter the magnetic field produced by the coils. This bends them into a circular orbit.
The magnetic force, F, on the electrons is given by
F = Bev where B is the magnetic field strength, e is the charge on the electron and v their speed
Given that the electrons are going in a circle, we know that this force must have a size of
mv2/r where m is the mass of the electron and r is the radius of the orbit.
So you can say that:
Bev = mv2/r
so Be = mv/r (equation 2)
There are two ways you could use these equations with students. The first is algebraic and the second uses values for the speed. The first is more complete. However, you might want to use the second so as not to distract students from the accomplishment of measuring properties of the electron.
Combining equations 1 and 2 will lead to the equation:
e/m = 2V/B2r2 from which students could calculate values of e/m
or r2 = 2Vm/eB2 from which students could plot a graph of r2 against V and find e/m from the gradient.
Values for speed
A simple rearrangement of equation 2 gives,
r = mv/Be
So a graph of r (radius) against v (speed) will yield a value for e/m. However, you cannot calculate the speed from equation 1 (without independently knowing e/m). However, you can use the table below to give students the (non-relativistic) speeds at different voltages. These are correct to within 1%.
|Gun Voltage/V||Speed of Electrons/m/s|
|100||6 x 106|
|140||7 x 106|
|180||8 x 106|
|230||9 x 106|
|285||10 x 106|
13 The accepted value of e/m is 1.76 x 1011 C/kg.
14 You will do well to get a result that is the correct order of magnitude. There are quite large uncertainties in the radius measurements and some uncertainty in the measurement of the magnetic field strength. Both of these quantities are squared in the algebraic method.
15 Knowing that the charge on an electron, e = 1.6 x 10-19 Coulomb from Millikan’s experiment, together with the value for e / m = 1.76 x 1011 C/kg leads to a value of 9 x 10-31 kg for the mass of the electron.
The same value of e / m is found for all electrons, whatever their source – hot filaments of metals, the photoelectric effect, bombarding gases by electrons, electric fields tearing atoms apart and even nuclei emitting beta particles. This evidence leads scientists to suppose that electrons are universal ingredients of matter, all identical.
16 A similar calculation for the proton, with charge e = 1.6 x 10-19 Coulomb and e / M = 9.6 x 107C/kg leads to a value of 1.67 x 10-27 kg for the mass of a proton.
This experiment was safety checked in March 2008