Turning Points Tutorial 1 – The Discovery of the Electron



Thermionic Emission

Charge to Mass Ratio



Thermionic Emission

If we heat a negatively charged piece of metal, we find that some of the conduction electrons have sufficient kinetic energy to escape from the surface of the wire.  It is quite easy to imagine this if we think about a metal wire as a lattice of ions in a sea of free electrons.  In effect we are boiling the electrons off.  This effect is called thermionic emission.  This phenomenon had been known about since the middle of the nineteenth century.  Experiments on gases at low pressure had revealed a glow around the negative terminal, the cathode, and these had been named cathode rays.  Some physicists had argued that the rays were waves and others had argued that they were negatively charged particles.  Indeed the particles had been named electrons by an Irish physicist, George Stoney.  Electron comes from the Greek word, elhktron, meaning "amber", a resinous material secreted by pine trees.


This was the starting point for Joseph John Thomson to produce his cathode ray tube (CRT) in 1897, the descendants of which we used to see every day, before TFT TV sets became more common.






Question 1

Explain what these different parts of the CRT do.

(a) Cathode

(b) Anode


Question 2

Why is there a vacuum in the CRT?



Notice that there is a hole in the anode.  While most of the electrons boiled off will hit the anode, some will pass through the hole.  Hence it is referred to as an electron gun.  The electrons striking both the anode and the glass at the end of the tube lose most of their energy as heat, but can give off some energy as X-rays.


The electrons leave the cathode with negligible speed.  They are accelerated by the attractive force of the positively charged anode.  By the time they leave the gun, the electrons have energy eV, where e is the charge on the electron (1.6 ´ 10-19 C) and V is the anode voltage. 


Energy = charge ´ voltage = eV


All the energy in the electron is kinetic, so we can say:


Kinetic energy = ½ mv2


So we can combine the equations to give:


eV = ½ mv2


Mass of an electron is 9.11 ´ 10-31 kg


Question 3

Calculate the kinetic energy and speed of an electron where the anode voltage is

(a) 400 V

(b) 400 kV



The last answer gives an electron speed above the speed of light.  This is impossible, because the electron gains mass as its speed approaches that of light.  This is a relativistic effect which we will look at later.



Charge to Mass Ratio

Thomson used apparatus like this in his classic experiment



In this experiment a beam of electrons was passed between a crossed magnetic and electric field.   We know:


Question 4

What shape is the path of a moving electron in:

(a) Magnetic Field

(b) Electric Field?



In this experiment we are not interested in the path.  Instead we adjust the values of the magnetic field and electric field so that they are equal in value and in opposite directions.



Question 5

What is the resultant force?  How can you tell?



Consider an electron of charge e that has been accelerated by a voltage V.  It passes into a magnetic field of flux density B which combined with an electric field of strength E.  The fields are set up so that the resultant force on the electron is zero.  This means that the force from the electric field is of equal value but opposite direction to the force from the magnetic field.


We know that:



Cancelling and rearranging to give us an expression for the speed:


From above:

We combine these two to give:


The e terms cancel to give:


The energy of the electron is given by:


All the energy in an electron is kinetic.  The kinetic energy is given by accelerating voltage Va:


We can rearrange to get the e/m ratio:

We can then get rid of the v term by substituting:

which gives us:


The terms in the equation are:


Things are simpler if we make the accelerating voltage the same as the voltage between the two plates.  If the accelerating voltage and the electric field voltage are the same, we can tidy up our expression to give:



Question 6

Which of these terms is:

(a) Easy to measure directly;

(b) Hard to measure directly;

(c) Of constant value?



There are two terms that need working out:


The term e/m is called the charge to mass ratio or specific charge of a particle.



Question 7

Show that the specific charge of an electron is -1.76 ´ 1011 C kg-1.  Does the value vary for an electron?  Would it be different for a positron?


Question 8

What is the specific charge of a proton?  How does it compare to an electron?

Mass of a proton = 1.67 ´ 10-27 kg.



The charge to mass ratio can be applied to any particle, although we have seen it applied to the electron and the proton (hydrogen ion).  It is constant for a given particle although it will vary between particles.


Always put the sign in for the value of the e/m ratio.



The Significance of this Discovery

Thomson then went on to show that the e/m ratio was the same whatever gas was used, and he concluded that all atoms contained electrons.


He then went on to propose that atoms were made up of electrons embedded in a uniform matrix of protons.  The total positive charge was balanced by the total negative charge.  If electrons were removed, the remaining ion was left with excess positive charge.


This became known as Thompson’s Plum Pudding model.



The Plum Pudding Model was the accepted theory for the structure of the atom until Ernest Rutherford’s alpha scattering experiments in 1911.  We make think it as rather twee nowadays, but the model worked well to explain things as far as physicists knew at the time.


Question 9

Is it possible to have anode rays?




Positive ions can be accelerated and deflected by a magnetic field.  The deflection will depend on the charge to mass ratio.  This is the basis of the mass spectrometer.


Question 10

A alpha particle is accelerated at a voltage V.  It passes into an electric field formed by the same voltage V across two plates that are 75.0 mm apart.  The alpha particle is not deflected when the magnetic field is 0.145 T.


(a) Calculate the charge to mass ratio of the alpha particle.

(b) Calculate the voltage, V, with which the alpha particle was accelerated.  Give your answer to an appropriate number of significant figures.


Mass of a proton = 1.67 ´ 10-27 kg.  Electronic charge = 1.60 × 10-19 C.



The Thomson Plum Pudding was disproved by Ernest Rutherford in his famous alpha scattering experiment.  You can find details of this HERE.  The idea of electrons orbiting the nucleus was put forward by Rutherford and Niels Bohr in 1913.  This is the model used in A-level physics text.


In those days, nobody knew about the neutron until James Chadwick discovered it in 1932, about 20 years after Rutherford discovered the nucleus.