Topic 1 – The Discovery of the Electron

In the exam you are expected to:

• Know about production of cathode rays in a discharge tube;

• Understand the principle of thermionic emission;

• Understand the work done on an electron through a p.d. and that eV = ½ mv²;

• Understand the significance of Thomson’s determination of e/m;

• Compare it to the specific charge of a hydrogen ion.

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. 

This was the starting point for Joseph John Thomson to produce his cathode ray tube (CRT) in 1897, the descendents of which we see every day.

Question 1 Explain what these different parts of the CRT do.

(a) Cathode

(b) Anode

ANSWER 

Question 2 Why is there a vacuum in the CRT?  ANSWER

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 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

ANSWER 

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:

• In an electric field the electron is attracted to the positive plate.  It will accelerate towards the positive plate;

• In a magnetic field the electron is deflected in a circular path at right angles to the magnetic field.

 Question 4  What shape is the path of a moving electron in:

(a) Magnetic Field

(b) Electric Field?  ANSWER

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? ANSWER  

We know that:

• The Force due to the magnetic field, F = Bev

• The electric field is given by E = V/d

• The electric field is force per unit charge E = F/e

• Force due to the electric field is F = eV/d

In this argument, we will make the accelerating voltage the same as the voltage between the two plates.  If they are different, the argument is more complex.

Since the force due to the electric field = force due to the magnetic field, we can write:

The kinetic energy supplied is eV (which is upstairs on the equation above).  We also know that:

eV = ½ mv²

So we can write:

We can do some cancelling:

This can be rearranged to give us;

The terms in the equation are:

• e – the electronic charge, 1.6 ´ 10^-19 C;

• m – the mass of an electron, 9.11 ´ 10^-31 kg;

• v – the velocity of the electron;

• B – the value of the magnetic field in T;

• d – the separation of the plates in m.

Question 6 Which of these terms is:

(a) Easy to measure directly;

(b) Hard to measure directly;

(c) Of constant value?

ANSWER

There are two terms that need working out:

• the velocity which we can work out from the kinetic energy;

• the B-field which uses the equation B = 0.716 m₀ NI/r.  This is not in the syllabus so we won’t pursue it any further.

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? ANSWER

Question 8

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

Mass of a proton = 1.661 ´ 10^-27 kg.  ANSWER

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.

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.

Question 9 Is it possible to have anode rays? ANSWER

  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.

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