148.the Discovery of the Electron

Physics Factsheet www.curriculum-press.co.uk

Number 148

The Discovery of the Electron Properties of Cathode Rays

In the nineteenth century, physicists were aware that electric currents could travel through solids, liquids, and even gases (e.g. lightning). The question arose – “Could electric current travel through a vacuum?”. It wasn’t until the mid-nineteenth century that a good enough vacuum could be produced to investigate this idea.

Some scientists hypothesized that cathode rays were a form of light rays. However Crookes then showed that he could deflect the cathode rays using a magnet.

The cathode ray tube The first vacuum tube experiments produced unexpected results.

•

Supply

• • A _

+ Anode

Crookes thought that the cathode rays could be a flow of negatively charged particles repelled from the cathode and attracted to the anode.

Cathode

Cathode ray to electron beam

A large potential difference was applied across the tube. A mysterious glow was observed in the tube, and the glass near the anode glowed green.

J.J.Thomson then took up the investigation of the cathode ray. He also thought that cathode rays were composed of negatively charged particles, but problems had arisen with this theory. The German physicist, Hertz, had performed two experiments with unexpected results: 1. A beam of charged particles would be deflected by an electric field as well as a magnetic field. But Hertz found no deflection. + Cathode Glow

Example 1 How could they be sure that a current was flowing between the anode and cathode in the tube?

Answer The ammeter indicated a current flow. As the vacuum inside the tube was improved, this glow faded, but the glass around the anode still glowed green.

Anode

Sir William Crookes hypothesized that rays were travelling from the cathode, and striking the glass near the anode, causing it to glow. Crookes verified this with a tube using a metal barrier (shaped as a Maltese cross) between the anode and the end of the tube. The cross caused a shadow to form on the end of the tube. The rays were obviously travelling in a straight line from the cathode .

Cathode

The results of Crookes’ work were: the cathode rays were not a form of light. (Light is not affected by magnetic fields.) cathode rays carried a charge – they were accompanied by current flow the cathode rays travelled from the cathode to the anode (negative to positive)

_ Deflection plates

2. When he put a sheet of gold foil inside the tube, the rays passed straight through the foil, still causing the glass to glow. Particles could not pass through solid matter. Thomson repeated Hertz’ experiment with the electric plates (in 1897), but he was able to produce a much better vacuum than Hertz. This time the cathode ray was deflected by the electric field between the plates. + Glow

Shadow Glow

_ Anode

Maltese Cross As expected for negatively charged particles (electrons), the deflection was towards the positive plate.

Example 2 What was surprising about the discovery that the rays travelled from the cathode to the anode?

(Where had Hertz gone wrong? Some of the air molecules remaining in the tube became ionised by the cathode ray. The positive and negative ions then travelled to the oppositely charged plates, considerable reducing the electric field between the plates.)

Answer The conventional assumption for many years among physicists was that the electric current flowed from positive to negative. Crookes showed that, in this example at least, the opposite was occurring.

Thomson was now convinced that the cathode ray was an electron beam.

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Physics Factsheet

148. The Discovery of the Electron

Exam Hint:- Be prepared to explain why the properties of the cathode ray led physicists to believe that they were actually dealing with a beam of charged particles (electrons).

Remember that both electric and magnetic forces depend on the applied fields, but only the magnetic force depends on the speed of the charged particle.

The specific charge of the electron

(b) Electric deflection only

There was no way (at that time) of detecting the charge or mass of the electron. But Thomson set out to find the specific charge (the charge-to-mass ratio for the electron). This was done in two steps.

The magnetic field was then turned off, and the beam deflected upwards by the electric field: V

(a) Balancing forces If an electron beam was travelling towards the right, and an electric field were to be applied downwards, the negative electrons would be deflected upwards:

e

h

θ

E e d

The force on each electron in the field would be F = eE. (where e is the charge on one electron and E is the value of electric field strength)

The expression for the charge-to-mass ratio of the electrons in the beam can be expressed as shown below

If the same beam (velocity, v) had a magnetic field into the page exerted on it, the deflection would be downwards:

e

2

e hvx tanθ Ckg-1 m= Vd

B

Remember, vx has already been determined. The other variables can be measured. Thomson obtained the value: e = 1.8×10-11 Ckg-1 ( to 2s.f) m This value was over 1800 times bigger than the charge-to-mass ratio (specific charge) of the hydrogen atom (ion). And the hydrogen atom was the smallest particle known at that time.

The magnetic force on each electron would be F = Bev. (where B is the magnetic field strength) If both fields were applied, and the magnitudes adjusted until the beam was undeflected, then the force exerted by the magnetic field must equal the force exerted by the electric field. Therefore: Bevx = Ee so vx =

Example 4 What are two most obvious possible conclusions that can be drawn from this result?

E B

Answer (a) The electron has the same mass as the hydrogen atom, and a charge 1800 times bigger.

By measuring the electric and magnetic fields, the horizontal speed, vx of the electron beam could be found.

(b) The charges are the same, and the electron is a tiny particle, with a mass 1800 times smaller than the hydrogen atom.

Example 3 Two plates with separation 25mm have a p.d. of 30V across them. ×104ms-1 is directed between the An electron beam with vx = 1.5× plates. Find the electric field between the plates, and the crossed magnetic field required to balance the electric force.

Example 5 Which of these explanations best explains Hertz’ observations with the gold foil?

Answer

Answer

E = V = 30 = 1200Vm-1 d 0.025

If the electron is an extremely tiny particle, then perhaps it is not surprising that an electron beam can travel through the gold foil.

vx = E B

B = E = 1200 4 = 0.008T vx 1.5 × 10

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Physics Factsheet

148. The Discovery of the Electron Millikan’s Experiment Robert Millikan worked out a way to determine the value of the charge on an electron. He used a fine spray of oil droplets, some of which contained a tiny unbalanced electric charge. He assumed that the electron had a fixed charge, and that these oil droplets held the charge from only a small number of electrons.

The upthrust of the air is often ignored in calculations, as it is often insignificant. But you should be able to show in any particular case that its value is very small.

First he used an electric field to balance the weight of the oil droplet.

Combining the equations allows the value of the electric charge, Q, on the oil droplet to be found.

+

+

+ + + QE

The upwards electric force = QE

droplet

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_

_

Therefore QE = 6πrηv

The downwards force = the weight of the oil drop = mg

mg _

QE = 4 πr3ρg and 4 πr3ρg = 6πrηv 3 3 and

Q = 6π r η v E

The oil droplets will have different numbers of electron charges on them, but Q is always found to be a simple multiple of the accepted electron charge, e = -1.6×10-19C. In this way, the electron charge could be determined.

_

The radius of the oil droplet could be measured so its volume could be calculated. The density, ρ, of the oil was known. Therefore the mass of the oil drop was calculated as 4 πr3ρ. 3 4 Therefore: QE = πr3 ρg 3

Example 5

The electric field was then turned off, and the terminal velocity of the oil droplet falling under gravity was measured. At terminal velocity, the weight of the droplet is balanced by the air resistance. (The upthrust, or buoyancy, from the air turns out to be insignificant, so is omitted from these calculations.)

Answer

Results from Millikan’s experiment on oil droplets give negative ×10-19C, 9.5× ×10-19C, and 6.3× ×10-19C. How charge values of 5.0× many electron charges are on each droplet? Droplet 1 2 3

Air resistance can be calculated using: air resistance = 6πrηv (where η is the viscosity of the air)

Q/C -5.0 × 10 -19 -9.5 × 10 -19 -6.3 × 10 -19

Number of electrons 3 electrons (3×1.6=4.8) 6 electrons (6×1.6=9.6) 4 electrons (4×1.6=6.4)

This gives the equation: 4 πr3ρg = 6πrηv 3

Practce Questions Electric field =

1. (a) Electrons are accelerated from rest across a vacuum tube (cathode to anode) through a p.d. of 5000V. Show that the resultant electron speed is about 4.2×107ms-1.

mg 1.37×10-11 = 2.85×107NC-1 (or Vm-1) = Q 3×1.6×10-19

3. Weight = mg = 1.4×10-12 × 9.81 = 1.37×10-11kg

(b) The electron beam then passes between horizontal parallel plates 6.0cm apart with a p.d. of 1500V between them. Calculate the electric field strength between the plates.

2. The downward force (weight minus upthrust) is: 4 F = πr3 (ρoil−ρair)g. 3 The density of the oil is hundreds of times greater than the density of the air.

(c) Determine the force applied to each electron by this field. (d) Describe the path of the electron before it reaches the plates, while it is between the plates, and after it leaves the plates.

(d) horizontal straight line, upward or downward curve (parabola), straight line in the direction determined by the end of the parabola.

2. Why can we ignore the upthrust of the air in Millikan’s experiment?

(c) 4.0×10-15N

3. An oil droplet has a mass of 1.4×10-6mg and holds 3 electron charges. Find the electric field required to balance its weight.

1. (b) 25 000 Vm-1 (the distance must be in metres)

Answers

Acknowledgements: This Physics Factsheet was researched and written by Paul Freeman The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

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