A LevelPhysicsAQARevision Notes12. Turning Points in Physics12.1 The Discovery of the Electron12.1.2 Thermionic Emission

Thermionic Emission (AQA A Level Physics)

Revision Note

Dan MG

Author

Dan MG

Last updated

Thermionic Emission

How do you Create a Cathode Ray?

  • We now know that cathode rays are electrons
  • The first discharge tubes produced cathode rays from the cathode using a strong electric field to "pull" electrons across the tube
  • Forming a cathode ray can be made easier by heating the cathode - this is called thermionic emission

Thermionic Emission

  • The electrons in the heated cathode (also called a heated filament) have more energy in their kinetic stores 
    • This is enough to leave the surface of the metal and move towards the anode

Cathode Ray Tubes

  • These are designed to "fire" the electrons emitted from the cathode towards a target - sometimes these are more dramatically called electron guns
  • Electrons are accelerated towards the anode, but pass through a hole in it and continue towards their target

Cathode Ray Tube Design

12-1-2-cathode-ray-tube-design

The filament is heated, giving electrons enough energy to be easily removed. All electrons are accelerated towards the anode's positive charge - some pass through a hole in the centre, forming a tight beam of electrons. 

Worked example

Figure 1 shows a thin filament placed near an anode with a hole in it. Passing current through the filament generates a beam of electrons.

Figure 1

12-1-2-worked-example-figure-1

Explain why passing a current through the filament causes the emission of electrons.

Answer:

Step 1: Recall that thermionic emission requires a heated cathode (filament)

  • The current heats the wire

Step 2: Explain why this higher temperature is needed

  • The electrons in the filament gain enough kinetic energy to escape the filament and accelerate towards the anode

Examiner Tip

Remember that current passing through a wire causes it to heat up - it is the higher temperature, not the current, that gives the electrons enough energy to be easily attracted from the cathode towards the anode. 

Work Done on an Electron

  • Using the concept of work done, the speed of electrons in a cathode ray tube can be calculated
  • Work done, W, by an electric field of potential difference, V, on a charge, q, is equal to:

W space equals space q V

  • For an electron:

W space equals space e V

  • In this instance, all work done on an electron is transferred to its kinetic store, and we can assume their initial velocity is near-zero so:

W space equals space E subscript k

e V space equals space 1 half m subscript e v squared

  • where me is electron mass and v is the speed of the electron

Worked example

An electron is accelerated through a vacuum by an electric field with a potential difference of 500 V. It exits through a small hole in a positive plate. 

Calculate its speed upon leaving the electric field. 

Answer:
Step 1: List the known quantities:

  • Electron mass, me = 9.11 × 10−31 kg
  • Magnitude of the charge of an electron, e =  1.60 × 10−19 C
  • Potential difference, V = 500 V

Step 2: Recall the equation relating work done on an electron and kinetic energy

  • To determine electron speed, v:

e V space equals space 1 half m subscript e v squared

Step 3: Rearrange this equation to make velocity the subject

  • Multiply both sides by 2, divide both by mass and square root both sides to get:

v space equals space square root of fraction numerator 2 e V over denominator m subscript e end fraction end root

Step 4: Substitute the known quantities

  • For the final answer:

v space equals space square root of fraction numerator 2 space cross times space open parentheses 1.60 space cross times space 10 to the power of negative 19 end exponent close parentheses space cross times space 500 over denominator 9.11 space cross times space 10 to the power of negative 31 end exponent end fraction end root space equals space 1.33 space cross times space 10 to the power of 7 space straight m space straight s to the power of negative 1 end exponent

Examiner Tip

This same concept of work done = kinetic energy is referred to again in Bertozzi's Experiment at the end of the Special Relativity section, so make sure you are familiar with this idea before moving on.

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Did this page help you?

    Previous:12.1.1 Cathode RaysNext:12.1.3 Specific Charge Experiments
    Dan MG

    Author: Dan MG

    Expertise: Physics

    Dan graduated with a First-class Masters degree in Physics at Durham University, specialising in cell membrane biophysics. After being awarded an Institute of Physics Teacher Training Scholarship, Dan taught physics in secondary schools in the North of England before moving to SME. Here, he carries on his passion for writing enjoyable physics questions and helping young people to love physics.