Thermionic Emission (Edexcel International A Level Physics)

Revision Note

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Katie M


Katie M



Thermionic Emission

  • When metals are heated, the conduction electrons within them gain energy
  • If these electrons gain sufficient energy, they are able to leave the surface of the metal
    • This is known as thermionic emission 
    • This is similar to the photoelectric effect, but the energy absorbed by electrons in this case is due to thermal energy, rather than the energy absorbed by incident photons

  • Once electrons are released from a metal surface they may be accelerated by electric or magnetic fields


Electrons are emitted from the (negative) cathode and accelerated to the (positive) anode

Worked example

Electron guns use electric fields to accelerate electrons to very high speeds.

Show that an electron accelerated from rest across a potential difference of 5.0 kV attains a speed of 4.2 × 107 m s–1

Use the following data: 

  • Mass of an electron me = 9.11 × 10–31 kg
  • Electron charge e = 1.6 × 10–19 C

Step 1: List the known quantities

    • Potential difference = 5 kV = 5000 V
    • Mass of an electron, me = 9.11 × 10–31 kg
    • Electron charge e = 1.6 × 10–19 C

Step 2: Equate the kinetic energy gained by the electron to the energy transferred across the potential difference

    • Potential difference V is the energy transferred W per unit of charge Q, or V equals W over Q
    • Since the charge in this case is an electron, Q = e and so W eV
    • Therefore, the kinetic energy gained is equal to eV so we can write:

1 half m v squared equals e V

Step 3: Make speed v the subject of the equation

    • Rearranging this equation for v gives:

m v squared equals 2 e V

v squared equals fraction numerator 2 e V over denominator m end fraction

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

Step 4: Substitute quantities and calculate the speed v

    • Substituting known quantities gives:

v equals square root of fraction numerator 2 cross times left parenthesis 1.6 cross times 10 to the power of negative 19 end exponent right parenthesis cross times 5000 over denominator left parenthesis 9.11 cross times 10 to the power of negative 31 end exponent right parenthesis end fraction end root= 41 908 313... = 4.2 × 107 m s–1

Exam Tip

Examiners commonly test if candidates can equate the energy gained across a potential difference with the kinetic energy of a particle, as we did here. Make sure you can combine the equations for kinetic energy and potential energy in order to calculate speed like we did here!   

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Katie M

Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.