AQA A Level Physics

Revision Notes

12.2.9 Relativistic Kinetic Energy

Electron Diffraction

Do Electrons Diffract?

  • To investigate whether electrons did exhibit wave-like properties, an electron diffraction tube was produced
  • The electrons were accelerated in an electron gun with a high potential difference, such as 5000 V, and were then directed through a thin film of graphite
    • The gaps between the carbon atoms were sufficiently small to cause diffraction, using the predicted DeBroglie wavelength of electrons 
    • Diffraction is a property of waves when passing through a small gap - if the electrons were seen to diffract then this was proof of their wave-like properties
  • The electrons were indeed seen to diffract from the gaps between carbon atoms and produce a circular pattern on a fluorescent screen made from phosphor

Diagram of the electron diffraction experiment

Investigating Electron Diffraction

A low voltage causes thermionic emission of electrons at the cathode, then these are accelerated towards anode directed through the graphite. The electrons diffract through the gaps between atoms in the graphite, forming a circular diffraction pattern made visible by the phosphor screen.

The Effect of Potential Difference on the Diffraction Pattern

  • Recall that DeBroglie's hypothesis predicted that, if more work was done on the electrons, their wavelength would be shorter
    • Recall from the diffraction grating subtopic of the Waves topic, that a shorter wavelength leads to a smaller diffraction angle θ  for a given gap width
  • Therefore, if DeBroglie was right, increasing the potential difference would cause the diffraction rings to move closer to the centre of the phosphor screen

Diagram showing the effect of increasing potential difference on diffraction

Electron Diffraction Experiment

The left of this diagram shows the equipment from another angle. The right of this diagram shows that increasing the voltage caused the diffraction rings to move closer to the centre of the screen, following DeBroglie's equation.

  • This was the first experimental proof of electrons exhibiting wave-like properties
    • The wavelength of electrons was significantly smaller than high energy forms of electromagnetic radiation, such as X-rays

Exam Tip

This topic builds on knowledge acquired from studying diffraction gratings. As is becoming a common theme with Turning Points, go back and make sure your understanding of diffraction is solid before getting stuck in with this topic.

Estimate of Anode Voltage

  • Changing the voltage across the anode (it is the voltage that causes the electrons to accelerate), scientists can manipulate the wavelength of the electron
  • A microscope's resolving power (its ability to differentiate between two nearby points on an object) depends on the wavelength of the radiation being used
    • A shorter wavelength means the microscope can be used to see finer details in an object
    • Electrons had wavelengths much shorter than X-rays
  • The size of an atom is roughly 10−10 m
    • By using DeBroglie's equation for an accelerated electron, researchers could find the accelerating voltage needed to make the electron's wavelength the same size as an atom

Worked example

An atom of carbon has a diameter of 1.5 × 10−10 m. Determine the anode potential difference needed in an electron diffraction tube to produce electrons with a wavelength equal to the diameter of a carbon atom.

Answer:

Step 1: Write down the equation for wavelength of accelerated electrons from the data and formulae booklet:

  • The wavelength of an accelerated electron is:

lambda space equals space fraction numerator h over denominator square root of 2 m e V end root end fraction

Step 2: List the known quantities:

  • Mass of an electron, m  = 9.11 × 10−31 kg
  • Magnitude of charge on an electron, e  = 1.60 × 10−19 C
  • Planck's constant, h  = 6.63 × 10−34 J s
  • Required wavelength of the electron, λ  = 1.5 × 10−10 m

Step 3: Rearrange the wavelength equation to make anode potential difference the subject:

  • Separate the V  term from the square root:

lambda space equals space fraction numerator h over denominator square root of 2 m e end root square root of V end fraction

square root of V space equals space fraction numerator h over denominator square root of 2 m e end root space cross times space lambda end fraction

  • Square the whole equation:

V space equals space fraction numerator h squared over denominator 2 m e lambda squared end fraction

Step 4: Substitute the known quantities:

V space equals space fraction numerator open parentheses 6.63 space cross times space 10 to the power of negative 34 end exponent close parentheses squared over denominator 2 space cross times space open parentheses 9.11 space cross times space 10 to the power of negative 31 end exponent close parentheses 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 open parentheses 1.5 space cross times space 10 to the power of negative 10 end exponent close parentheses squared end fraction

V space equals space 67.0 space straight V

Exam Tip

In a high pressure situation like an exam, it's easy to make mistakes on your calculator when substituting values in standard form. Use brackets carefully and retype your equation to ensure you get the same value when checking over your answers.

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Dan MG

Author: Dan MG

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.