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De Broglie's Hypothesis (AQA A Level Physics)

Revision Note

Dan Mitchell-Garnett

Last updated

De Broglie's Hypothesis of Wave-Particle Duality

What was DeBroglie's hypothesis?

  • Louis DeBroglie hypothesised that all particles can behave both like waves and like particles, following Einstein's work with photons

  • By equating two equations from Einstein, he derived an equation for the momentum of a photon:

E space equals space m c squared (more on this in Special Relativity)

E space equals space h f (the energy of a photon)

m c squared space equals space h f

m c space equals space h f over c space equals space h over lambda

  • Where h  is Planck's constant, c  is the speed of light, m  is mass, λ  is wavelength and f  is frequency

  • mc  is the momentum, p, of a photon - DeBroglie extended this idea to particles with mass to obtain the relation you should recall from Particles & Radiation:

p space equals space h over lambda

Finding the Wavelength of Accelerated Particles

  • This idea can be applied to accelerated electrons to find their wavelength

    • Finding their momentum directly is difficult, but recall from The Discovery of the Electron that the work done on an electron by an electric field (eV) is equal to its kinetic energy - this can be used to find the electron's speed:

e V space equals space 1 half m v squared

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

  • This can be substituted into the momentum term in DeBroglie's hypothesis to then find wavelength:

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

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

  • The wavelength of the electron depends on the work done on it by the electric field, eV

    • From this equation, as eV  increases, λ  decreases

    • When the electron is accelerated to a higher speed, its DeBroglie wavelength decreases

Worked Example

An electron is accelerated through an electric field and is found to have a DeBroglie wavelength of λ. The potential difference across the electric field then increases by a factor of 25. Write the new wavelength of the electron in terms of λ.

Answer:

Step 1: Write out the equation for an accelerated particle's wavelength from your data and formulae sheet:

  • The wavelength of an accelerated particle is:

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

Step 2: Label the new wavelength and substitute the new potential difference:

  • We call label the new wavelength lambda with tilde on top and substitute the new potential difference, 25V :

lambda with tilde on top space equals space fraction numerator h over denominator square root of 2 m e cross times space 25 V end root end fraction

  • Now we will manipulate this expression until we can pull out the original expression for λ :

lambda with tilde on top space equals space fraction numerator h over denominator square root of 25 space cross times space 2 m e V end root end fraction space equals space fraction numerator h over denominator square root of 25 space cross times space square root of 2 m e V end root end fraction

lambda with tilde on top space equals space fraction numerator 1 over denominator square root of 25 space end fraction space cross times space fraction numerator h over denominator square root of 2 m e V end root end fraction space equals space 1 fifth space cross times space fraction numerator h over denominator square root of 2 m e V end root end fraction space equals space 1 fifth space cross times space lambda

  • Therefore the new wavelength is:

lambda with tilde on top space equals space lambda over 5

  • This checks out with common sense - the particle is moving faster under a stronger potential difference so, as was mentioned above, its new wavelength should be smaller

Examiner Tips and Tricks

This equation requires some confidence in algebra involving square roots. Remembering that you can combine square roots when multiplying or combining will help a great deal:

square root of m space end root space cross times space square root of n space equals space square root of m space cross times space n end root

fraction numerator square root of m over denominator square root of n end fraction space equals space square root of m over n end root.

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    Dan Mitchell-Garnett

    Author: Dan Mitchell-Garnett

    Expertise: Physics Content Creator

    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 Save My Exams. Here, he carries on his passion for writing challenging physics questions and helping young people learn to love physics.