12.2.8 De Broglie's Hypothesis

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:

$E = m c^{2}$ (more on this in Special Relativity)

$E = h f$ (the energy of a photon)

$m c^{2} = h f$

$m c = h \frac{f}{c} = \frac{h}{λ}$

- 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:

$p = \frac{h}{λ}$

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:

$e V = \frac{1}{2} m v^{2}$

$v = \sqrt{\frac{2 e V}{m}}$

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

$p = m v = m \sqrt{\frac{2 e V}{m}} = \sqrt{m^{2}} \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{m^{2} 2 e V}{m}} = \sqrt{2 m e V}$

$λ = \frac{h}{\sqrt{2 m e V}}$

- 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:

$λ = \frac{h}{\sqrt{2 m e V}}$

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

We call label the new wavelength $\tilde{λ}$ and substitute the new potential difference, 25V :

$\tilde{λ} = \frac{h}{\sqrt{2 m e \times 25 V}}$

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

$\tilde{λ} = \frac{h}{\sqrt{25 \times 2 m e V}} = \frac{h}{\sqrt{25} \times \sqrt{2 m e V}}$

$\tilde{λ} = \frac{1}{\sqrt{25}} \times \frac{h}{\sqrt{2 m e V}} = \frac{1}{5} \times \frac{h}{\sqrt{2 m e V}} = \frac{1}{5} \times λ$

Therefore the new wavelength is:

$\tilde{λ} = \frac{λ}{5}$

- 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

Exam Tip

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:

$\sqrt{m} \times \sqrt{n} = \sqrt{m \times n}$

$\frac{\sqrt{m}}{\sqrt{n}} = \sqrt{\frac{m}{n}}$ .

AQA A Level Physics

Revision Notes

De Broglie's Hypothesis of Wave-Particle Duality

What was DeBroglie's hypothesis?

Finding the Wavelength of Accelerated Particles

Worked example

Exam Tip

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Author: Dan MG