The de Broglie Wavelength (AQA AS Physics)

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The de Broglie Wavelength

  • The de Broglie equation relates wavelength (a property of waves) to momentum (a property of matter)
  • In the electron diffraction experiment:
    • Increasing the speed of electrons, and hence their momentum, causes the angle of diffraction to decrease (as seen by the decrease in the diameter of the diffraction rings)
  • This is in agreement with wave theory, where the angle of diffraction decreases as the wavelength decreases
    • Therefore, the greater the momentum of the particle, the smaller the de Broglie wavelength
  • Using ideas based on quantum theory and Einstein’s theory of relativity, de Broglie suggested that the momentum (p) of a particle and its associated wavelength (λ) are related by the equation:

lambda space equals space h over p

  • Using momentum space p space equals space m v, the de Broglie wavelength can be written in terms of the speed of a moving particle:

lambda space equals space fraction numerator h over denominator m v end fraction

  • Using kinetic energy E space equals space 1 half m v squared, momentum and kinetic energy can be related by:

Energy:  E space equals space 1 half m v squared space equals space fraction numerator space m v squared over denominator 2 end fraction space equals space fraction numerator space p squared over denominator 2 m end fraction 

Momentum:  space p space equals space square root of 2 m E end root

  • Combining this with the de Broglie equation gives a form which relates the de Broglie wavelength of a particle to its kinetic energy:

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

  • Where:
    • λ = the de Broglie wavelength (m)
    • h = Planck’s constant (J s)
    • p = momentum of the particle (kg m s-1)
    • E = kinetic energy of the particle (J)
    • m = mass of the particle (kg)
    • v = speed of the particle (m s-1)

Worked example

A proton and an electron are each accelerated from rest through the same potential difference.

Determine the ratio: fraction numerator d e space B r o g l i e space w a v e l e n g t h space o f space t h e space p r o t o n over denominator d e space B r o g l i e space w a v e l e n g t h space o f space t h e space e l e c t r o n end fraction

  • Mass of a proton = 1.67 × 10–27 kg
  • Mass of an electron = 9.11 × 10–31 kg

2.5.4 De Broglie Wavelength Worked Example

Exam Tip

Particles with a greater mass, such as a proton, have a greater momentum. The greater the momentum, the smaller the de Broglie wavelength. Always perform a logic check on your answer to check that makes sense.

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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.