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The de Broglie Wavelength (AQA A Level Physics)

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Katie M

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Katie M

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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

Answer:

Step 1: Determine how the proton and electron can be related via their mass

  • The only information we are given is the mass of the proton and the electron

  • When the proton and electron are accelerated through a potential difference, their kinetic energy will increase

  • Therefore, we can use kinetic energy to relate them via their mass

Step 2: Write out the equation for the de Broglie wavelength in terms of the kinetic energy of the particle

  • The de Broglie wavelength

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

  • Kinetic energy

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

  • Kinetic energy in terms of momentum

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

  • Substitute the expression for momentum into the de Broglie wavelength equation

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

Step 3: Find the proportional relationship between the de Broglie wavelength and the mass of the particle

lambda space equals space fraction numerator h over denominator square root of 2 m E end root end fraction space equals space k fraction numerator 1 over denominator square root of m end fraction

  • Where k is a constant

    • Since h is constant, and E is equal, then:

lambda space proportional to space fraction numerator 1 over denominator square root of m end fraction

Step 4: Calculate the ratio

lambda subscript p over lambda subscript e space equals space fraction numerator 1 over denominator square root of m subscript p end root end fraction space divided by space fraction numerator 1 over denominator square root of m subscript e end root end fraction space equals space fraction numerator 1 over denominator square root of m subscript p end root end fraction space cross times space fraction numerator square root of m subscript e end root over denominator 1 end fraction space equals space fraction numerator square root of m subscript e end root over denominator square root of m subscript p end root end fraction

square root of m subscript e over m subscript p end root space equals space square root of fraction numerator 9.11 cross times 10 to the power of negative 31 end exponent over denominator 1.67 cross times 10 to the power of negative 27 end exponent end fraction end root space equals space 0.0234 space open parentheses 3 space straight s. straight f. close parentheses

  • This means that the de Broglie wavelength of the proton is 0.0234 times smaller than that of the electron

  • Or that the de Broglie wavelength of the electron is about 40 times larger than that of the proton

Examiner Tips and Tricks

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

    Expertise: Physics

    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.