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The de Broglie Wavelength (HL) (HL IB Physics)

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Katie M

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Katie M

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The de Broglie Wavelength

  • Louis de Broglie thought that if waves can behave like particles, then perhaps particles can also behave like waves

  • He proposed that electrons travel through space as waves
    • This would explain why they can exhibit wave-like behaviour such as diffraction
  • De Broglie suggested that electrons must also hold wave properties, such as wavelength
    • This came to be known as the de Broglie wavelength
  • However, he realised that all particles can show wave-like properties, not just electrons
    • He hypothesised that all moving particles have a matter wave associated with them
  • The definition of a de Broglie wavelength is:

The wavelength associated with a moving particle

  • 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

  • Since momentum p = mv, the de Broglie wavelength can be related to the speed of a moving particle v by the equation:

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

Kinetic Energy & de Broglie Wavelength

  • Kinetic energy is defined as

E subscript K space equals space 1 half m v squared

E subscript K space equals space fraction numerator p squared over denominator 2 m end fraction space space space space space rightwards double arrow space space space space space p space equals space square root of 2 m E subscript K end root

  • Combining this with the previous equation gives relationship between the de Broglie wavelength of a particle to its kinetic energy:

lambda space equals space h over p space equals space fraction numerator h over denominator square root of 2 m E subscript K 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)
    • EK = kinetic energy of the particle (J)
    • m = mass of the particle (kg)
    • v = speed of the particle (m s1)

Worked example

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

Determine the ratio: fraction numerator de space Broglie space wavelength space of space the space proton over denominator de space Broglie space wavelength space of space the space electron end fraction

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


Answer:

2.5.4 De Broglie Wavelength Worked Example

Evidence for the Wave Nature of Matter

  • The majority of the time, and for everyday objects travelling at normal speeds, the de Broglie wavelength is far too small for any quantum effects to be observed
    • A typical electron in a metal has a de Broglie wavelength of about 10 nm
  • Therefore, the quantum effects of diffraction will only be observable when the width of the aperture is of a similar size to the de Broglie wavelength

Electron Diffraction Experiment

Electrons accelerated through a high potential difference demonstrate wave-particle duality 

  • The electron diffraction tube can be used to investigate how the de Broglie wavelength of electrons depends on their speed

  • To observe the diffraction of electrons, they must be focused through a gap similar to their de Broglie wavelength, such as an atomic lattice
    • Graphite film is ideal for this purpose because of its crystalline structure
    • The gaps between neighbouring planes of the atoms in the crystals act as slits, allowing the electron waves to spread out and create a diffraction pattern
    • This phenomenon is similar to the diffraction pattern produced when light passes through a diffraction grating
    • If the electrons acted as particles, a pattern would not be observed, instead, the particles would be distributed uniformly across the screen. The diffraction pattern is observed on the screen as a series of concentric rings
  • It is observed that:
    • a larger accelerating voltage reduces the diameter of a given ring
    • a lower accelerating voltage increases the diameter of the rings
  • As the voltage is increased:
    • The speed of the electrons is increased, therefore:
      • the momentum of the electrons is increased
      • the kinetic energy of the electrons is increased
      • the de Broglie wavelength is decreased
    • The angle of diffraction is decreased, therefore:
      • the radius of the diffraction pattern is decreased

  • Electron diffraction was the first clear evidence that matter can behave like light and has wave properties
  • The de Broglie wavelength can be used to calculate the angle of the first maximum in the diffraction pattern

n lambda space equals space d sin theta space space rightwards double arrow space space sin theta space equals space lambda over d

  • Where:
    • theta = angle of diffraction of the first maximum
    • lambda = de Broglie wavelength
    • d = spacing between atoms used for diffraction

 

  • The first minimum:
    • will be half the value of the first maximum
    • will not be zero
  • Subsequent minima reduce in intensity until reaching zero

Worked example

Electrons are accelerated through a film of graphite. The electrons are accelerated through a potential difference of 40 V. The spacing between the graphite atoms is 2.1 × 10−10 m.

Calculate the angle of the first minimum of the diffraction pattern.

Answer:

Step 1: Determine the kinetic energy gained by an electron

  • Kinetic energy gained through a potential difference of 40 V = 40 eV

E subscript k space equals space 40 space cross times space open parentheses 1.6 cross times 10 to the power of negative 19 end exponent close parentheses space equals space 6.4 cross times 10 to the power of negative 18 end exponent space straight J

Step 2: Determine the speed of the electron

E subscript k space equals space 1 half m v to the power of 2 space end exponent space space rightwards double arrow space space v space equals space square root of fraction numerator 2 E subscript k over denominator m end fraction end root

v space equals space square root of fraction numerator 2 open parentheses 6.4 cross times 10 to the power of negative 18 end exponent close parentheses over denominator 9.11 cross times 10 to the power of negative 31 end exponent end fraction end root space equals space 3.748 cross times 10 to the power of 6 space straight m space straight s to the power of negative 1 end exponent

Step 3: Determine the de Broglie wavelength of the electron

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

lambda space equals space fraction numerator 6.63 cross times 10 to the power of negative 34 end exponent over denominator open parentheses 9.11 cross times 10 to the power of negative 31 end exponent close parentheses open parentheses 3.748 cross times 10 to the power of 6 close parentheses end fraction space equals space 1.942 cross times 10 to the power of negative 10 end exponent space straight m

Step 4: Determine the angle of the first maximum

sin space theta space equals space lambda over D

sin space theta space equals space fraction numerator 1.942 cross times 10 to the power of negative 10 end exponent over denominator 2.1 cross times 10 to the power of negative 10 end exponent end fraction space equals space 0.9248

theta space equals space sin to the power of negative 1 end exponent open parentheses 0.9248 close parentheses space equals space 68 degree space open parentheses 2 space straight s. straight f. close parentheses

Step 5: Determine the angle of the first minimum

68 over 2 space equals space 34 degree

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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.