Diffraction Effects of Momentum (AQA A Level Physics)

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Katie M

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Diffraction Effects of Momentum

  • When electrons pass through a slit similar in size to their de Broglie wavelength, they exhibit diffraction, a property of waves

  • The regular spacing of atoms in a crystalline solid acts as a diffraction grating, scattering the electrons in a predictable manner

  • The observed diffraction pattern can be used to deduce the structure of the crystal producing that pattern

  • High energy electrons have a shorter wavelength and can therefore be used to look at the size of the nucleus of an atom (as opposed to the arrangement of atoms in a crystal)

  • The de Broglie wavelength tells us about the wave-particle relationship:

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

  • Where:

    • λ = the de Broglie wavelength (m)

    • h = Planck’s Constant (J s)

    • m = mass of the electron (kg)

    • v = velocity of the electron (m s–1)

Diffraction Effects of Momentum, downloadable AS & A Level Physics revision notes

Comparison of electron diffraction patterns at different values of momentum

Momentum of electrons

  • Momentum is equal to p = mv, so, from de Broglie's equation:

    • A smaller momentum will result in a longer wavelength

    • A larger momentum will result in a shorter wavelength

Kinetic energy of electrons

  • The speed of an electron can be increased by increasing the accelerating voltage (or potential difference)

  • If the electron speed, and therefore kinetic energy is increased, then:

    • The wavelength of the wave will decrease

    • The diffraction rings will appear closer together

  • The higher the kinetic energy of the electron, the higher its momentum hence the shorter its de Broglie wavelength

Radius of the diffraction pattern

  • The radius of the diffraction pattern depends on the wavelength:

    • The longer the wavelength, the more the light spreads out hence a larger radius is produced

    • The shorter the wavelength, the smaller the radius produced

  • Therefore, electrons with smaller momentum will produce a more diffuse diffraction pattern

Worked Example

Electrons are accelerated through a film of graphite. The electrons are accelerated through a potential difference of 4 kV. The spacing between the graphite atoms is 1.4 × 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 4 kV = 4 keV = 4000 eV

E subscript k space equals space 4000 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 16 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 cross times open parentheses 6.4 cross times 10 to the power of negative 16 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 7 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 space 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 7 close parentheses end fraction space equals space 1.942 cross times 10 to the power of negative 11 end exponent space straight m

Step 4: Determine the angle of the first minimum

  • The diffraction grating equation is given by

d space sin space theta space equals space n lambda

  • For the first minimum, n space equals space 1 half

sin space theta space equals space fraction numerator lambda over denominator 2 d end fraction

sin space theta space equals space fraction numerator 1.942 cross times 10 to the power of negative 11 end exponent over denominator 2 cross times open parentheses 1.4 cross times 10 to the power of negative 10 end exponent close parentheses end fraction space equals space 0.0694

theta space equals space sin to the power of negative 1 end exponent space open parentheses 0.0694 close parentheses space equals space 4.0 degree space open parentheses 2 space straight s. straight f. close parentheses

Examiner Tips and Tricks

Take a look at the revision note on diffraction gratings if you aren't sure where the equation used in the final step of the worked example comes from

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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.