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:

$λ = \frac{h}{m v}$

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

$E_{k} = 4000 \times (1.6 \times 10^{- 19}) = 6.4 \times 10^{- 16} J$

Step 2: Determine the speed of the electron

$E_{k} = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{\frac{2 E_{k}}{m}}$

$v = \sqrt{\frac{2 \times (6.4 \times 10^{- 16})}{9.11 \times 10^{- 31}}} = 3.748 \times 10^{7} m s^{- 1}$

Step 3: Determine the de Broglie wavelength of the electron

$λ = \frac{h}{p} = \frac{h}{m v}$

$λ = \frac{6.63 \times 10^{- 34}}{(9.11 \times 10^{- 31}) (3.748 \times 10^{7})} = 1.942 \times 10^{- 11} m$

Step 4: Determine the angle of the first minimum

The diffraction grating equation is given by

$d \sin θ = n λ$

For the first minimum, $n = \frac{1}{2}$

$\sin θ = \frac{λ}{2 d}$

$\sin θ = \frac{1.942 \times 10^{- 11}}{2 \times (1.4 \times 10^{- 10})} = 0.0694$

$θ = \sin^{- 1} (0.0694) = 4.0 ° (2 s . f .)$

Examiner Tip

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

Diffraction Effects of Momentum (AQA AS Physics)

Revision Note