3.8.1.5
Electron diffraction gives a more accurate measurement of nuclear radius
Nuclear Structure & Radiation — AQA A-Level Physics
- High-speed electrons have a de Broglie wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m).The wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). associated with a moving particle, demonstrating wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m).; photons exhibit particle behaviour in the photoelectric effectThe emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). is incident on it... comparable to the size of a nucleus.
- When a beam of electrons hits a thin target, each electron diffracts around the nuclei.
- The diffraction pattern is a central bright spot with dimmer concentric rings.
- The angle of the first minimum intensityThe powerThe rate of energy transfer. Measured in watts (W). transmitted per unit area perpendicular to the wave direction. Measured in W m⁻². Proportional to amplitude squared. is used to calculate the nuclear radius.
$$\sin \theta = 1.22 \frac{\lambda}{2R}$$
- $θ$: angle of the first minimum (°)
- $λ$: de Broglie wavelengthThe wavelength associated with a moving particle, demonstrating wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelength; photons exhibit particle behaviour in the photoelectric effectThe emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). is incident on it... of the electrons (m)
- $R$: radius of the nucleus (m)
Advantages over closest approach
- Gives a direct measurement of nuclear radius, not just an upper limit.
- Much more accurate — electrons are leptons, not affected by the strong nuclear force.
- Electrons can be treated as point particles.
Disadvantages
- Electrons must be accelerated to very high speeds (close to c) to achieve a short enough de Broglie wavelengthThe wavelength associated with a moving particle, demonstrating wave-particle dualityThe concept that all matter and radiation exhibit both wave-like and particle-like properties. Particles have a de Broglie wavelength; photons exhibit particle behaviour in the photoelectric effectThe emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). is incident on it....
- Excessive scattering can make the first minimum difficult to identify.
Common Mistake
MEDIUM
Students often: Don't say electron diffraction occurs because the electrons pass through the gap between nuclei.
Instead: The de Broglie wavelength is comparable to the nucleus, not the gap between nuclei. Electrons diffract AROUND a nucleus, like a sound wave around a barrier.
Instead: The de Broglie wavelength is comparable to the nucleus, not the gap between nuclei. Electrons diffract AROUND a nucleus, like a sound wave around a barrier.
Worked Example
Electrons with de Broglie wavelength 3.35 × 10⁻¹⁵ m are diffracted by oxygen-16 nuclei. The first minimum occurs at θ = 42°. Calculate the nuclear radius.
Show Solution
1
Write the equation
$$\sin \theta = 1.22 \frac{\lambda}{2R}$$
2
Rearrange for R
$$R = \frac{1.22 \lambda}{2 \sin \theta}$$
3
Substitute and calculate
$$R = \frac{1.22 \times 3.35 \times 10^{-15}}{2 \times \sin 42°} = \frac{4.087 \times 10^{-15}}{1.338} = 3.05 \times 10^{-15} \text{ m}$$
Answer
$R = 3.05 \times 10^{-15}$ m