Nuclear Radius (AQA A Level Physics)

• Nuclear radius can be measured experimentally using one of two methods
  • Rutherford scattering - the closest approach method
  • Electron scattering

Closest Approach Method

• In Rutherford scattering:
  • Alpha particles were fired at a thin sheet of gold foil
  • Some of the alpha particles were found to rebound from the gold foil by 180°

• Rutherford scattering indicates that there must be an electrostatic repulsion between the alpha particles and the gold nucleus
• The initial kinetic energy of an alpha particle is:

• At the point of closest approach, r, the repulsive force reduces the speed of the alpha particles to zero momentarily
• At this point, the initial kinetic energy of an alpha particle, E_k, is equal to electric potential energy, E_p
• The radius of the closest approach can be found be equating the initial kinetic energy to the electric potential energy

• Equating the two equations gives:

An alpha particle transfers its maximum kinetic energy to maximum potential energy at the distance of closest approach to a gold nucleus

Advantages of the Closest Approach Method

1. Alpha scattering gives a good estimate of the upper limit for a nuclear radius
2. The mathematics behind this approach are very simple
3. The alpha particles are scattered only by the protons and not all the nucleons that make up the nucleus

Disadvantages of the Closest Approach Method

1. Alpha scattering does not give an accurate value for nuclear radius as it will always be an overestimate
   • This is because it measures the smallest separation between the alpha particle and the gold nucleus, not its radius
2. Alpha particles contain hadrons which are affected by the strong nuclear force
   • This affects high-energy alpha particles which get very close to the nucleus (0.5 to 3 fm)
   • The mathematics in this method cannot account for the effects due to the strong nuclear force
3. The gold nucleus will recoil as the alpha particle approaches
4. Alpha particles have a finite size and mass whereas electrons can be treated as a point mass
5. Very few alpha particles rebound at exactly 180°, so to detect these, a small collision region is required
6. The alpha particles in the beam are assumed to have the same initial kinetic energy which may not be realistic
7. The foil target must be extremely thin to prevent multiple scattering

Electron Scattering

• Electrons accelerated to close to the speed of light are found to have wave-like properties, such as the ability to diffract
• The de Broglie wavelength of an electron is equal to:

• Where:
  • h = Planck's constant
  • m = mass of an electron (kg)
  • v = speed of the electrons (m s⁻¹)

• The diffraction pattern forms a central bright spot with dimmer concentric circles around it
• From this pattern, a graph of intensity against diffraction angle can be used to find the diffraction angle of the first minimum
• Using this, the size of the atomic nucleus, R, can be determined from:

• Where:
  • θ = angle of the first minimum (degrees)
  • λ = de Broglie wavelength (m)
  • R = radius of the nucleus (m)

The diffraction pattern produced by passing high-energy electrons through a target foil can be used to determine the nuclear radius

Pros & Cons of Electron Diffraction Method

Advantages

• Electron diffraction is much more accurate than the closest approach method
• This method gives a direct measurement of the radius of a nucleus
• Electrons are leptons; therefore, they will not interact with nucleons in the nucleus through the strong nuclear force as an alpha particle would

Disadvantages

• Electrons must be accelerated to very high speeds to minimise the de Broglie wavelength and increase resolution
  • This is because significant diffraction takes place when the electron wavelength is similar in size to the nuclear diameter

• Electrons can be scattered by both protons and neutrons
  • If there is an excessive amount of scattering, then the first minimum of the electron diffraction can be difficult to determine

• The graph of intensity against angle obtained through electron diffraction is as follows:

Step 1: Identify the first minimum from the graph

• Angle of first minimum, θ = 42°

Step 2: Write out the equation relating the angle, wavelength, and nuclear radiusStep 3: Calculate the nuclear radius, R