Nuclear Radius (AQA A Level Physics): Revision Note

Estimating Nuclear Radius

• 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 are fired at a thin sheet of gold foil

  • Some of the alpha particles are 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 alpha particle is fired with an initial kinetic energy $E_k$

• At the point of closest approach $r$, the repulsive force reduces the speed of the alpha particles to zero momentarily

• The initial kinetic energy of the alpha particle is transferred to electric potential energy, which is given by:

$E_p=\frac{Qq}{4\mathrm{\pi }{\epsilon }_{0}r}$

• Where:

  • Charge of an alpha particle, Q = 2e

  • Charge of a gold nucleus, q = 79e

  • e = elementary charge (C)

  • r = the radius of closest approach (m)

  • ε₀ = permittivity of free space

• At this point, the initial kinetic energy $E_k$ of the alpha particle is equal to the electric potential energy $E_p$ of the gold nucleus:

$E_k=E_p=\frac{\left(2e\right)\left(79e\right)}{4\mathrm{\pi }{\epsilon }_{0}r}=\frac{158e^2}{4\mathrm{\pi }{\epsilon }_{0}r}$

• Rearranging for the distance of closest approach $r$:

$r=\frac{158e^2}{4\mathrm{\pi }{\epsilon }_0{E}_p}=\frac{158e^2}{4\mathrm{\pi }{\epsilon }_0{E}_k}$

• This gives a value for the radius of the nucleus, assuming the alpha particle is fired at a high energy

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

• In Rutherford scattering using alpha particles of initial energy 5 MeV:

$r=\frac{158e^2}{4\mathrm{\pi }{\epsilon }_0{E}_k}=\frac{158\times {\left(1.6\times {10}^{-19}\right)}^2}{4\mathrm{\pi }\left(8.85\times {10}^{-12}\right)\times \left(5\times {10}^6\right)\left(1.6\times {10}^{-19}\right)}=4.546\times {10}^{-14}$ m

• The distance of closest approach is 4.5 × 10⁻¹⁴ m, or 45 fm

• For comparison, the actual radius of a gold nucleus is about 6.6 × 10⁻¹⁵ m, or 6.6 fm

  • Therefore, the closest approach method gives an estimate of the upper limit of the radius of 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:

$\lambda =\frac{h}{mv}$

• Where:

  • h = Planck's constant

  • m = mass of an electron (kg)

  • v = speed of the electrons (m s⁻¹)

• This equation shows that as the speed of an electron increases, the smaller its de Broglie wavelength becomes

• When a beam of electrons is directed at a thin target, each electron will diffract around a nucleus

  • This happens because the de Broglie wavelength of a high-speed electron is similar to the size of a nucleus

• The resulting diffraction pattern that forms is a central bright spot with dimmer concentric circles around it

• From this pattern, a graph of intensity against diffraction angle can be plotted

• The size of the nucleus can be determined using the angle of the first minimum intensity

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

Advantages of Electron Scattering

1. Electron scattering is much more accurate than the closest approach method

2. This method gives a direct measurement of the radius of a nucleus

3. Electrons are leptons, so they are not affected by the strong nuclear force

Disadvantages of Electron Scattering

1. Electrons must be accelerated to very high speeds to maximise the resolution

   • This is because significant diffraction takes place when the electron wavelength is similar in size to the nucleus

   • Higher speeds are needed to shorten the de Broglie wavelength, since $\lambda \propto \frac{1}{v}$

2. 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

Electron Diffraction by a Nucleus

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

The first minimum of the intensity-angle graph can be used to determine nuclear radius

• The pattern formed by this diffraction has a predictable minimum which forms at an angle θ to the original direction according to the equation

$\sin \theta =\frac{\lambda }{d}$

• To determine an accurate value for nuclear radius R, a multiplication factor of 1.22 is required, as follows:

$\sin \theta =1.22\frac{\lambda }{d}=1.22\frac{\lambda }{2R}$

• Where:

  • θ = angle of the first minimum (degrees)

  • λ = de Broglie wavelength (m)

  • d = diameter of the nucleus (m)

  • R = radius of the nucleus (m)