Rutherford Scattering & Nuclear Radius

In the Rutherford scattering experiment, alpha particles are fired at a thin gold foil
Initially, before interacting with the foil, the particles have kinetic energy,

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

Some of the alpha particles are found to come straight back from the gold foil
This indicates that there is electrostatic repulsion between the alpha particles and the gold nucleus

WE - Rutherford scattering question image 1

Experimental set up of the Rutherford alpha scattering experiment

At the point of closest approach, d, the repulsive force reduces the speed of the alpha particles to zero momentarily, before any change in direction
- 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

$E_{p} = k \frac{Q q}{d}$

Where:
- Charge of an alpha particle, Q = 2e
- Charge of a target nucleus, q = Ze
- Z = proton number
- e = charge on an electron (or proton)
Substituting into the equation:

$E_{p} = k \frac{(2 e) (Z e)}{d}$

This gives an expression for the potential energy at the point of repulsion:

$E_{p} = k \frac{2 Z e^{2}}{d}$

Due to conservation of energy:
- This expression also gives the initial kinetic energy possessed by the alpha particle

Rearranging and calculating for the distance, d, gives a value for the radius of the nucleus when the alpha particle is fired with high energy

$d = k \frac{2 Z e^{2}}{E_{p}}$

Closest Approach Method, downloadable AS & A Level Physics revision notes

The closest approach method of determining the size of a gold nucleus

Nuclear Radius

The radius of nuclei depends on the nucleon number, A of the atom
This makes sense because as more nucleons are added to a nucleus, more space is occupied by the nucleus, hence giving it a larger radius
The exact relationship between the radius and nucleon number can be determined from experimental data
By doing this, physicists were able to deduce the following relationship:

Where:
- R = nuclear radius (m)
- A = nucleon / mass number
- R₀ = constant of proportionality = 1.20 fm

Nuclear Density

Assuming that the nucleus is spherical, its volume is equal to:

Where R is the nuclear radius, which is related to mass number, A, by the equation:

Where R₀ is a constant of proportionality

Combining these equations gives:

Therefore, the nuclear volume, V, is proportional to the mass of the nucleus, A

Mass (m), volume (V), and density (ρ) are related by the equation:

The mass, m, of a nucleus is equal to:

m = Au

Where:
- A = the mass number
- u = atomic mass unit

Using the equations for mass and volume, nuclear density is equal to:

Since the mass number A cancels out, the remaining quantities in the equation are all constant

Therefore, this shows the density of the nucleus is:
- Constant
- Independent of the radius

The fact that nuclear density is constant shows that nucleons are evenly separated throughout the nucleus regardless of their size

The accuracy of nuclear density depends on the accuracy of the constant R₀, as a guide nuclear density should always be of the order 10¹⁷ kg m^–3
Nuclear density is significantly larger than atomic density, this suggests:
- The majority of the atom’s mass is contained in the nucleus
- The nucleus is very small compared to the atom
- Atoms must be predominantly empty space

Worked example

Determine the value of nuclear density.

You may take the constant of proportionality, R₀, to be 1.20 × 10^–15 m.

Step 1: Derive an expression for nuclear density

- Using the equation derived above, the density of the nucleus is:

$ρ = \frac{3 u}{4 π {R_{0}}^{3}}$

Step 2: List the known quantities

Atomic mass unit, u = 1.661 × 10^–27 kg
Constant of proportionality, R₀ = 1.20 × 10^–15 m

Step 3: Substitute the values to determine the nuclear density

$ρ = \frac{3 \times (1.661 \times 10^{- 27})}{4 π {(1.20 \times 10^{- 15})}^{3}} = 2.3 \times 10^{17}$ kg m⁻¹

Examiner Tip

Make sure you're comfortable with the calculations involved with the alpha particle closest approach method, as this is a common exam question.

You will be expected to remember that the charge of an α is the charge of 2 protons (2 × the charge of an electron)

Rutherford Scattering & Nuclear Radius (DP IB Physics: HL)

Revision Note