IB DP Physics: HL复习笔记12.2.1 Rutherford Scattering & Nuclear Radius

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,
  • 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


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, Ek, is equal to electric potential energy, Ep
  • The radius of the closest approach can be found be equating the initial kinetic energy to the electric potential energy


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:

Exam 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)