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

• The radius of a nucleus depends on the nucleon number A of the atom
  • The greater the number of nucleons a nucleus has, the greater the space the nucleus occupies, 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:

• Combining this with the expression for nuclear radius gives:

• This tells us that the nuclear volume V is proportional to the mass of the nucleus m, which is equal to

• Where u = atomic mass unit (kg)

• Using the definition for density, nuclear density is equal to:

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

• Therefore, this shows the density of the nucleus is:
  • The same for all nuclei
  • 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 which 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

• 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 equal to

• 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 the alpha particle is equal to the electric potential energy  of the target nucleus:  

• Where the electric potential energy is given by

• Where:
  • Charge of an alpha particle, Q = 2e
  • Charge of a target nucleus, q = Ze
  • Z = proton (atomic) number
  • e = elementary charge (C)
  • k = Coulomb constant

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

• Which, due to the conservation of energy also gives the initial kinetic energy possessed by the alpha particle
• Rearranging for the distance of closest approach d

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

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