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

Revision Note

Katie M

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Katie M

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Nuclear Radius

  • 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, such as Rutherford scattering
  • By doing this, physicists were able to deduce the following relationship:

R space equals space R subscript 0 A to the power of 1 third end exponent

  • Where:
    • R = nuclear radius (m)
    • A = nucleon / mass number
    • R0 = Fermi radius
  • The constant of proportionality R0 = 1.20 × 10−15 m is known as the Fermi radius
  • This is the radius of a hydrogen nucleus which contains only one proton (A = 1)

Nuclear Density

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

V space equals space 4 over 3 straight pi R cubed

  • Combining this with the expression for nuclear radius gives:

V space equals space 4 over 3 straight pi open parentheses R subscript 0 A to the power of 1 third end exponent close parentheses cubed space equals space 4 over 3 straight pi R subscript 0 cubed A

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

m space equals space A u

  • Where u = atomic mass unit (kg)
  • Using the definition for density, nuclear density is equal to:

rho space equals space m over V

rho space equals space fraction numerator A u over denominator 4 over 3 straight pi R subscript 0 cubed A end fraction space equals space fraction numerator 3 u over denominator 4 straight pi R subscript 0 cubed end fraction space

  • 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

Worked example

Determine the value of nuclear density.

You may take the constant of proportionality R0 to be 1.20 fm.

Answer:

Step 1: Derive an expression for nuclear density

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

rho space equals space fraction numerator 3 u over denominator 4 straight pi R subscript italic 0 cubed end fraction

Step 2: List the known quantities

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

Step 3: Substitute the values to determine the nuclear density

rho space equals space fraction numerator 3 space cross times space left parenthesis 1.661 cross times 10 to the power of negative 27 end exponent right parenthesis over denominator 4 pi space open parentheses 1.20 cross times 10 to the power of negative 15 end exponent close parentheses cubed end fraction space equals space 2.3 space cross times space 10 to the power of 17 space kg space straight m to the power of negative 3 end exponent

Examiner Tip

You do not need to remember the value of the Fermi radius R0 as it is included in the data booklet in the 'fundamental constants' section.

Rutherford Scattering Experiment

  • 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

E subscript k space equals space 1 half m v squared

  • 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 E subscript k of the alpha particle is equal to the electric potential energy E subscript p of the target nucleus:  

E subscript k space equals space E subscript p

  • Where the electric potential energy is given by

E subscript p space equals space k fraction numerator Q q over denominator d end fraction

  • 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:

E subscript p space equals space k fraction numerator open parentheses 2 e close parentheses open parentheses Z e close parentheses over denominator d end fraction space equals space k fraction numerator 2 Z e squared over denominator d end fraction

  • 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

d space equals space k fraction numerator 2 Z e squared over denominator E subscript p end fraction space equals space k fraction numerator 2 Z e squared over denominator E subscript k end fraction

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

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

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

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)

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.