# Nuclear Physics Tutorial 6 - Nuclear Radius

 Contents

The key idea to the use of particles and radiation to investigate the structure of matter lies in the de Broglie (pronounced ‘de Broy’) relationship, which states that particles have wave properties.  It is the logical extension of the particulate nature of electromagnetic wave phenomena (see Quantum Physics Tutorial 6).

When we investigate large objects, light is a good way to do this.  We have eyes to see with, and we can make direct observations with a magnifying lens or a microscope.  On the atomic and nuclear level, microscopic objects like fleas and bacteria are very large objects indeed.  However the light is limited by its wavelength to resolving objects about 1 mm across.  Much less than that, then diffraction becomes important.  Waves will not travel through a gap less than a wavelength.

Electrons can be shown to have wave properties by the simple use of an electron diffraction tube.  A slice of carbon is placed in a beam of electrons so that the electrons diffract. This has led to the development of the electron microscope, which allows magnifications much bigger than was ever possible with the light microscope.  A good light microscope can magnify up to 1000 times.  The electron microscope can magnify up to about 1 million times, and can reveal the existence of individual atoms.  The electron beams are focused by magnets just like the lenses on a microscope.

The existence of the nucleus was determined about 110 years ago by Ernest Rutherford (1871 – 1937) in his famous alpha scattering experiment. The idea that inspired the experiment was that the alpha particles were considered to be nuclear sized bullets that would smash the atoms in the gold foil like water melons.  Instead they found that many alpha particles were deflected, while a few came back in the direction they came from.  It led to two important conclusions:

• There is a positively charged nucleus

• The nucleus is very tiny compared with the rest of the atom; most of the atom is just empty space.

Further calculations led to an estimate of nuclear size of about 1 × 10-15 m or 1 femtometre (fm).

 Question 1 Use Coulomb’s Law (Module 4) to work out the force of repulsion between two protons 1 × 10-15 m apart.  Comment on your result. Use the value you worked out in Question 1 to comment on the magnitude of the strong force.  Explain your answer.

X ray diffraction has been a useful tool to discover the structure of solid materials.  It was perfected by the father and son team of William and Lawrence Bragg.  X rays are of course electromagnetic waves, which are scattered by diffraction by the crystal lattices of materials.  A sample of the material is placed in the beam of X-rays, and the resulting scattering pattern is picked up on a photographic plate.  The X rays are diffracted in a cone: By use of a simple equation we can determine the separation of layers of atoms.  The equation (which you won't be asked about in the AQA exam) is:

nl = 2d sin q

There is a simple student experiment in which microwaves are used in conjunction with a lattice of 4 cm polystyrene spheres.  It gives students experience in the use of this equation to determine the layer separation.

More complex analysis is needed for determination of the structure of the crystal lattices, and this is covered in the discipline of X-ray crystallography.  Some very complicated structures indeed have been worked out using such techniques.

To get the resolutions required to look at the nucleus, we need de Broglie wavelengths of 10-15 m, and this is not possible with electrons.  If we use more massive particles, we can obtain much shorter de Broglie wavelengths, hence more resolution.  That’s the theory.  In practice, the nuclei bombarded with high energy particles have tended to break up.  It has been described as finding out how a watch works by smashing it with another watch, and guessing how the pieces fit together.

Particle physics experiments have been able to interpret the sub nuclear particles that make up the nucleons within the nucleus.  They cannot be seen directly of course, but inference has been drawn from the patterns observed in collisions between particles.  Physicists use powerful computers to explain these phenomena.

Rutherford estimated the radius of a nucleus as 3.0 × 10-14 m from the data obtained in alpha scattering experiments.  He used Coulomb’s law in his calculations: The point P is the closest that the alpha particle gets to the nucleus before being repelled.  Its kinetic energy is 0 because it is stationary.  All its energy is potential.  So we can use our knowledge of electrostatic potential energy to calculate the distance.

Ep =  potential at P × charge of the alpha particle Rearranging gives us: This calculation is based on alpha particles of 7.68 MeV energy which converts to 1.23 × 10-12 J.   Therefore: rc = 2.96 × 10-14 m = 3.0 × 10-14 m (2 s.f.)

There are one or two points to bear in mind from this calculation:

• The nucleus is treated as a point charge.  At this level it is not.

• The alpha particles are stopped some distance away from the nucleus.

• It takes higher energy alpha particles to penetrate the nucleus.

• The values for the nuclear radius given by other particles such as protons, neutrons and electrons are slightly different.

 Question 3 Describe the principal features of the nuclear model of the atom suggested by Rutherford.

A more accurate estimate of the nuclear radius has been determined by the use of a technique called electron scattering.  The electrons interact with the nucleus entirely by the electromagnetic interaction whereas the alpha particles interact by the strong nuclear interaction, which is not well understood.

The scattering of electrons is treated like the diffraction of waves around a spherical object.  The in depth analysis of the results is quite complicated, but a reasonable estimate can be obtained with a relatively simple equation (which you don't need for the exam): The term l is the de Broglie wavelength of the high-energy electrons, q is the angle of diffraction, and R is the nuclear radius.

This gives a result of the nuclear radius being 2.65 × 10-15 m.

We need to note the following:

• To get an appreciable electron scattering effect, we need to have electrons with a de Broglie wavelength of about the nuclear diameter.  This requires very high energies.

• The electron diffraction minima are not zero, indicating that the boundary of the nucleus is fuzzy, not sharp.

• Since the boundary is not sharp, various methods of determining the radius give rather variable results, from 1.2 fm to 1.5 fm.

The radius depends on the nucleon number through a simple relationship: [The term A1/3 means the cube root of A, the nucleon number.  The term r0 is a constant with the value 1.4 × 10-15 m.  R is the nuclear radius.]

 Worked Example What is the nuclear radius of gold with a nucleon number of 197? Answer R = r0A1/3 = 1.4 × 10-15 m × 3Ö(197) = 1.4 × 10-15 m × 5.82 = 8.15 × 10-15 m

We can plot the radius against the cube root of the nucleon number, which gives us a straight line plot as shown on the graph. So we can see a nice linear progression.  We need to note one or two points.

• The constant r0 is variable depending on the type of particle used.

• The relationship R = r0A1/3 is always true, whatever the method used.

• Be careful not to confuse the nuclear radius with the atomic radius.  The atomic radius is remarkably similar, whether the element is light or heavy.  This would make sense as the nucleus occupies such a small fraction of the space in an atom.

• Remember to use the nucleon number, not the proton number.

If we plot ln R against ln A, we get a straight line. The intercept is r0: Question 4 What is the nuclear radius of an iron atom, of which the nucleus contains 56 nucleons?  Would it be any different to a cobalt-56 nucleus? We are used to the density (mass per unit volume) of solid materials being in the order of 1000 to 10000 kg m-3.  Since most of the atom is empty space, and the vast majority of the matter in an atom is contained in the tiny space of a nucleus, it makes sense that the density of nuclear material is very high indeed.

 Question 5 What is the nuclear density of an iron atom, of which the nucleus contains 56 nucleons?  Mass of a nucleon = 1.67 × 10-27  kg This calculation gives you a really big number, which works out at about 100 million tonnes per cubic centimetre.  Drop that on your foot, and you really will have tears in your eyes.

Put another way, the mass of the Earth would fit in a sphere of radius 100 metres; it would fit on top of your school.

When large stars die, they can do so in a cataclysmic explosion called a supernova.  Strange things happen: the remains of the star collapse in on themselves under gravity, and the material is squashed so much that electrons are absorbed by protons to make neutrons. The process of electron degeneracy stops further collapse.  Neutron stars are the result, which have an intense gravity field.  If the star is bigger, then the neutrons are squashed by the force of gravity into a really tiny space until stopped by neutron degeneracy.  Really big stars collapse in such a way that even neutron degeneracy will stop it.  Here the laws of physics no longer apply, and even light cannot escape.  We have a black hole.