Nuclear Physics Tutorial 2 - Evidence for the Nucleus

 

Contents

Rutherford Scattering

Limitations

 

Rutherford Scattering

In the early part of the last century, the accepted model of the atom was proposed by J J Thompson in his plum pudding model.  This consisted of a matrix of protons in which were embedded electrons.

 

Ernest Rutherford (1871 – 1937) used alpha particles in a vacuum to study the nature of atomic structure with the following apparatus:

 

 

Rutherford was using alpha particles (helium nuclei) as nuclear bullets to smash up the atoms; he wanted to see atoms bursting like watermelons.  But…

 

His observations are best illustrated with this diagram:

 

 

Instead of bits of atom, Rutherford found that a small proportion of the alpha particles were deflected, while an even smaller proportion bounced right back.  From analysis of these observations he concluded:

Rutherford’s estimates were not far out.  Later research has shown the nuclear radius to be in the order of 1.5 ´ 10-14 m.  However the boundary is not sharp, but rather fuzzy, as the nucleus is a very dynamic entity.

 

Question 1

What led Rutherford to conclude that the nucleus was very tiny and had a positive charge? 

Answer

 

We need to remember that most of the volume of an atom is empty space.  The general size of an atom is about 10-10 m (0.1 nm) and that is determined by the electron clouds.  The size does not vary much across the elements.  A calcium atom is about the same size as a gold atom.  Even the heavy gold nucleus has a tiny diameter compared to the diameter of the atom.

 

Limitations

There are limitations to Rutherford's calculations:

 

We can use electrical potential energy to get an estimate of the closest approach.  From Electric Fields, we saw:

 

 

In the calculation you are going to do, the alpha particle will have an energy of 5 MeV.  We can assume that it is kinetic energy.  In reality, there is a certain amount of vibrational energy, but this is small compared to the kinetic energy.  The proton number for gold is 79 and the proton number for the alpha particle is 2.   (You knew that, didn't you?)

 

Question 2

A 5 MeV alpha particle approaches a gold nucleus.

(a) What is the charge carried by the alpha particle?

(b) What is the charge carried by the gold nucleus?

(c) What is the energy in joules of the alpha particle?

(d) What is the minimum approach distance?

Answer


The nuclear radius is actually rather smaller than this, about 1 × 10-14 m.  Most atoms have a radius of about 1 × 10-10 m, and that is remarkably consistent between atoms.  This is because the mass is contained in the nucleus, which is 10 000 times smaller.  As each electron has a mass of 1/1800 the mass of a proton, the mass contributed by the electrons in their shells is very small.

 

 

Rutherford Scattering at High Energy

If the alpha particles have a very high energy, their behaviour deviates from what is predicted by the Rutherford Equation (not on the syllabus).  The deflection angle in this case is set at 60o, rather than being varied.   The idea is shown below:

 

 

Alpha particles of different energies are used.  This can be achieved by accelerating the alpha particles using an electric field.  The only alpha particles measured are those that are deflected by 60o.  The relative intensity of the deflected alpha particles is measured by the detector.  The data are recorded and are shown on the graph below:

 

The Rutherford Equation predicts the number of alpha particles per unit area striking the detector is inversely proportional to the square of the kinetic energy, i.e.:

 

 

This graph shows that the model is followed until the alpha particles have an energy of about 27.5 MeV.  Then the relative intensity falls much more rapidly that what is suggested by the model.  The dotted line shows the expected results if the Rutherford Equation were followed.

 

The conclusion from high energy alpha particle experiments was that the nucleus had a radius of about 7.1 fm (7.1 × 10-15 m).