##### Nuclear Physics Tutorial 2 - Evidence for the Nucleus

 Contents

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 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:

• Most of the atom was empty space.

• The positive charge was concentrated in a very small space

• The radius of the nucleus was in the order of 3 ´ 10-14 m.

• The alpha particles that were deflected back had to be travelling in a line with the nucleus.

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?

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.

There are limitations to Rutherford's calculations:

• The distance is the distance of the closest approach.

• This is determined by the energy of the alpha particle.

• The higher the energy, the closer it will get.

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?

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.