Nuclear Physics

The nucleus
Applications
Particle accelerators

# The nucleus

### 1.Definitions

In the first of the last century, Ernest Rutheford and his students Hans Geiger and Ernest Marsden conducted some experiments to understand the behavior of alpha particles scattering by the thin gold-leaf. Rutheford concluded that matter contains positive charges among electrons and stated a planetary model for the atom. These positive charge constitute the nucleus of the atom, the negative charges electrons circulate around the nucleus. Twenty years after (in 1931), Chadwick discovered the neutron that completed the definition of the nucleus.

### 2.The composition of the nucleus

The nucleus is composed of Z protons and N neutrons. The atomic number Z defines the chemical properties of the atom. The mass number A=Z+N determines approximately the mass of the nucleus in atomic mass units. Nuclei with the same Z but different N are known as isotopes. These have different masses but the same chemical properties. Protons and neutrons are each called nucleons.

### 3.Atomic mass unit

Nuclear energies are very high (order of one million) compared to atomic energies, and need larger units Nuclear masses are measured in terms of atomic mass units. The carbon-12 nucleus weights exactly 12 amu.

1 uma = 12x 10-3 /12 Na , where Na is the Avogadro number

1 uma = 1.66054x10-27 kg = 931.494 MeV

MeV = 106 eV = 1.6x10-19 joules

A nucleus is extremely small in weight and size, but extremely dense. For the Uranium (238,92)U, the volume is (8.7 x 10-15m)3 = 6.58 x 10-43 m3, and its mass is 238.0508 uma; then its density is: ρ = 238.051 x 1.66 x 10-27/6.58 x 10 -43 = 6 x 1014g/cm3 !

### 4. The strong nuclear force and binding energies

Since the protons inside a nucleus have postive charge, how are they maitained all together inside a nucleus? The fact is the presence of an extra energy.
Let's consider the example of the (40,20)Ca. Its mass is 39.951 uma. But the masses of the separated 20 protons and 20 neutrons are respectively: 20 x 1.007276 + 20 x 1.008665 = 40.319 uma. The difference is then equal to 0.368 uma = 0.61 x 10 -27 kg. This extra mass is responsible of the stability of the nucleus. In other words, when these 40 particles are liberated, they will give an energy of 0.368 x c2 = 0.61 x 10 -27 x 9 x 10 16 = 5.5 x 10 -11 joules.
This value is, of course, related to only one nucleus. For a mol of atoms, that is 40 grams of Ca, we have 5.5 x 10 -11 x 6.023 x 1023 = 3.3 x 1013 joules ! ( we need just 10 joules to lift 1 kg over 1 meter).
This energy 3.3 x 1013 joules is also the energy to supply to the Ca nucleus to separate the 40 particles.. It is called the binding energy of the nucleus.

To compare the binding energies for nucleus, we use the ratio:
r = (nucleons mass - nucleus mass)c2/number of nucleons = Binding energy/number of nucleons

For Ca, we have r = 5.5 x 10 -11 /40 = 1.375 x 10 -12 joules. With 1 eV = 1.6 x 10-19 joules, r = 8.6 MeV.

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