
The atom has a large nucleus with protons and neutrons that is surrounded by very small electrons. The fundamental charged particles for everyday considerations are the proton (positive) and the electron (negative). Chemistry is mostly concerned with the electrons because they are always involved with keeping atoms together. Nuclear physics is concerned with the nucleus with its protons and neutrons. The nucleus is much harder to understand. In fact, we don't have a good theory that explains nuclear reactions. We can very accurately describe what happens, but we don't have any idea why it happens. We can say, for example, how long radioactive atoms will be around statistically, but we don't know when any individual atom with decay. It is a baffling, and thus very interesting, area of study.

Not all combinations of protons and neutrons are stable. The number of protons identifies the element, but the nucleus can have a wide range of neutrons and still be the same element. For the lighter elements the stable nuclides (in chemistry we call them isotopes) generally have one neutron per proton, but in the heavier elements isotopes with more neutrons than protons are more stable.

The mass of a nuclide would be the sum of the masses of all of the particles in the neutral atom with the number of proton, neutrons, and electrons for that particular nuclide. Each element could have several nuclides (isotopes) and each nuclide would have its own mass. The mass on the periodic table is the weighted average of all of the naturally occuring isotopes, so a different table must be consulted when we need the mass of a particular nuclide. Most nuclides on the "Chart of the Nuclides" do not exist naturally, but many of them do play a role in our study of nuclear physics.

The energies associated with the nucleus are very small and so a special unit is often used when doing nuclear physics. The unit is MeV, or million electron volts. 1 Mev = 1.6 x 10^{13} J.

One unit for the mass of an atom is the unified atomic mass unit, u (often it is labeled amu). In this system the mass of a carbon atom with 12 protons, 12 neutrons, and 12 electrons is defined as exactly 12 u. This unit works well in chemistry if you realize that the mass of a mole of cabon atoms is 12 g. We say that that carbon isotope has an atomic mass of 12 g/mole.

According to Einstein mass and energy can be converted between each other. The relationship is E = mc². The mass unit and the energy unit can then be related. The relation is approximately 1 u = 931 MeV.

The binding energy would the the amount of energy that binds two objects together. Gravity binds us to earth and the amount of energy needed to separate us from the earth could be calculated using Newton's Laws. That would correspond to the binding energy holding us to earth. The energy needed to separate an electron from the proton in a hydrogen nucleus is experimentally found to be about 13.6 eV. In organic molecules is is between 2 and 20 eV.

At the nuclear level it is very difficult to experimentally determine binding energies. They simply aren't understood well enough. We can, however, get the binding energy from our nuclide masses.
Example: What is the binding energy of the least strongly bound neutron in a ^{14}N nucleus?
This question is really asking what is the difference between ^{14}N and ^{13}N plus a neutron. The masses of these particles can be found in a table. The masses are: ^{14}N = 14.003074 u; ^{13}N = 13.005739 u; and the mass of the neutron, n = 1.008665 u.
^{13}N + n  ^{14}N = 13.005739 u + 1.008665 u  14.003074 u = 0.11330 u (This is the mass difference.)
Binding Energy = (0.11330 u)(931 MeV/u) = 10.5 MeV
It appears that 0.11330 u of mass has been converted into binding energy that holds the neutron in the nucleus.

Many physics books have a graph of the nuclear binding energy. Those graphs show that the most stable nuclide is ^{56}Fe, which is often found in meteorites. These tables predict that separation of carbon into its individual parts would require energy, while combining parts would give off energy. The opposite is true for ^{200}Hg or ^{235}U. That is why nuclear bombs work.

Radioactive decay is a spontaneous breaking apart of a nucleus. This process doesn't depend on external conditions like temperature and pressure. It also doesn't matter what the atom is combined with. For example, UCl_{3}, UO_{2}, and UF_{6} all have the same radioactive activity for the same amount of compound.

The rate of decay only depends on the number of nuclei, N. The change in the number of nuclides with respect to time is directly proportional to the number of nuclides: dN/dt = λN, where λ is the proportionality constant. A little math results in the following equation for the decay of nuclei.
ln(N/N_{0}) = λt
Where ln is the natural log, N is the number of nuclides at time t, N_{0} is the initial number of nuclides, λ is the decay constant, and t is the time.

The activity, A, is defined as the rate constant times N: A = λN. Substituting into the above equation gives the new relationship where the natural log of the final activity divided by the initial activity is equal to the negative of the rate constant times the time. Often the unit of activity is counts/sec (or disintegrations/sec). Another common unit is the Curie, Ci. 1 Ci = 3.70 x 10^{10} disintegrations per second.
ln(A/A_{0}) = λt

The halflife of a nuclide is the amount of time needed for half of the nuclides to decay. After one halflife there are only half of the original nuclides remaining. Letting N = ½N_{0} in the above equation and solving for t gives:
t_{½} = ln(2)/λ = 0.693/λ
Example
(a) What is the halflife of a nuclide that has λ = 0.0693 sec^{1}? (b) If the inital activity is 2000 counts/sec, what will the activity be after 20 minutes? (c) When will the activity be 125 counts/sec?
(a) t_{½} = ln(2)/λ = 0.693/(0.0693 sec^{1}) = 10 sec
(b) A = A_{0}e^{λt} = (2000 counts/sec)e^{(0.0693 sec1)(20 min)(60 sec/min)}
A = 1.53 x 10^{33} counts/sec
(c) t = ln(A/A_{0})/λ = ln(125/2000)/(0.0693 sec^{1}) = 40 sec
Modes of Decay
Radioactive decay occurs in at least one of the following ways.

Electron emission (β^{})
In this mode of decay a neutron is converted into a proton and an electron. An example would be ^{55}Cr being converted to ^{55}Mn and an electron.

Electron capture (ε)
In this mode of decay an electron and a proton form a neutron. An example would be ^{51}Cr plus an electron being converted to ^{51}V.

Positron emission (β^{+})
In this mode of decay a proton is converted into a neutron and a positron. A positron is just like an electron, except it has a positive charge. An example would be ^{49}Cr being converted to ^{49}V and a positron.

Alphadecay ()
In this mode of decay an αparticle (two protons and two neutrons) are emitted from the nucleus. An example would be ^{226}Rn being converted to ^{222}Rn, an αparticle, and two electrons.

γray emission (γ)
In this mode of decay a γray (high energy light) is emitted. γrays often accompany the other modes of decay.
The mode of decay that any particular nuclide undergoes can be found on the chart of the nuclides. From this chart a radioactive decay series can be written. A radioactive decay series shows all of the daughter nuclides going from the original nuclide until it ends up at a stable nuclide. This link gives an example for ^{222}Rn (radon222). Below is the decay series for bismuth212 showing two paths that both end up at lead208, which is not radioactive.
Decay Series for Bismuth212
β^{}
β^{}
For a long time experimenters couldn't account for all of the energy in these radioactive decays. In the end they had to postulate that there must be another particle given off. They called the new particle the neutrino, which are neutral particles with extremely small (approximately zero) rest mass. They have now been experimentally verified. It turns out that every βdecay simultaneously emits both an electron and a neutrino. A β^{+} emits a neutrino and β^{} emits an antineutrino.
Elementary Particles
Here is a short video of the standard model produced by CERN.