Nuclear reactions may be shown in a form similar to chemical equations, for which invariant mass (the mass not considering the mass defect) must balance for each side of the equation, and in which transformations of particles must follow certain conservation laws, such as conservation of charge and baryon number (total atomic mass number, number of protons and neutrons). An example of this notation follows:
To balance the equation above for mass, charge and mass number, the second nucleus to the right must have atomic number 2 and mass number 4; it is therefore also helium-4. The complete equation therefore reads:
Or, more simply:
Graphically, this reaction would look like this:
Instead of using the full equations in the style above, in many situations a compact notation is used to describe nuclear reactions. This style of the form A(b,c)D, which is equivalent to A + b gives c + D. Common light particles are often abbreviated in this shorthand, typically p for proton, n for neutron, d for deuteron, α representing an alpha particle or helium-4, β for beta particle or electron, γ for gamma photon, etc. The reaction above would be written as Li-6(d,α)α.
In balancing a nuclear equation, it is useful to remember that the sum of all the top numbers (mass number) must be equal on both sides of the equation, as well as the sum of all the bottom numbers (the atomic number). In addition, problems will also often be given as word problems, so it is useful to know the various names of radioactively emitted particles.
Here's a sample problem:
This could be written out as Uranium-235 gives Thorium-231 plus what? In order to solve, we find the difference between the atomic masses and atomic numbers in the reactant and product. The result is an atomic mass difference of 4 and an atomic number difference of 2. This fits the description of an alpha particle. Thus, we arrive at our answer:
Here's another sample problem:
This could also be written out as Polonium-214 plus two alpha particles plus two electrons gives what? In order to solve this equation, we simply add the mass numbers, 214 for Polonium plus 8 (two times four) for helium (two alpha particles), plus zero for the electrons gives a mass number of 222. For the atomic number, we take 84 for Polonium, add 4 (two times two) for helium, then subtract two (two times -1 for two electrons, or beta emission) to give 86, the atomic number for Radon, Rn. Therefore, the equation should read: