 Another common way to describe nuclear masses is in terms of the unified atomic mass unit. The symbol for this unit is a lowercase letter U. As we will see shortly, this is a convenient unit because the mass of an atom in atomic mass units is approximately equal to the total number of nucleons in the nucleus, adding both the protons and the neutrons together. The reason this happens is obvious from the definition of the atomic mass unit, where we define the mass of a carbon-12 atom to be exactly 12U. The experimentally measured mass of a carbon-12 atom is about 11,000 MeV over C squared. Dividing this by 12, we can see that one atomic mass unit is equal to 931.494 MeV over C squared. Let us now look at an example of how we can solve nuclear problems using these atomic mass units. Consider the alpha decay of Polonium-212 that we looked at earlier in this lecture, and let's try to work out the energy of the emitted alpha particle. We will need to know the masses of the particles involved, and we can look these up in a table of atomic masses. The numbers we need for Polonium-212, lead-208 and helium-4 are shown here. You can see that the masses in atomic mass units are all very close to the number of nucleons for each case. Now, if we write the alpha decay as a process, we can see that we start with Polonium-212 on the left-hand side and end up with lead-208 and helium-4 on the right-hand side. The total mass on the left-hand side is 211.9885 atomic mass units, while the total mass on the right-hand side is 211.97924 atomic mass units. There is a change in mass, with the right-hand side being 0.00961 atomic mass units smaller than the left-hand side. Since mass has been lost, E equals MC squared tells us that energy has been liberated in the reaction, and this must be shared as kinetic energy of the reaction products. We can write that the emitted energy is equal to the change in mass times C squared, and hence the energy emitted is easily calculated using the definition of an atomic mass unit. We write 0.00961 times 931.494 MeV ever C squared times C squared. The C squared on the top line cancels with the C squared on the bottom line, and we obtain that the energy is 8.95 MeV. Since we also need to conserve momentum, and since the mass of the lead 208 nucleus is so much heavier than the mass of the helium-4 nucleus, the vast majority of this energy is given to the helium-4 nucleus. Now that we've looked at nuclear energies and masses in a quantitative fashion, we can begin to understand a range of nuclear reactions and processes, both those that are naturally occurring and those that are used in industries, such as medicine, manufacturing, and power production. In the next lecture, we will investigate nuclear reactions In the one after that, we will look in detail at neutron induced fission reactions and how they are used in nuclear power plants.