This example problem demonstrates how to write a nuclear reaction process involving alpha decay.
An atom of 241Am95 undergoes alpha decay and produces an alpha particle.
Write a chemical equation showing this reaction.
Nuclear reactions need to have the sum of protons and neutrons the same on both sides of the equation. The number of protons must also be consistent on both sides of the reaction.
Alpha decay occurs when the nucleus of an atom spontaneously ejects an alpha particle. The alpha particle is the same as a helium nucleus with 2 protons and 2 neutrons. This means the number of protons in the nucleus is reduced by 2 and the total number of nucleons is reduced by 4.
241Am95 → ZXA + 4He2
A = number of protons = 95 - 2 = 93
X = the element with atomic number = 93
According to the periodic table, X = neptunium or Np.
The mass number is reduced by 4.
Z = 241 - 4 = 237
Substitute these values into the reaction:
241Am95 → 237Np93 + 4He2