Nothing needs to really be conserved in a nuclear equation: let me just illustrate one #alpha# and one #beta# equation to emphasise this.
#alpha#
#Pu#-238 (Plutonium, 238) decays by #alpha# emission to form an atom, which atom is this?
In an #alpha# decay equation, we lose an atomic number of #2# an a mass number of #4# - this is the equivalent of a Helium (#He#) atom. So,
#Pu-238 -> U-234 + ##alpha#
Uranium is formed because it is element number #92# - Plutonium is element number #94#, so if we take two away from #94# we get #92# which is the atomic number of #U#. There is nothing conserved in this reaction.
#beta#
When writing a #beta# equation, remember that in the nucleus, a neutron (#n#) decays into a proton (#p^+#) and a high energy electron which is known as the beta (#beta#) particle. Because a new proton has formed, the atomic number of the original atom will increase by #1#.
#I-131 -> Xe-131 +##beta#
Nothing is being conserved in this equation.
