Nuclear Reaction Question - radioactive decay?

when an isotope undergoes alpha decay, it loses the nucleus of the He atom. 4 mass and 2 protons (atomic number)

when an isotope undergoes beta decay, it gains 1 in atomic number because a beta particle is an electron with 0 mass and -1 for atomic number and the loss of a beta particle converts a neutron to a proton, thereby increasing the atomic number by 1

90Th-238 --> 88Ra-234 + 4/2He.....the 4 = mass, the 2 = atomic number

88Ra-234 --> 89 Ac-234 + 0/-1e + neutrino.....the 0 = mass, the -1 = atomic number

88 - beta particle = 88 - (-1) = 89

continue

Radioactive decay is the process in which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. This decay, or loss of energy, results in an atom of one type, called the parent nuclide transforming to an atom of a different type, called the daughter nuclide. For example: a carbon-14 atom (the "parent") emits radiation and transforms to a nitrogen-14 atom (the "daughter"). This is a random process on the atomic level, in that it is impossible to predict when a particular atom will decay, but given a large number of similar atoms, the decay rate, on average, is predictable. In nuclear physics, a nuclear reaction is a process in which two nuclei or nuclear particles collide to produce products different from the initial particles. In principle a reaction can involve more than two particles colliding, but because the probability of three or more nuclei to meet at the same time at the same place is much less than for two nuclei, such an event is exceptionally rare. While the transformation is spontaneous in the case of radioactive decay, it is initiated by a particle in the case of a nuclear reaction. If the particles collide and separate without changing, the process is called an elastic collision rather than a reaction. A nuclear reaction can be written in terms of a formula just like a chemical reaction. Nuclear decays can be written in a similar way, but with only one nucleus on the left side. Every particle partaking in the reaction is written with its chemical symbol, with the mass number at the upper left and the atomic number at the lower left. The neutron is written "n"; the proton can be written "1H" or "p". The equation is correct only if the sums of the mass numbers on both sides are identical (as required by the conservation law for baryon number), and if the sums of the atomic numbers on both sides are identical (as required by the conservation law for electric charge). In the example shown above, this leads to (assuming we would know only one particle to the right): {}^{6}_{3}\mathrm{Li}+{}^{2}_{1}\mathrm{... +\ ?. To make the sums correct, 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: {}^{6}_{3}\mathrm{Li}+{}^{2}_{1}\mathrm{... or more simply: {}^{6}_{3}\mathrm{Li}+{}^{2}_{1}\mathrm{... {}^{4}_{2}\mathrm{He}. I hope it helps!