Write a balanced net ionic equation for the following reaction?

Write a balanced nuclear equation for the following reaction. Include both mass numbers and atomic numbers?

• write a balanced nuclear equation for the following reaction. Include both mass numbers and atomic numbers with all atomic symbols. Alpha decay of 234 92 U

• Answer:

  I'm just an organic chemist who had a radioisotope class, so take what I with a grain of salt. The way I write these transitions, which may or may not be the actual way that radiochemists write them, is to have the atomic mass (A) and atomic number (Z) in front of the element and have them separated by a slash (/) sign, e.g. 20/10Ne For the decay particles, we have alpha = 4/2a = 4/2He (equivalent to a helium nucleus) beta = 0/-1ß = 0/-1e (equivalent to an electron) For decay schemes, I use the indicated notation and calculate the change in atomic mass (A) and in atomic number (Z). For example, 20/10Ne → A/Z(unknown) + 0/-1e ∆A = 20 - 0 = 20 ∆Z = 10 - (-1) = 10 +1 = 11 Looking up element 11 in a periodic chart, we find that it is sodium, so the complete equation for beta-decay of 20/10Ne is 20/10Ne → 20/11Na + 0/-1e For your problem, alpha-decay is set up this way: 234/92U → A/Z(unknown) + 4/2He ∆A = 234 - 4 = 230 ∆Z = 92 - (2) = 90 Looking up element 90 in a periodic chart, we find that it is thorium, so the complete equation for alpha-decay of 234/92U is 234/92U → 230/90Th + 4/2He