Final answer:
Balance the nuclear equation for the alpha decay of 225/80 Ac, we subtract the mass number and atomic number of the emitted alpha particle from the original actinium nucleus, resulting in the product 221/78 Pt plus a 4/2 He nucleus.
Step-by-step explanation:
The question involves completing a nuclear equation for alpha decay. In an alpha decay, an alpha particle, which is a Helium-4 nucleus (4/2 He), is emitted from the nucleus of an atom. The mass number (A) and atomic number (Z) both decrease in this process. The mass number decreases by 4, and the atomic number decreases by 2.
Given the nuclear equation 225/80 Ac --> ??? + 4/2 He, we must identify the product of the alpha decay. Since the helium nucleus has a mass number of 4 and an atomic number of 2, we subtract these from the original actinium nucleus to find the unknown product:
Mass number (A) of unknown product: 225 - 4 = 221
Atomic number (Z) of unknown product: 80 - 2 = 78
Therefore, the completed nuclear equation becomes:
225/80 Ac --> 221/78 ??? + 4/2 He
The missing element with atomic number 78 is Platinum (Pt), resulting in the final balanced nuclear equation:
225/80 Ac --> 221/78 Pt + 4/2 He