Final answer:
The decay of the gold isotope is an example of alpha decay, where an α-particle (helium nucleus) is emitted, leading to a decrease in both atomic number and mass number.
Step-by-step explanation:
The equation 174Au → 170Ir + ? represents the radioactive decay of a gold isotope. In this case, the atomic number decreases from 79 to 77, and a gold (Au) atom becomes an iridium (Ir) atom. Since the atomic number is reduced by two, this is characteristic of an alpha decay process. In alpha decay, a helium nucleus, which consists of two protons and two neutrons (α-particle), is emitted from the nucleus of the radioactive atom. Therefore, the missing product in the nuclear equation is an α-particle, confirming that the decay mode is alpha decay.