Final answer:
The nucleus of a radon atom loses two protons and two neutrons during the emission of an alpha particle, resulting in a new element with a reduced atomic number and mass number.
Step-by-step explanation:
When the nucleus of a radon atom undergoes the emission of an alpha particle, which is composed of two protons and two neutrons, it loses four nucleons. This results in the reduction of the radon atom's mass number by four and its atomic number by two. Consequently, the nucleus transforms into a different element with a lower atomic number and mass number. For example, when radium-226 undergoes alpha decay, it forms radon-222. This is because the atomic number decreases from 88 (Ra) to 86 (Rn), and the mass number from 226 to 222, accounting for the two protons and two neutrons lost as the alpha particle. These changes are depicted in a nuclear equation, showing the relationship between the parent nucleus, the alpha particle, and the daughter nucleus that results from the decay.