22688Ra  → ___ + 42HeA. 22285RnB. 22286RnC. 23086RnD. 22086Rn

The given equation represents a radioactive decay process, specifically alpha decay. In alpha decay, an unstable parent nucleus emits an alpha particle (42He) and transforms into a daughter nucleus.

The parent nucleus in this case is 22688Ra (Radium-226).

The atomic number (the subscript) of the daughter nucleus is found by subtracting the atomic number of the alpha particle (2) from the atomic number of the parent nucleus (88). So, 88 - 2 = 86.

The mass number (the superscript) of the daughter nucleus is found by subtracting the mass number of the alpha particle (4) from the mass number of the parent nucleus (226). So, 226 - 4 = 222.

Therefore, the daughter nucleus is 22286Rn, which corresponds to option A. 22285Rn.