____  →  42He +  20881Th A. 21282PoB. 21283PoC. 21282BiD. 21283Bi

The question seems to be about nuclear decay, specifically alpha decay. In alpha decay, an atomic nucleus emits an alpha particle and transforms into a nucleus with mass number 4 less and atomic number 2 less.

Given the equation: "____ → 42He + 20881Th"

We can see that the product of the decay is Thorium (Th) with atomic number 81 and mass number 208, and an alpha particle (He) with atomic number 2 and mass number 4.

To find the original nucleus before decay, we add the atomic numbers and the mass numbers of the products:

Atomic number: 81 (Th) + 2 (He) = 83 Mass number: 208 (Th) + 4 (He) = 212

So, the original nucleus before decay has atomic number 83 and mass number 212.

Looking at the options, the correct answer is:

B. 21283Po (Polonium-212)