Question

212 84Po→208 82Pb+AZX

What would the value of A be here?

212 84Po→208 82Pb+AZX

what would the value of z be here?

Answer 1

Here is the way I write these. The atomic number goes on the left, the mass number on the right. So the Po would be
84Po212 ==> 82Pb208 + AXZ
where A is atomic number, Z is mass number, X is the element.

The left side must add up and the right side add up; i.e., atomic numbers must be the same on both sides and mass number must be the same.
On the left then we have 84 = 82 + A. A must be 2
On the right we have 212 = 208 + Z. Z must be 4
And X must be which element. Atomic number 2 must be He so that's an alpha particle that was ejected.