Answer:
[tex]_{89}^{227}\text{Ac} \longrightarrow _{87}^{223}\text{Fr} + _{2}^{4}\text{He}[/tex]
Explanation:
The most stable isotope of actinium, Ac-227, undergoes alpha decay.
Actinium is element 89, so the unbalanced nuclear equation is
[tex]_{89}^{227}\text{Ac} \longrightarrow ? + _{2}^{4}\text{He}[/tex]
It is convenient to replace the question by an atomic symbol, [tex]_{x}^{y}\text{Z}[/tex], where x = the atomic number, y = the mass number, and Z = the symbol of the element .
Then your equation becomes
[tex]_{89}^{227}\text{Ac} \longrightarrow _{x}^{y}\text{Z} + _{2}^{4}\text{He}[/tex]
The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.
Then
89 = x + 2, so x = 89 - 2 = 87
227 = y + 4, so y = 223
Element 87 is francium, so the nuclear equation becomes
[tex]_{89}^{227}\text{Ac} \longrightarrow _{87}^{223}\text{Fr} + _{2}^{4}\text{He}[/tex]
