Answer
$^{225}Ac$
Work Step by Step
From the figure, we understand that $\Delta Z_{\alpha} = – 2 $ and $\Delta A_{\alpha} = – 4$ and $\Delta Z_\beta = +1$ from the short lines toward the right are beta decays.
Note that the figure shows three alpha decays and two beta decays. Thus,
$Z_f = Z_i + 3(\Delta Z_{\alpha}) + 2(\Delta Z_\beta)$
Where $Zi = 93$ for neptunium, so
$Z_f = 93 + 3(-2) + 2(1)$
$Z_f = 89$
$A_f = A_i + 3\Delta A_{\alpha} $
Where $Ai = 237$ for neptunium, so
$A_f =237 + 3(-4) $
$A_f = 225$
Therefore, the final element that has $A= 225$ and $Z=89$ is Actinium isotope, $^{225}Ac$
