Answer
${\;^{228}\text{Th}\;}$
Work Step by Step
We need to find the mass of the parent nucleus so we can determine the element.
Let's assume that the released energy during the alpha decay is the kinetic energy of the alpha particle.
So this energy released is given by
$$
K_\alpha = (m_X - m_Y - m_{\text{He}}) c^2= 5.52 \, \text{MeV}
$$
where:
- $ m_X $: Mass of the parent nucleus
- $ m_Y $: Mass of the daughter nucleus
- $ m_{\text{He}} $: Mass of the alpha particle ($^4\text{He}$)
where $ 1 \, \text{u} = 931.49 \, \text{MeV}/c^2 $, so
$$
m_X - m_Y - m_{\text{He}} = \frac{5.52 \, \text{MeV}}{931.49 \, \text{MeV/u}}
$$
$$
m_X - m_Y - m_{\text{He}} =0.00593 \, \text{u}
$$
$$
m_X - m_Y =0.00593 \, \text{u} + m_{\text{He}}
$$
From appendix c, we can see that $ m_{^4\text{He}}=4.002602\;\rm u$
$$
m_X - m_Y =0.00593 \, \text{u} + 4.002602 \, \text{u}$$
$$
m_X - m_Y=\bf 4.00853\;\rm u\tag 1
$$
We are given that the total combined mass of the daughter nucleus and alpha particle is
$$
m_X+ m_Y = 452 \, \text{u}\tag 2
$$
Now we have two equations and two unknowns.
Adding these two equations gives:
$$
2m_X = 456.00853
$$
Thus,
$$m_X=\bf 228.004265\;\rm u$$
Now we can see, from Appendix C, that the nucleus of a mass of about 228 u and has an alpha decay is $^{228}\text{Th}$.
Thus, the parent nucleus is
$$\boxed{\;^{228}\text{Th}\;}$$
