Which of the following is the correct nuclear equation for the alpha decay of Th-227?

• A. Th-227 → Pa-227 + γ
• B. Th-227 → Ra-223 + α
• C. Th-227 → Ac-227 + β
• D. Th-227 → Th-223 + α

Answer: (B)

Verified step by step guidance

1. Identify the type of decay: Alpha decay involves the emission of an alpha particle, which consists of 2 protons and 2 neutrons, equivalent to a helium nucleus (He-4).
2. Write the general form of an alpha decay equation: \( _{Z}^{A}X \rightarrow _{Z-2}^{A-4}Y + _{2}^{4}\text{He} \), where \( _{Z}^{A}X \) is the parent nucleus and \( _{Z-2}^{A-4}Y \) is the daughter nucleus.
3. Apply the alpha decay process to Th-227: The parent nucleus is Thorium-227 (Th-227), so the daughter nucleus will have an atomic number reduced by 2 and a mass number reduced by 4.
4. Calculate the daughter nucleus: Thorium (Th) has an atomic number of 90, so the daughter nucleus will have an atomic number of 88, which corresponds to Radium (Ra). The mass number will be 223.
5. Write the complete nuclear equation: The correct nuclear equation for the alpha decay of Th-227 is \( _{90}^{227}\text{Th} \rightarrow _{88}^{223}\text{Ra} + _{2}^{4}\text{He} \).