Answer
Partial decay series for Th-232 is as follows:
Work Step by Step
At first, Th-232 (atomic number 90) undergoes alpha decay. So, the atomic number will be decreased by 2 units and mass number will be decreased by 4 units. Atomic number of the daughter nuclide will be 90-2 = 88 and mass number of the daughter nuclide will be 232-4= 228. The daughter nuclide is Ra (Atomic number 88). When writing the nuclear equation, every element is represented in symbolic form (Write the symbol of the element, write atomic number on the left subscript and write mass number on the left superscript). So, on the left side of the equation we write Th-232 in symbolic form and on the right side of the equation we write Ra-228 in symbolic form. Also, Alpha decay can be represented by writing He atom in symbolic form on the right side of the equation.
After that, Beta decay occurred on Ra-228 (Atomic number 88). Here, parent nuclide is Ra-228 and after beta decay, atomic number will be increased by 1 unit but mass number will remain same. In this case, the daughter nuclide will have the atomic number 88+1 = 89 . The daughter nuclide is Ac (Atomic number 89). Parent nuclide will be on the left side of the equation and daughter nuclide will be on the right side of the equation. Every nuclide including beta particle can be represented in symbolic form while writing the equation. Beta particle can be represented in symbolic form by writing the chemical symbol 'e' and put '-1' on left subscript and '0' on left superscript. Beta particle is emitted from parent nucleus so it is on the right side of the equation.
Another beta decay is occurred on Ac-228 (Atomic number 89). In this case, parent nuclide is Ac and daughter nuclide will have atomic number 89+1 = 90 and mass number will be the same as parent nuclide (228). So, the new element form will be Th-228. Now, we can write the nuclear equation by writing every element including beta particle in symbolic form. Parent nuclide will be on left side whereas daughter nuclide and beta particle will be on the right side of the equation.
After that, Th-228 will undergo alpha emission. So, the atomic number of the daughter nuclide will be 90-2 = 88 and mass number will be 228-4= 224. Now, the daughter nuclide is Ra-224 (atomic number 88). we can write the nuclear equation by writing every element including alpha particle in symbolic form. Parent nuclide will be on left side whereas daughter nuclide and alpha particle will be on the right side of the equation.
