Chapter 27, Problem 96QAP

To determine

Trace out the pathway that terminates with a stable nucleus for each of the radioactive decay.

Explanation of Solution

Calculation:

A. Thorium Series: A = 4n
  $Th\to \alpha +Ra\to \beta +Ac \to \beta +Th\to \alpha +Ra\to \alpha +Rn\to \alpha +Po\to \alpha +Pb\to \beta +Bi\to \beta +Po\to \alpha +Pb$

There is another path from Bi-212
  $Bi\to \alpha +Th\to \beta +Pb$

B. Uranium Series: A = 4n+2
  $U\to \alpha +Th\to \beta +Pa\to \beta +U\to \alpha +Th\to \alpha +Ra\to$

  $\alpha +Rn\to \alpha +Po\to \alpha +Pb\to \beta +B\to \beta +Po\to \alpha +Pb\to$

  $\beta +Bi\to \beta +Po\to \alpha +Pb$

C. Actinium Series: A = 4n+3
  $U\to \alpha +Th\to \beta +Pa\to \alpha +Ac\to$

The above then breaks into two pathways, a and b, which eventually converge
  $Ac\to \alpha +Fr\to \beta +Ra$

  $Ac\to \beta +Th\to \alpha +Ra\to \alpha +Rn\to \alpha +Po\to \alpha +Pb\to \beta +Bi\to \alpha +Tl\to \beta +Pb$ D. Neptunium Series: A = 4n+1
  $Np-\to \alpha +Pa-\to \beta +U\to \alpha +Th\to \alpha +Ra\to$

  $\beta +{}_{}^{225}Ac\to \alpha +Fr\to \alpha +At\to \alpha +Bi\to$

  $Np\to \alpha +Pa\to \beta +U\to \alpha +Th\to \alpha +R\to$

  $\beta +Ac\to \alpha +Fr\to \alpha +At\to \alpha +Bi\to$

At this point the chain diverges
  $Bi\to \alpha +Tl\to \beta +Pb$

  $Bi\to \beta +Po\to \alpha +Pb\to \beta +Bi$