Editing Radioactive decay (section)

=====Chain of any number of decays=====

For the general case of any number of consecutive decays in a decay chain, i.e. {{math|A<sub>1</sub> → A<sub>2</sub> ··· → A<sub>''i''</sub> ··· → A<sub>''D''</sub>}}, where {{math|''D''}} is the number of decays and {{math|''i''}} is a dummy index ({{math|''i'' {{=}} 1, 2, 3, ..., ''D''}}), each nuclide population can be found in terms of the previous population. In this case {{math|''N''<sub>2</sub> {{=}} 0}}, {{math|''N''<sub>3</sub> {{=}} 0}}, ..., {{math|''N<sub>D</sub>'' {{=}} 0}}. Using the above result in a recursive form:

:<math> \frac{\mathrm{d}N_j}{\mathrm{d}t} = - \lambda_j N_j + \lambda_{j-1} N_{(j-1)0} e^{-\lambda_{j-1} t}. </math>

The general solution to the recursive problem is given by '''Bateman's equations''':<ref name="general solution of Bateman">{{cite journal|last=Cetnar|first=Jerzy|title=General solution of Bateman equations for nuclear transmutations|journal=Annals of Nuclear Energy|date=May 2006|volume=33|issue=7|pages=640–645|doi=10.1016/j.anucene.2006.02.004|bibcode=2006AnNuE..33..640C }}</ref>

{{Equation box 1
|indent=:
|title='Bateman's equations'
|equation=<math>\begin{align}
  N_D &= \frac{N_1(0)}{\lambda_D} \sum_{i=1}^D \lambda_i c_i e^{-\lambda_i t} \\[3pt]
  c_i &= \prod_{j=1, i\neq j}^D \frac{\lambda_j}{\lambda_j - \lambda_i}
\end{align}</math>
|cellpadding
|border
|border colour = #0073CF
|background colour=#F5FFFA
}}