I have been trying to obtain an analytic solution for a daughter radionuclide's activity (or just the number of daughter atoms), as a function of $t\geq0$, resulting from the decay of a parent radionuclide specifically for the case where $\lambda_{parent}=\lambda_{daughter}$ (and no alternative decay modes so branching ratio=1), subject to initial conditions $\left.{N_{daughter}(t) }\right|_{\;t=0}^{\;}=0$ and $\left.{N_{parent}(t) }\right|_{\;t=0}^{\;}=C\neq0$. This is typically a pretty easy problem, I know, you just solve


to get parent solution (exponential decay with initial value of C and decay const $\lambda_{parent}$) then solve


to get daughter solution, which is (skipping to multiplying solution for $N_{daughter}$(t) by $\lambda_{daughter}$ to get daughter activity):

$$A_{daughter}=\lambda_{daughter}N_{daughter}(t)=\frac{C\lambda_{parent}\lambda_{daughter}}{\lambda_{daughter}-\lambda_{parent}}(e^{-\lambda_{parent}t}-e^{-\lambda_{daughter}t})+ [\left.{N_{daughter}(t) }\right|_{\;t=0}^{\;}]e^{-\lambda_{daughter}t}$$


which is obtained by applying initial cond's.

However, if you note what happens in the case $\lim_{\lambda_{daughter} \to \lambda_{parent}}[A_{daughter}(t)]$, you will see that $A_{daughter}(t)$ seems to go to an indeterminate form (0/0) if you apply direct substitution. In light of this, I wanted to try to apply L'Hopitals rule to this via:





then applying

$$\lim_{\lambda_{daughter} \to \lambda_{parent}}[\frac{g(t,\lambda_{daughter})}{h(\lambda_{daughter})}]=[\frac{\lim_{\lambda_{daughter} \to \lambda_{parent}}\frac{dg(t,\lambda_{daughter})}{d\lambda_{daughter}}}{\lim_{\lambda_{daughter} \to \lambda_{parent}}\frac{dh(\lambda_{daughter})}{d\lambda_{daughter}}}]$$

which gives me a solution that doesn't seem unreasonable


but I'm not absolutely 100% sure that this approach is mathematically sound. Can anyone confirm or deny the validity of this approach, and if it is not valid could you help by suggesting an alternative (and valid) approach?

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    $\begingroup$ $p$ and $d$ would work nicely as subscripts rather than writing out “parent” and “daughter”. $\endgroup$
    – Ghoster
    Mar 4 at 18:12
  • $\begingroup$ lol, fair enough. I have a tendency to be a bit odd with my notation (overly specific) but i usually do just use d and p (on paper). Just didn't want there to be any room for miscommunication but in hindsight I went a bit overkill for sure. If it bothers everyone a lot I'll change it, but for now I'll just leave it as is because I don't want to go and delete all the "daughter" and "parent" subscripts right this second (about to be busy lol) $\endgroup$ Mar 4 at 18:17
  • $\begingroup$ No problem. BTW, the usual convention in physics typography is that subscripted words are not italicized. $\endgroup$
    – Ghoster
    Mar 4 at 18:19
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    $\begingroup$ @AlbertusMagnus Physics journals have style guides so that their articles have a consistent look. If you look at articles in Physical Review, for example, I think you’ll find that subscripted words are not in italic case. $\endgroup$
    – Ghoster
    Mar 4 at 18:55
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    $\begingroup$ @AlbertusMagnus In general, italics are for variables and $abc$ means the product of $a$, $b$, and $c$. That why we write $\sin(x)$ and not $sin(x)$. $\endgroup$
    – Ghoster
    Mar 4 at 19:02

1 Answer 1


$$ \begin{align} \tag1N_p&=Ce^{-\lambda_pt}\\ \tag2\frac{\mathrm d\ }{\mathrm dt}\left(e^{\lambda_dt}N_d\right) &=\lambda_pCe^{(\lambda_d-\lambda_p)t}\\ \tag3A_d=\lambda_dN_d&=\frac{\lambda_d\lambda_pC}{\lambda_d-\lambda_p}\left(e^{-\lambda_pt}-e^{-\lambda_dt}\right) \end {align} $$ This last equation is in agreement with what you have.

However, it is extremely clear that Equation (2) has a totally different qualitative behaviour when $\lambda_d=\lambda_p=\lambda$; in that case, we would obtain that $$\tag4A_d=\lambda N_d=\lambda^2C\,t\,e^{-\lambda t}$$ which agrees with your L'Hôpital's approach (if you didn't have a typo here), but with manifest correctness. It is probably easier to see that this last equation is dimensionally correct by writing it as $\lambda C\cdot(\lambda t)e^{-\lambda t}$, where it is obvious that $\lambda t$ is dimensionless.


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