Problem:Nuclei of a radioactive element $\Bbb X$ having decay constant $\lambda$ , ( decays into another stable nuclei $\Bbb Y$ ) is being produced by some external process at a constant rate $\Lambda$.Calculate the number of nuclei of $\Bbb X$ and $\Bbb Y$ at $t_{1/2}$

I tried to create an equation for rate of change of the number of nuclei a:

$$\dfrac{dN_{X}}{dt}=\Lambda-N_X\lambda $$

I did that because in simple decay $\dfrac{dN}{dt}=-\lambda N$ holds and here it's also being produced by rate. But after integration should we write $$ln\Bigg(\dfrac{\lambda N_X-\Lambda}{\lambda N_0-\Lambda}\Bigg)=-\lambda t$$ or $$ln\Bigg(\dfrac{\lambda N_X-\Lambda}{N_0}\Bigg)=-\lambda t$$ First one because limit was on $N: (N_0\to N)$ And next what to substitute for $t$ (ie. what is $t_{1/2}$? $ln2/\lambda$ or something else?)

Also how to do it for $\Bbb Y$? Just write $$\dfrac{dN_Y}{dt}=\lambda N_x $$?

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    $\begingroup$ BTW--There is a reason for this problem. When you have a decay chain $W \to X \to Y$ where the $\tau_W \gg \tau_X$ measurements on the time scale of $\tau_X$ look like they have a constant rate refill of species $X$. This is effectively the case, for instance at the bottom of the Radon chain where $W$ is Pb-210 and $X$ is Po-210 (actually there is also an intervening species with even shorter half-life, but if you work it out you'll see you can effectively ignore that). $\endgroup$ May 8, 2013 at 14:03
  • $\begingroup$ The differential equation is correct, but the solution is not. You need to use an integrating factor to solve the differential equation. The short answer to your question is that after one half-life, the value of $N_x$ will be $\Lambda/2$. This assumes that $N_x$ started at zero. $\endgroup$ Dec 21, 2021 at 20:49

2 Answers 2


The first of your equations is correct. You can see this in two ways. First, just look at the dimensions. In general, the argument of a logarithm should be dimensionless; only your first option is. Second, and maybe more convincingly, look at what you get when you take $\Lambda \to0$. You should be able to reproduce the standard decay equation: \begin{equation} N_X(t) = N_0\, e^{-\lambda\, t}~. \end{equation} In your first equation, the factors of $\lambda$ on the left-hand side cancel, and you get this result. With your second equation, you would get $N_X(t) = \frac{N_0}{\lambda}\, e^{-\lambda\, t}$. So that must be wrong.

As for what $t_{1/2}$ is, surely it must just be the half-life of $\mathbb{X}$ (with no creation). In particular, if $\Lambda$ is large enough, $N_X$ will actually grow, so there is no time at which half of the material is left. Since $\mathbb{Y}$ is stable, you can assume there's no relevant half-life there.

Also, your expression for $N_Y$ is correct. It's a slightly harder integration, but not too bad.

  • $\begingroup$ See here , (part of this answer is wrong I think) en.wikipedia.org/wiki/Half-life $\endgroup$
    – user23503
    May 8, 2013 at 13:27
  • $\begingroup$ Hi @nonagon. I understand the definition of half-life, but the question is slightly ambiguous as to which meaning is to be attached to the symbol $t_{1/2}$. As I argued above, and 007 argued in response to your answer, if $\Lambda$ is large enough, there will never be a time at which $N_X = N_0/2$. And even if there is such a time, the meaning of $t_{1/2}$ would then depend on the circumstances. A different reasonable interpretation of the symbol $t_{1/2}$ is to consider it a fixed characteristic of a certain isotope. In that case, the question actually makes sense in general. $\endgroup$
    – Mike
    May 8, 2013 at 14:47
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    $\begingroup$ Also, I think treating $t_{1/2}$ as a fixed characteristic of an isotope is just a more common convention. Part of taking physics classes is deciphering the meaning of questions. :) $\endgroup$
    – Mike
    May 8, 2013 at 14:48
  • $\begingroup$ If the production rate is large enough, the population will grow to some equilibrium larger population. $\endgroup$
    – user22620
    Nov 21, 2014 at 17:41

So here's what I did, I discretized the problem and deduced $N_x$ and $N_y$ at some $t_n$. These happened to include sums that could easily be turned into integrals. Namely: $$N_x(t_n)=N_0e^{-\lambda t_n}+\Lambda\sum_{i=0}^n\Delta t_ie^{-\lambda(t_n-t_i)}$$ $$N_y(t_n)=N_0(1-e^{-\lambda t_n})+\Lambda\sum_{i=0}^n\Delta t_i(1-e^{-\lambda(t_n-t_i)})$$

As a result, I found the following:

$$N_x(t)~=~N_0e^{-\lambda t}+{\Lambda\over\lambda}(1-e^{-\lambda t})$$ $$N_y(t)~=~N_0(1-e^{-\lambda t})-{\Lambda\over\lambda}(1-e^{-\lambda t})+\Lambda t$$

At $t_{1\over2}={ln(2)\over\lambda}$, $e^{-\lambda t}={1\over2}$, therefore: $$N_x(t_{1\over2})~=~{N_0+{\Lambda\over\lambda}\over2}$$ $$N_y(t_{1\over2})~=~{N_0\over2}+{\Lambda\over\lambda}(ln(2)-{1\over2})$$

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    $\begingroup$ Okay, so I'm willing to accept a deserved -1, but common courtesy would have whoever it was leave a comment about why they disliked my answer. It is consistent with what mike said and it fully answers the question as far as I know. Have I made a mistake somewhere? $\endgroup$
    – Jim
    May 8, 2013 at 18:48
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    $\begingroup$ I don't really understand that -1 either. This isn't the way I would solve this, but it appears to be correct. $\endgroup$ May 8, 2013 at 20:11
  • $\begingroup$ +1 . Correct answers. With full integration. :). Well I'll do that tomorrow. My daily quota ended.:p $\endgroup$
    – ABC
    May 9, 2013 at 14:30
  • $\begingroup$ @007 Your daily quota ended? You are a credit to the site, sir. $\endgroup$
    – Jim
    May 9, 2013 at 14:56
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    $\begingroup$ I know, I was serious. If you're maxing out your quota, you're clearly dedicating ample time to the site in the effort to improve it for all of us. Thus a credit, the "sir" was a proffered title of respect $\endgroup$
    – Jim
    May 9, 2013 at 15:00

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