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 $$\dfrac{dN_Y}{dt}=\lambda N_x$$?

• 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). – dmckee --- ex-moderator kitten May 8 '13 at 14:03

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.

• See here , (part of this answer is wrong I think) en.wikipedia.org/wiki/Half-life – user23503 May 8 '13 at 13:27
• 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. – Mike May 8 '13 at 14:47
• 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. :) – Mike May 8 '13 at 14:48
• If the production rate is large enough, the population will grow to some equilibrium larger population. – user22620 Nov 21 '14 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: $$N_x(t_{1\over2})~=~{N_0+{\Lambda\over\lambda}\over2}$$ $$N_y(t_{1\over2})~=~{N_0\over2}+{\Lambda\over\lambda}(ln(2)-{1\over2})$$

• 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? – Jim May 8 '13 at 18:48
• I don't really understand that -1 either. This isn't the way I would solve this, but it appears to be correct. – Colin McFaul May 8 '13 at 20:11
• +1 . Correct answers. With full integration. :). Well I'll do that tomorrow. My daily quota ended.:p – ABC May 9 '13 at 14:30
• @007 Your daily quota ended? You are a credit to the site, sir. – Jim May 9 '13 at 14:56
• 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 – Jim May 9 '13 at 15:00