homework and exercises - Radioactive Decay

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

Answer

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.