# Radioactive Tracer - calculate fraction that decay before excretion

I was reading about the use of Technetium-99m as a radioactive tracer, how it decays via gamma emission but is also excreted by the body. Assuming the body handles Tc-99 and Tc-99m the same, some will be excreted in the excited state before it decays. How could you calculate what proportion decay while inside the body? (e.g. to calculate the dose received)

I made an attempt at calculating:

If the technetium has a biological half life of ~1 day (half of the amount in the body is excreted in 1 day), I assume it can be modelled with an exponential law. This can be combined with the exponential decay law to find an amount remaining in the body, and technetium has a half life of ~6 hours:

$$N = N_0 e^{-\lambda_1t} e^{-\lambda_2 t}$$

with $$\lambda = \frac{ln 2}{T_{1/2}}$$ for the radioactive and biological half lives.

I could not think how I could translate this into a fraction of decaying before or after excretion.

The cumulated probability density function that the decay happens before time $$t$$ is given by $$F_{\textrm{decay}}(t) = P(T\le t) = 1 - e^{- \lambda_1 t}$$ Thus, the probability density function of the decay at time $$t$$ is given by $$f_{\textrm{decay}}(t) = \frac{dF}{dt} = \lambda_1 e^{- \lambda_1 t} ~~~~ \textrm{, for t\ge 0}$$ An analog relation holds for the probability density function of being outside of the body at time $$t$$. Thus, $$f_{\textrm{outside}}(t) %= f_{\textrm{outside body}}(t) = \frac{dF}{dt} = \lambda_2 e^{- \lambda_2 t} ~~~~ \textrm{, for t\ge 0}$$ Since these two processes are linear independent, the joint probability density function is the product of both, $$f_{\textrm{decay}}(t_1) \cdot f_{\textrm{outside}}(t_2)$$. Note, that I included two different times. I agree, this is weird. Nevertheless, we have to allow question such as "what is the probability that the tracer leaves the body at time $$t_2$$ and decays at time $$t_1$$?".
Actually, this is exactly the question we are interested in. In particular, we are interested in $$0\le t_1 < t_2$$, which gives that probability that the tracer decays before leaving the body. Thus \begin{align} P(t_1 < t_2) &= \int_0^\infty dt_2 \int_0^{t_2} dt_1 \; \lambda_2 e^{- \lambda_2 t_2} \cdot \lambda_1 e^{- \lambda_1 t_1} \\ %%% &= \ldots = \frac{\lambda_1}{\lambda_1 + \lambda_2} \end{align} To check this equation we consider special cases:
• $$\lambda_1 \gg \lambda_2$$: Here the decay rate is "large". Thus, we expect that almost all nuclei decay inside the body. The upper formula yields $$1$$.
• $$\lambda_1 \ll \lambda_2$$: Here the decay rate is "small". Thus, we expect that almost all nuclei decay outside the body. The upper formula yields $$0$$.
• $$\lambda_1 = \lambda_2$$: Here the decay rate is and the extraction rate are equal. Thus, we know that there is a $$50\%:50\%$$ chance for each particle to decay inside or outside the body. The upper formula yields $$0.5$$.