3

If a population is subject to two simultaneous, independent exponential decay processes ($x_1$ and $x_2$), the decay rate is the sum of the two processes and the proportion remaining after time $t$ becomes:

$$N_t = N_0 e^{-(x_1+x_2)t}$$

But what proportion of the leaving fraction leaves via each process?

I've attempted some Google searches from which I gather this problem crops up a lot in particle physics where two or more decay modes occur, but I'm not familiar enough with the field to know what the answer is (and can't find an entry-level explanation for a non-physicist).

(An answer which can be generalised to $n$ processes would be even better.)

arboviral
  • 83
  • Is this just (x1/(x1+x2))*(1-(exp(-(x1+x2)))), for process x1, or is that too obvious? – arboviral Dec 19 '17 at 15:07
  • "the proportion remaining after time $t$ becomes $N_t = N_0 e^{-(x_1 + x_2) }.$" Where is the time $t$ on the right side of this equality? Are $x_1$ and $x_2$ functions of $t$? What sort of functions? I find this sloppily expressed. – Michael Hardy Jan 09 '18 at 19:21
  • @MichaelHardy Thanks for the welcome! No need to worry, someone else worked it out. – arboviral Jan 10 '18 at 08:54

1 Answers1

1

In fact what we usually measure is the total decay rate and the fraction that decay by each process. We then calculate the decay rate for each process. The combination is given in Wikipedia. As you say, if you have two decays with time constants $\lambda_1$ and $\lambda_2$ the total decay rate is $\lambda_c=\lambda_1+\lambda_2$ and the amount remaining is $N_t=N_0e^{-(\lambda_1+\lambda_2)t}$ You are looking for the branching fraction into each mode. For mode $1$ it is given by $\frac {\lambda_1}{\lambda_c}$

Ross Millikan
  • 374,822