How can you model half life decay?
The equation would be:
##[A] = 1/(2^(t”/”t_”1/2″))[A]_0##
Read on to know what it means.
Just focus on the main principle:
The upcoming concentration of reactant ##A## after half-life time ##t_”1/2″## becomes half of the current concentration.
So, if we define the current concentration as ##[A]_n## and the upcoming concentration as ##[A]_(n+1)##, then…
##[A]_(n+1) = 1/2[A]_n## ##” “mathbf((1))##
We call the (1) the recursive half-life decay equation for one half-life occurrence, i.e. when ##t_”1/2″## has passed by only once. This isn’t very useful though, because half-lives can range from very slow (thousands of years) to very fast (milliseconds!).
Let’s go through another half-life, until we’ve gone through ##mathbf(n)## half-lives. For this, we rewrite ##[A]_n## as ##[A]_0## (the initial concentration), and ##[A]_(n+1)## as ##[A]## (the upcoming concentration).
Notice how ##[A]_0## will always be the same, but ##[A]## will keep changing over time.
##[A] = (1/2)(1/2)cdots(1/2)[A]_0##
##= (1/2)^n[A]_0##
##=> [A] = 1/(2^n)[A]_0## ##” “mathbf((2))##
Now we have (2), the equation for any number of half-life decays… once we know how many half-lives passed by.
However, (2) can be made more convenient since we know that each half-life takes ##t_”1/2″## time to occur. When ##n## half-lives occur, each one taking ##t_”1/2″## to occur, it must occur over a set amount of time ##t##. So:
##nt_”1/2″ = t## ##” “mathbf((3))##
That means ##n = t/t_”1/2″##, which is saying that we can divide the total time passed during the process by the time it takes to lose half of ##A## again to get the number of half-lives that passed by.
Therefore:
##color(blue)([A] = 1/(2^(t”/”t_”1/2″))[A]_0)## ##” “mathbf((4))##
So, we can use (4) to determine half-lives of any typical radioactive element for which we know ##t##, the time passed during the half-life decay(s) AND:
##[A]_0##, the initial concentration, and ##[A]##, the upcoming concentration, OR
##([A])/[A]_0##, the fraction of the element left after time ##t## passes.