I will gladly donate to anyone that helps me figure this out.

There are '8' of element 1, '0' of element 2, and '0' of element 3 in a sample.

The half-life of element 1 is 1 year, the half life of element 2 is 2 years, and element 3 is stable.

Create a table showing how much of each element there is for each year for 12 years.

At what point does the sample become stable?

^ isn't the answer never because an element never completely goes away?

Please help, I am confused..this is what I was doing so far but it is clearly wrong after the 7th year.

You seem to have left out the instructions that element 1 decays into element 2? and element 2 decays into element 3?

Well the problem is that on the 4th year, you predict 0.5 atoms of element 1. But that is a fiction, because elements can not exist in fractions (you can't have 1/2 of an oxygen atom, since it would not have the properties of oxygen!). So I'm assuming you are leaving out other parts of the question as well. Perhaps something like, "how long does it take for there to be a 99% chance there is only element 3 present"?

Additionally, you need to double check your formula for element 2, since there shouldn't be any negative entries...(what would it mean to have negative 3 atoms of oxygen?!?)

This post was edited by Azrad on Oct 10 2013 10:44am

The instructions do not say that these values are atoms, but it is implied in a hint. Also it does not say explicitly that element 1 decays into element 2, that is also implied. There is nothing saying 99% chance or anything like that though...and I understand that there cannot be .5 of an atom, that is why I know I am wrong and am asking for help. Thank you for the assistance thus far.

So would I instead go from element 1 = 1 to element 1 = 0? I think that would make my life a lot easier...it just seemed to simple.

Well perhaps it isn't 8 atoms but instead 8 moles of atoms. Then 0.5 is fine, but sooner or later, the function will produce results that will mean less than an atom, at which point it should be interpreted as the probability of having 1 or more atoms remaining. Making it impossible to say when the sample will be "stable" (unless there is some unspoken convention on what is stable, like my 99% example above). The amount of material 1 left (or the probability of having any material 1 left) will asymptotically approach 0 as time increases, but it will never be 0. So in its present form, this problem is fucked from the start, IMO.

But the part of the problem that says "Create a table showing how much of each element there is for each year for 12 years" is quite doable. So you need to address the error in your formula for calculating element 2 (and I assure you there is an error somewhere). I see you fixed that problem, nice.

To rephrase it another way, it is entirely possible that all 8 atoms of element 1 break down into element 2 during year one, and that all 8 atoms of elements 2 breakdown into element 3 during year 2. So it could become stable in 2 years, or 4, or 10, or 100, for 10 seconds. You can not predict when it will become stable. You can however predict when it will have a 99% chance of being stable.

This post was edited by Azrad on Oct 10 2013 11:17am

The error that I was running into is that element 2's half life is 2 years, and you are adding element 1's decay to it every year (while element 1 is active). So if I have 0 of element 2 in year 0, 4 of element 2 in year 1, and 6 of element 2 in year 2, i only take half of 4 to get what I am dividing out for year 3, correct? Lol this whole thing's confusing me. Does this chart look any better? I still feel like it's wrong




I guess one main question is: Do I only divide out element 2 every 2 years? i.e. years 3, 5, 7, ... ?

This post was edited by shinigamiapple777 on Oct 10 2013 11:17am

element 3, year 3 =/= 0?

This post was edited by saber_x3 on Oct 10 2013 11:32am

I am not positive of my work, but... If during year 1 element 2 has 4 atoms, and the half life of element 2 is 2 years, wouldn't that mean that element 3 has some atoms for year 3?

To avoid issues with discrete values like this (and wondering at what precise time we should apply the disintegration), I would rather use continuous functions :

a(t) = quantity of element a in terms of t (time)
b(t) = ...
c(t) = ...

So, a(t) = 8 * exp ( - t * ln2 )
since half life of element a is 1 year (express t in years).

b' (t) = - a' (t) - b(t) * ln2 / 2
since element b appears with the disintegration of a, and disappears from its own disintegration (half life = 2 years),

and c (t) = 8 - a(t) - b(t).

What do you think ?

Those are your functions, you will always be setting t = 1 (because you are making a table with entries in 1 year increments). Of course, as you've seen, to get N_0 for g(t) you have to add in the decay from the previous year of f(t)

import means, "the value from the difference column from the chart on the left" in this case. I highlighted the most complicated function so you can read it in the formula bar.

/e and of course this is just an approximation, as feanur pointed out. If you think about it, we are only processing the decay every year: essentially we are saying that on the last instant of the year, all the decays happen (which isn't right but I think it is the best we can do without calculus).

This post was edited by Azrad on Oct 10 2013 12:23pm