[Kealan]

My chemistry teacher just moved us into nuclear physics today. I usually grasp whatever she teaches instantly, if I didn't already know it from reading it somewhere else, but I just don't get this, in particular I mean half-lifes. That is the time it takes for 1/2 of a sample of the element to decay. It is supposed to be very simple to calculate, but I don't see how it can be. For example:

Say we have 100 grams of a substance that has a 5 day halflife, ok? Now, going with what she said, that would mean there is 50 grams left after 5 days, 25 after 10, ect. But I don't see how this is possible. It would seem like the halflife would be impossible to calculate, since with every atom that decays the rate slows. See, if you use the above example of 100 grams with a 5 day halflife and do it like this, it is completely different. Start with 100 grams, but do it day-by-day, instead of 5 days at a time. So, after the first day, 20% of the halflife, 1/10th of it would be gone, leaving 90 grams. The next day, another 1/10th gone, leaving 81 grams this time. Next day, 72.9. Next, 65.61, and on the 5th day, 59.049. See the problem I am having? And it would be an even more extreme differance if there was a way to re-calculate it every instant, as it should be for an accurate number. I don't see how it could be an average of the decay over that time, either, since a larger number of atoms would change the time it takes for half to decay. Anyone understand the problem I am having, and have a way to explain it correctly?

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OK, this is something I can help you out with. I am finishing up my masters in chemistry.

Here's something to remember. This is an exponential type of decay not a linear decay. Basically, every 5 days the value is multiplied by 1/2.

So lets just look at the problem and see what the value is as the half-lives go on.

We will call the original value N₀ (this is called N sub zero, N initial or N starting)

So at the first half-life we have (1/2)N₀
At the second half-life we have (1/2)(1/2)N₀
At the third half-life we have (1/2)(1/2)(1/2)N₀

This means we have an exponential of 1/2 type of decay

So we get N = (1/2)^xN₀, where x is the number of half-lives and N is the resulting amount.

Now to prove to you this works I will show you the result of the above three (plus the starting value).

At the starting point our value for half-lives is 0. This results in (1/2)⁰N₀ which results in (1)*N₀ because any number to the zeroth power = 1.

Now at the first half-life (1/2)¹N₀ = (1/2)N₀
For the second half-life (1/2)²N₀ = (1/2)(1/2)N₀
For the third half-life (1/2)³N₀ = (1/2)(1/2)(1/2)N₀

Fractional Half-lives

So now that I have shown how to come to the equation N = (1/2)^xN₀. HOw do we deal with fractional half-lives. Simple, we substitue them in for x. Thats it.

The most common way to do this is to create a fraction with the half-life on the bottom and the amount of time that has passed on the top and substitute that for x.

This sounds confusing, but I think if you see it written out it will make more sense.

The common one is T/T_1/2, where T_1/2 = the half-life and T = the amount of time that has passed.

The only caveat is these need to be in the SAME units, so for example, your half-life is 5. Your fraction would be x = T/5.

So for 1 day, x = 1/5, 2 days, x = 2/5...........5 days x= 5/5 or 1 half-life.

Now if you wanted to do it in minutes, you would have to change the bottom value to be 5-days but in minutes, so 5 days * 24 hrs * 60mintes = 7200 minutes so your x = T/7200

SO our final equation is N = (1/2)^(T/T_1/2N₀.

Though the fraction can be substituted for a numberical value of half-lives as well, if the teacher asks how much is left after 4.5 half-lives, you simply use 4.5 for your x value instead of the fraction.


I hope this helps.

Edited by PedroDaGR8, 16 January 2009 - 12:26 PM.

[PedroDaGR8]

OK, this is something I can help you out with. I am finishing up my masters in chemistry.

Here's something to remember. This is an exponential type of decay not a linear decay. Basically, every 5 days the value is multiplied by 1/2.

So lets just look at the problem and see what the value is as the half-lives go on.

We will call the original value N₀ (this is called N sub zero, N initial or N starting)

So at the first half-life we have (1/2)N₀
At the second half-life we have (1/2)(1/2)N₀
At the third half-life we have (1/2)(1/2)(1/2)N₀

This means we have an exponential of 1/2 type of decay

So we get N = (1/2)^xN₀, where x is the number of half-lives and N is the resulting amount.

Now to prove to you this works I will show you the result of the above three (plus the starting value).

At the starting point our value for half-lives is 0. This results in (1/2)⁰N₀ which results in (1)*N₀ because any number to the zeroth power = 1.

Now at the first half-life (1/2)¹N₀ = (1/2)N₀
For the second half-life (1/2)²N₀ = (1/2)(1/2)N₀
For the third half-life (1/2)³N₀ = (1/2)(1/2)(1/2)N₀

Fractional Half-lives

So now that I have shown how to come to the equation N = (1/2)^xN₀. HOw do we deal with fractional half-lives. Simple, we substitue them in for x. Thats it.

The most common way to do this is to create a fraction with the half-life on the bottom and the amount of time that has passed on the top and substitute that for x.

This sounds confusing, but I think if you see it written out it will make more sense.

The common one is T/T_1/2, where T_1/2 = the half-life and T = the amount of time that has passed.

The only caveat is these need to be in the SAME units, so for example, your half-life is 5. Your fraction would be x = T/5.

So for 1 day, x = 1/5, 2 days, x = 2/5...........5 days x= 5/5 or 1 half-life.

Now if you wanted to do it in minutes, you would have to change the bottom value to be 5-days but in minutes, so 5 days * 24 hrs * 60mintes = 7200 minutes so your x = T/7200

SO our final equation is N = (1/2)^(T/T_1/2N₀.

Though the fraction can be substituted for a numberical value of half-lives as well, if the teacher asks how much is left after 4.5 half-lives, you simply use 4.5 for your x value instead of the fraction.


I hope this helps.

Edited by PedroDaGR8, 16 January 2009 - 12:26 PM.

[BHowett]

Typically we don’t help with homework type questions, if you have questions you should really ask them in class. Lucky for you PedroDaGR8 posted prior to me reading it.

[PedroDaGR8]

Ooops, I didn't see him asking a specific question so I assumed it wasn't HW. Oh well.

[jaxisland]

nevermind

Edited by jaxisland, 16 January 2009 - 01:10 PM.

[Kealan]

It isn't homework, I don't need to know for any specific assignment. I was just having trouble understanding the theory itself, I tried asking my teacher but she couldn't really explain it to me in a way that made sense. I think I get it now, though, thanks for the help, and sorry about asking here, didn't think it would hurt anything.

[BHowett]

No worries... it just appeared as it might be homework with the whole example question, and to stay consistent I just want to mention that we don’t typically help with school work.

[hfcg]

A conversation on the theorys of a scientific subject would not violate the TOU, but help with a specific home work assignment would defeat the purpose of you going to school (to learn).
While I understand the theory of half life, this subject is over my head!