Anyone in chemistry?

Uh, this may seem like a stupid question but...

how do you calculate the half-life of a radioactive element, when given the initial amount, remaining amount, and the time it has been sitting there? Thanks in advance.

 

Never done chemistry.
Do they give you like a problem or something? Or is that all they ask? This would be so much easier if it was MATH!​

 

Haha yeah, it is sort of like math. There are actually some problems I need to do but here is a formula:

Amount remaining = (initial amount)(1/2)^(time elapsed/half-life)

 

Oh wow. It looks like you're gonna have to use logarithms... That's harsh.

 

That's the problem too. I don't know how to use them. D:

 

ahh the horrible memories of chemistry. stick to saying it is all a lie.

 

Okay, well let me explain how to use them. Just a quick note in case you didn't know, log( is a function, not a number, so it's not something you can multiply out. Just keep that in mind. You get to this point:

(Amount remaining)/(initial amount) = (1/2)^(time elapsed/half-life)

You can then do this:

log((Amount remaining)/(initial amount)) = log((1/2)^(time elapsed/half-life))

And by one of the log rules, you can take the exponent and put it in front
like this:

log((Amount remaining)/(initial amount)) = (time elapsed/half-life) * log(1/2)

So then you put log((Amount remaining)/(initial amount)) and log(1/2) into your calculator, find out what they are, and then solve for half-life. It should look like this in the end:

log((Amount remaining)/(initial amount))/log(1/2) = (time elapsed/half-life)

 

Okay, so then if I have 150.0 mg of something and after 11.4 days, the sample remaining is 18.7 mg, is the half-life 3.8?

 

I don't know what I'm doing. Just what I'm reviewing.

P E MD AS

Amount remaining= initial amount times one-half doubled-over, I think? times time divided by half-life

I have to think about it. D:​

 

That sounds about right. =D

 

Yayz~ As long as I get an answer, tis fine. =D

 

Well plug your number back into the equation and see if you get the correct answer.

 

For me, science like this is...well..a little...awkward to learn? xD

Martial Arts ftw!

 

O.o

Oh right. xDDD I completely forgot. I don't remember if I ever got it, but I've gotten the combustion analysis down pretty well now. xD