anonymous
  • anonymous
Somebody please help meee!!! Suppose that U^238 has a half life of 4.5 billion years, decaying (through a series of relatively short lived intermediate atoms) to Pb^206. In a certain mineral sample there are .31 times as many Pb^206 atoms as there are of U^238. If one assumes that the mineral deposit contained no Pb^206 when it was formed and that no lead or uranium have been added to or escaped from the sample (except through the natural decay process) how old is the sample?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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JamesJ
  • JamesJ
*bookmark
anonymous
  • anonymous
thanks (:
anonymous
  • anonymous
JAMES CAN YOU HELP <3

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JamesJ
  • JamesJ
Right. Having a half life of time T means that after time t, a material will be \[ d(t) = 2^{-t/T} \] of what it was originally. For example at time t= 0, it should be all itself, so \[ d(0) = 2^0 = 1.\] At time t= T, it should have half decayed, so it's half of what it was, \[ d(T) = 2^{-T/T} = 2^{-1} = \frac{1}{2} \] After time t=2T, it should be half of half of the original substance, i.e., 1/4. And that's what our formula gives us, as \[ d(2T) = 2^{-2T/T} = 2^{-2} = \frac{1}{4}. \] So are you convinced this formula d(t) makes sense?
anonymous
  • anonymous
yes the formula makes sense, i just dont understand how i'm suppose to use it with lead and uranium together ..
JamesJ
  • JamesJ
Ok. In this problem, what is the half time, T?
JamesJ
  • JamesJ
It's right there at the beginning of your problem: "Suppose that U^238 has a half life of 4.5 billion years, decaying (through a series of relatively short lived intermediate atoms) to Pb^206."
JamesJ
  • JamesJ
Hence T = ... what? Talk to me.
anonymous
  • anonymous
ok so t = 4.5 billion years right?
JamesJ
  • JamesJ
The half life T = 4,500,000,000 years, yes. What we want to find is the time t. So let's be careful to differentiate between little t and capital T.
anonymous
  • anonymous
ok, i'm listening.
JamesJ
  • JamesJ
Now, let U denote the amount of Uranium and let Pb denote the amount of iron. What do we know about the ratio: \[ \frac{Pb}{U} \] What is it equal to?
anonymous
  • anonymous
.31/1
JamesJ
  • JamesJ
Right. Now, in terms of U and Pb, what fraction of the original U is still Uranium?
anonymous
  • anonymous
uhm, after the half life none of it.
JamesJ
  • JamesJ
No. The total amount of material now at time t is U + Pb. Therefore what fraction of the total is Uranium?
JamesJ
  • JamesJ
If there C cats and D dogs, what fraction of all the animals are cats?
anonymous
  • anonymous
uhm the ratio multiplied by t right?
JamesJ
  • JamesJ
\[ \frac{C}{C+D} \] Therefore what fraction of the total of U + Pb is Uranium?
anonymous
  • anonymous
okay so itd be C / C+D
anonymous
  • anonymous
okay so U / U + Pb
JamesJ
  • JamesJ
Right, and \[ \frac{U}{U + Pb} = \frac{1}{1 + Pb/U} \] Now, what's that equal to?
anonymous
  • anonymous
honestly, i wish i could tell you
JamesJ
  • JamesJ
We just calculated Pb/U above
anonymous
  • anonymous
Oh! .31/1
JamesJ
  • JamesJ
Pb/U = 0.31 Therefore \[ \frac{U}{U+Pb} = \frac{1}{1+Pb/U} = \frac{1}{1.31} \] Yes?
anonymous
  • anonymous
ok ya that makes sense.. sorry i'm such a retard btw aha
JamesJ
  • JamesJ
Now, what we have is that \[ d(t) = \frac{U}{U + Pb} \] You have a formula for the left-hand side (LHS) and now you know the value of the RHS. What formula do you now have then for t?
anonymous
  • anonymous
do we have to differentiate?
JamesJ
  • JamesJ
Nope. \[ d(t) = 2^{-t/T} \] remember.
anonymous
  • anonymous
ok so that is equal to what you said above, whats the different between the big t and the little t?
JamesJ
  • JamesJ
T = half life t = elapsed time
anonymous
  • anonymous
because one of them is 4.5 billion years right? and then we just have to solve for the other one? but i dont understand what we sub in for the different elements then
JamesJ
  • JamesJ
We have after a time t, \( d(t) \) of the Uranium is still Uranium. We need to solve for t. Now, another expression for the amount of material that is still Uranium is \[ \frac{U}{U + Pb} \] because Pb is what the U decays into. Therefore if we know the value of \( \frac{U}{U + Pb} \) we can find the time t that has passed in order that \[ d(t) = \frac{U}{U + Pb} \]
anonymous
  • anonymous
and we do know that, so we can solve for t?
JamesJ
  • JamesJ
Now that last equation is equivalent to \[ 2^{-t/T} = \frac{1}{1.31} \] You know the value of the half life T. Now solve for elapsed time t.
anonymous
  • anonymous
ok, when solving should i put the whole 4.5 billion years in or just keep it at 4.5?
JamesJ
  • JamesJ
You can use whatever units you like. Time units of billion years is perfectly good.
anonymous
  • anonymous
so then if i solve for t as 4.5 that should be fine as long as my answer is in billion years?
JamesJ
  • JamesJ
T = 4.5. NOT, NOT t. You're solving for t.
anonymous
  • anonymous
ok right. can i ask one more question? how would i go about solving for the variable i know that if its e with an exponent you just ln it, but i'm not sure if its an actual number.
JamesJ
  • JamesJ
\[ 2^{-t/T} = 1/1.31 \] take the log base 2 of both sides and you have \[ \frac{-t}{T} \log_2 2 = \log_2 (1/1.31) \]
anonymous
  • anonymous
ah ok thank you so much!!!
JamesJ
  • JamesJ
What answer do you get for t?
anonymous
  • anonymous
i got 1.75
anonymous
  • anonymous
right or wrong?
JamesJ
  • JamesJ
Me too. t = 1.75 billion years.
anonymous
  • anonymous
oh yay i did it right!! ok thank you so much!
anonymous
  • anonymous
i know i've already asked you a billion questions because your so smart, but do you know how to change basis of transformation matricies?
JamesJ
  • JamesJ
Let T be your original matrix with respect to a basis E = {e1, e2, ... , en}. Let F = {f1, f2, ..., fn} be another basis and U the matrix that transforms E to F. Then if you want T in basis F, then you want to use \[ UTU^{-1} \] I think the logic for this is clear. \( U^{-1} \) takes a vector in basis F, puts it in terms of the basis of E. \( T \) then operates on it. Then \( U \) transforms that back into the basis F.
anonymous
  • anonymous
ok thanks! i'm trying to get something from its basis in F, back to the standard basis!

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