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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0JAMES CAN YOU HELP <3

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Right. 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
 4 years ago
Best ResponseYou've already chosen the best response.0yes the formula makes sense, i just dont understand how i'm suppose to use it with lead and uranium together ..

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Ok. In this problem, what is the half time, T?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1It'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
 4 years ago
Best ResponseYou've already chosen the best response.1Hence T = ... what? Talk to me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so t = 4.5 billion years right?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1The 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.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Now, 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?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Right. Now, in terms of U and Pb, what fraction of the original U is still Uranium?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhm, after the half life none of it.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1No. The total amount of material now at time t is U + Pb. Therefore what fraction of the total is Uranium?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1If there C cats and D dogs, what fraction of all the animals are cats?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhm the ratio multiplied by t right?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1\[ \frac{C}{C+D} \] Therefore what fraction of the total of U + Pb is Uranium?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay so itd be C / C+D

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Right, and \[ \frac{U}{U + Pb} = \frac{1}{1 + Pb/U} \] Now, what's that equal to?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0honestly, i wish i could tell you

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1We just calculated Pb/U above

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Pb/U = 0.31 Therefore \[ \frac{U}{U+Pb} = \frac{1}{1+Pb/U} = \frac{1}{1.31} \] Yes?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok ya that makes sense.. sorry i'm such a retard btw aha

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Now, what we have is that \[ d(t) = \frac{U}{U + Pb} \] You have a formula for the lefthand side (LHS) and now you know the value of the RHS. What formula do you now have then for t?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do we have to differentiate?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Nope. \[ d(t) = 2^{t/T} \] remember.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so that is equal to what you said above, whats the different between the big t and the little t?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1T = half life t = elapsed time

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because 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
 4 years ago
Best ResponseYou've already chosen the best response.1We 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
 4 years ago
Best ResponseYou've already chosen the best response.0and we do know that, so we can solve for t?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Now 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok, when solving should i put the whole 4.5 billion years in or just keep it at 4.5?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1You can use whatever units you like. Time units of billion years is perfectly good.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so then if i solve for t as 4.5 that should be fine as long as my answer is in billion years?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1T = 4.5. NOT, NOT t. You're solving for t.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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
 4 years ago
Best ResponseYou've already chosen the best response.1\[ 2^{t/T} = 1/1.31 \] take the log base 2 of both sides and you have \[ \frac{t}{T} \log_2 2 = \log_2 (1/1.31) \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah ok thank you so much!!!

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1What answer do you get for t?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Me too. t = 1.75 billion years.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yay i did it right!! ok thank you so much!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.1Let 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok thanks! i'm trying to get something from its basis in F, back to the standard basis!
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