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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?

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  1. JamesJ
    • 4 years ago
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    *bookmark

  2. anonymous
    • 4 years ago
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    thanks (:

  3. anonymous
    • 4 years ago
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    JAMES CAN YOU HELP <3

  4. JamesJ
    • 4 years ago
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    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?

  5. anonymous
    • 4 years ago
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    yes the formula makes sense, i just dont understand how i'm suppose to use it with lead and uranium together ..

  6. JamesJ
    • 4 years ago
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    Ok. In this problem, what is the half time, T?

  7. JamesJ
    • 4 years ago
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    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."

  8. JamesJ
    • 4 years ago
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    Hence T = ... what? Talk to me.

  9. anonymous
    • 4 years ago
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    ok so t = 4.5 billion years right?

  10. JamesJ
    • 4 years ago
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    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.

  11. anonymous
    • 4 years ago
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    ok, i'm listening.

  12. JamesJ
    • 4 years ago
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    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?

  13. anonymous
    • 4 years ago
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    .31/1

  14. JamesJ
    • 4 years ago
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    Right. Now, in terms of U and Pb, what fraction of the original U is still Uranium?

  15. anonymous
    • 4 years ago
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    uhm, after the half life none of it.

  16. JamesJ
    • 4 years ago
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    No. The total amount of material now at time t is U + Pb. Therefore what fraction of the total is Uranium?

  17. JamesJ
    • 4 years ago
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    If there C cats and D dogs, what fraction of all the animals are cats?

  18. anonymous
    • 4 years ago
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    uhm the ratio multiplied by t right?

  19. JamesJ
    • 4 years ago
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    \[ \frac{C}{C+D} \] Therefore what fraction of the total of U + Pb is Uranium?

  20. anonymous
    • 4 years ago
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    okay so itd be C / C+D

  21. anonymous
    • 4 years ago
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    okay so U / U + Pb

  22. JamesJ
    • 4 years ago
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    Right, and \[ \frac{U}{U + Pb} = \frac{1}{1 + Pb/U} \] Now, what's that equal to?

  23. anonymous
    • 4 years ago
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    honestly, i wish i could tell you

  24. JamesJ
    • 4 years ago
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    We just calculated Pb/U above

  25. anonymous
    • 4 years ago
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    Oh! .31/1

  26. JamesJ
    • 4 years ago
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    Pb/U = 0.31 Therefore \[ \frac{U}{U+Pb} = \frac{1}{1+Pb/U} = \frac{1}{1.31} \] Yes?

  27. anonymous
    • 4 years ago
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    ok ya that makes sense.. sorry i'm such a retard btw aha

  28. JamesJ
    • 4 years ago
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    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?

  29. anonymous
    • 4 years ago
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    do we have to differentiate?

  30. JamesJ
    • 4 years ago
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    Nope. \[ d(t) = 2^{-t/T} \] remember.

  31. anonymous
    • 4 years ago
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    ok so that is equal to what you said above, whats the different between the big t and the little t?

  32. JamesJ
    • 4 years ago
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    T = half life t = elapsed time

  33. anonymous
    • 4 years ago
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    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

  34. JamesJ
    • 4 years ago
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    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 \[ d(t) = \frac{U}{U + Pb} \]

  35. anonymous
    • 4 years ago
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    and we do know that, so we can solve for t?

  36. JamesJ
    • 4 years ago
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    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.

  37. anonymous
    • 4 years ago
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    ok, when solving should i put the whole 4.5 billion years in or just keep it at 4.5?

  38. JamesJ
    • 4 years ago
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    You can use whatever units you like. Time units of billion years is perfectly good.

  39. anonymous
    • 4 years ago
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    so then if i solve for t as 4.5 that should be fine as long as my answer is in billion years?

  40. JamesJ
    • 4 years ago
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    T = 4.5. NOT, NOT t. You're solving for t.

  41. anonymous
    • 4 years ago
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    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.

  42. JamesJ
    • 4 years ago
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    \[ 2^{-t/T} = 1/1.31 \] take the log base 2 of both sides and you have \[ \frac{-t}{T} \log_2 2 = \log_2 (1/1.31) \]

  43. anonymous
    • 4 years ago
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    ah ok thank you so much!!!

  44. JamesJ
    • 4 years ago
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    What answer do you get for t?

  45. anonymous
    • 4 years ago
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    i got 1.75

  46. anonymous
    • 4 years ago
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    right or wrong?

  47. JamesJ
    • 4 years ago
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    Me too. t = 1.75 billion years.

  48. anonymous
    • 4 years ago
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    oh yay i did it right!! ok thank you so much!

  49. anonymous
    • 4 years ago
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    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?

  50. JamesJ
    • 4 years ago
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    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.

  51. anonymous
    • 4 years ago
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    ok thanks! i'm trying to get something from its basis in F, back to the standard basis!

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