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kkid Group Title

How can I calculate the time taken for an element to decay by a certain amount?

  • one year ago
  • one year ago

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  1. kkid Group Title
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    I have the decay constant and need to find out how long it will take for the element to decay from 5ug to 1ug

    • one year ago
  2. henpen Group Title
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    Do you know differential equations or calculus?

    • one year ago
  3. kkid Group Title
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    Yes, but I don't think this question requires it

    • one year ago
  4. kkid Group Title
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    The question I'm tryingt o answer is: TA sample of 24Na has a half-life of 234 hours, How much time elapses before a 5μg sample contains 1μg of undecayed atoms?

    • one year ago
  5. kkid Group Title
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    A sample of*

    • one year ago
  6. henpen Group Title
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    \[\frac{1}{c}e^{-t \lambda}=ke^{-t \lambda}=N\]

    • one year ago
  7. kkid Group Title
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    but I want to know how to do it myself

    • one year ago
  8. kkid Group Title
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    whats c?

    • one year ago
  9. henpen Group Title
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    Sorry, it deleted half of my answer

    • one year ago
  10. kkid Group Title
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    lol, ok

    • one year ago
  11. kkid Group Title
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    I have used a similar equation \[N = N _{0}e ^{-\lambda t}\]

    • one year ago
  12. kkid Group Title
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    ...in the past

    • one year ago
  13. henpen Group Title
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    http://upload.wikimedia.org/math/a/8/a/a8a0bd0a12874474352b307d9e919076.png \[\frac{-1}{\lambda}(lnN+lnk)=\frac{-1}{\lambda}(\ln(kN))=t\] \[(\ln(kN))=-\lambda t\] \[kN=e^{-\lambda t}\] \[N=ce^{-\lambda t}\]

    • one year ago
  14. henpen Group Title
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    Set c= sample at t=0 That is, \[ c= N_0 \]

    • one year ago
  15. henpen Group Title
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    That's the derivation of it

    • one year ago
  16. kkid Group Title
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    ah, so its the same as my equation?

    • one year ago
  17. henpen Group Title
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    You're given the half life, so I'd convert it into 2^... instead of e^....

    • one year ago
  18. henpen Group Title
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    Yes

    • one year ago
  19. kkid Group Title
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    I don't know how to get the values of N and N0 though, because that is the number of atoms, not the mass

    • one year ago
  20. kkid Group Title
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    hendce why I haven't already used that equation

    • one year ago
  21. henpen Group Title
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    \[N=N_0 (2^{\log_2(e)})^{-\lambda t}=N_02^{-\log_2(e) \lambda t}\]

    • one year ago
  22. henpen Group Title
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    N_0= 5μg Just call it '5'- It doesn't matter which units you use (I suppose moles or no. of atoms would be most natural, though) as long as you're consistent

    • one year ago
  23. kkid Group Title
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    but N refers to atoms, not mass (which is the 5 and the 1)

    • one year ago
  24. henpen Group Title
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    So obviously the half-life \[= \log_2(e) \lambda\]

    • one year ago
  25. henpen Group Title
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    Which element is it?

    • one year ago
  26. kkid Group Title
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    Na (Sodium)

    • one year ago
  27. kkid Group Title
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    More specifically Sodium-24

    • one year ago
  28. henpen Group Title
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    \[Moles= \frac{mass_{grams}}{A_r}\] \[Atoms= Moles* Avogadro's-number\]

    • one year ago
  29. henpen Group Title
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    A.N.= 6.0221415 *10^23

    • one year ago
  30. kkid Group Title
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    I thought about that, but in my questions, avogadros number isn't mentioned until the next part (therefore, due to my experience with exam paper questions, it isn't needed until then)

    • one year ago
  31. henpen Group Title
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    So initially the mass (N_0)\[=\frac{5*10^{-6}}{24}6.0221415 *10^{23}\]

    • one year ago
  32. henpen Group Title
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    I'm not sure if there's any other (fundamental- i.e. not derived from it) way of working it out. Is this Homework or an exam paper?

    • one year ago
  33. kkid Group Title
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    both, set an exam question for HW

    • one year ago
  34. kkid Group Title
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    I'll try working it out with the arogarrdo number (or whatever it is called lol)

    • one year ago
  35. henpen Group Title
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    Anyway, \[\huge N=125461281250000000*2^{-\frac{t_{hours}}{234}}\]

    • one year ago
  36. henpen Group Title
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    Sorry, you're right, you don't need AN

    • one year ago
  37. henpen Group Title
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    You just need\[\large 2^{-\frac{t}{234}}=\frac{1}{5}\]

    • one year ago
  38. henpen Group Title
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    \[t=-234(\log_2(0.2))\]

    • one year ago
  39. kkid Group Title
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    I got t = 1955669

    • one year ago
  40. kkid Group Title
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    :/

    • one year ago
  41. henpen Group Title
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    =543.33 hours

    • one year ago
  42. kkid Group Title
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    is my answer right?

    • one year ago
  43. kkid Group Title
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    Its in seconds

    • one year ago
  44. henpen Group Title
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    I got 1955992.2

    • one year ago
  45. kkid Group Title
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    close enough (rounding errors)

    • one year ago
  46. henpen Group Title
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    So yes. Did you do it using the equation 8 posts ago?

    • one year ago
  47. henpen Group Title
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    *9

    • one year ago
  48. kkid Group Title
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    no, I did this...

    • one year ago
  49. kkid Group Title
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    |dw:1351694706420:dw|

    • one year ago
  50. kkid Group Title
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    sorry about my terrible handwriting

    • one year ago
  51. henpen Group Title
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    I suppose that's possible, but it's not necessary to use Avoardo's number. All you need is \[N=N_0 2^{-t/k}\] and\[N=N_0 \frac{1}{5}\]

    • one year ago
  52. henpen Group Title
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    k=234

    • one year ago
  53. kkid Group Title
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    ah, i think I see what you have done now

    • one year ago
  54. kkid Group Title
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    the N0 will cancel

    • one year ago
  55. kkid Group Title
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    where has half-life came into it?

    • one year ago
  56. kkid Group Title
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    (it seems a way better way to use it)

    • one year ago
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