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kkid

  • 2 years ago

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

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  1. kkid
    • 2 years ago
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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

  2. henpen
    • 2 years ago
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    Do you know differential equations or calculus?

  3. kkid
    • 2 years ago
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    Yes, but I don't think this question requires it

  4. kkid
    • 2 years ago
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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?

  5. kkid
    • 2 years ago
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    A sample of*

  6. henpen
    • 2 years ago
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    \[\frac{1}{c}e^{-t \lambda}=ke^{-t \lambda}=N\]

  7. kkid
    • 2 years ago
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    but I want to know how to do it myself

  8. kkid
    • 2 years ago
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    whats c?

  9. henpen
    • 2 years ago
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    Sorry, it deleted half of my answer

  10. kkid
    • 2 years ago
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    lol, ok

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

  12. kkid
    • 2 years ago
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    ...in the past

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

  14. henpen
    • 2 years ago
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    Set c= sample at t=0 That is, \[ c= N_0 \]

  15. henpen
    • 2 years ago
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    That's the derivation of it

  16. kkid
    • 2 years ago
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    ah, so its the same as my equation?

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

  18. henpen
    • 2 years ago
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    Yes

  19. kkid
    • 2 years ago
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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

  20. kkid
    • 2 years ago
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    hendce why I haven't already used that equation

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

  22. henpen
    • 2 years ago
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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

  23. kkid
    • 2 years ago
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    but N refers to atoms, not mass (which is the 5 and the 1)

  24. henpen
    • 2 years ago
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    So obviously the half-life \[= \log_2(e) \lambda\]

  25. henpen
    • 2 years ago
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    Which element is it?

  26. kkid
    • 2 years ago
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    Na (Sodium)

  27. kkid
    • 2 years ago
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    More specifically Sodium-24

  28. henpen
    • 2 years ago
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    \[Moles= \frac{mass_{grams}}{A_r}\] \[Atoms= Moles* Avogadro's-number\]

  29. henpen
    • 2 years ago
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    A.N.= 6.0221415 *10^23

  30. kkid
    • 2 years ago
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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)

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

  32. henpen
    • 2 years ago
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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?

  33. kkid
    • 2 years ago
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    both, set an exam question for HW

  34. kkid
    • 2 years ago
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    I'll try working it out with the arogarrdo number (or whatever it is called lol)

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

  36. henpen
    • 2 years ago
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    Sorry, you're right, you don't need AN

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

  38. henpen
    • 2 years ago
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    \[t=-234(\log_2(0.2))\]

  39. kkid
    • 2 years ago
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    I got t = 1955669

  40. kkid
    • 2 years ago
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    :/

  41. henpen
    • 2 years ago
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    =543.33 hours

  42. kkid
    • 2 years ago
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    is my answer right?

  43. kkid
    • 2 years ago
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    Its in seconds

  44. henpen
    • 2 years ago
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    I got 1955992.2

  45. kkid
    • 2 years ago
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    close enough (rounding errors)

  46. henpen
    • 2 years ago
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    So yes. Did you do it using the equation 8 posts ago?

  47. henpen
    • 2 years ago
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    *9

  48. kkid
    • 2 years ago
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    no, I did this...

  49. kkid
    • 2 years ago
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    |dw:1351694706420:dw|

  50. kkid
    • 2 years ago
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    sorry about my terrible handwriting

  51. henpen
    • 2 years ago
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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\[N=N_0 \frac{1}{5}\]

  52. henpen
    • 2 years ago
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    k=234

  53. kkid
    • 2 years ago
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    ah, i think I see what you have done now

  54. kkid
    • 2 years ago
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    the N0 will cancel

  55. kkid
    • 2 years ago
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    where has half-life came into it?

  56. kkid
    • 2 years ago
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    (it seems a way better way to use it)

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