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Derivation for this equation: N = N0(1/2)^t/t1/2?

  1. May 11, 2008 #1
    Anyone know the derivation for this equation: N = N0(1/2)^t/t1/2? I can understand it plugging numbers in but I don't really know how it was derived in the first place.

    Thanks!
     
  2. jcsd
  3. May 11, 2008 #2

    malty

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    Well, we know that the rate of radioactive decay is proportional to the Number of particles at a time t.
    So:

    [tex] -\frac{dN_{(t)}}{dt}=\lambda N_{(t)} [/tex]

    Now can you derive it?
     
  4. May 11, 2008 #3
    Using what you gave, I am able to derive N = N0e^-kt but the equation I provided is without the decay constant...
     
  5. May 11, 2008 #4

    malty

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    Well the half life will be when [tex]N_{(t)}=\frac{N_0}{2}[/tex]. and when [tex]t=T_{\frac{1}{2}}}[/tex].

    /Not sure how the decay constant disappears tbh.

    /No wait I see now.
     
    Last edited: May 11, 2008
  6. May 11, 2008 #5
    Can you explain it a bit more clearly? Thanks.
     
  7. May 11, 2008 #6

    malty

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    Sure no problem.

    Half life is the when the number of particles is reduced by half.
    Hence this occurs when [tex] N_{(t)}=\frac{N_o}{2})[/tex] No is the original number of particles.
    The time which this occurs will be the half life and we call it [tex] T_{\frac{1}{2}}[/tex]

    So we have:
    [tex]\frac{N_o}{2}=N_o e^{-\lambda T_{\frac{1}{2}}}[/tex]
    [tex]\frac{1}{2}=e^{-\lambda T_{\frac{1}{2}} }[/tex]

    We already had:
    [tex] N_{(t)}=\frac{N_o}{2})[/tex]
    but
    [tex] e^{-\lambda T_{\frac{1}{2}}}=\frac{1}{2}[/tex]

    so [tex] -\lambda T_{\frac{1}{2}}= L_n(\frac{1}{2}}) [/tex]

    [tex] e^{-\lambda}=\frac{e^{L_n(\frac{1}{2})}}{e^{T_{\frac{1}{2}}}}= \frac{1}{2}e^{-T_{\frac{1}{2}}}[/tex]

    Sub that into the equation [tex]N = N_0e^{-\lambda t} [/tex]
    and you got it :D

    *Phew that took some time to type*

    :)
     
    Last edited: May 11, 2008
  8. May 11, 2008 #7
    Thanks for the reply but I'm not entirely sure how you got the above lines. My exponent rules might be a big hazy but I don't think e^a/b is the same as e^a/e^b if you know what I'm saying.
     
  9. May 11, 2008 #8

    malty

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    Not 100% sure what you are refering to.

    Here's how I got that line:

    [tex] -\lambda T_{\frac{1}{2}}= L_n(\frac{1}{2}}) [/tex]
    and
    [tex] {-\lambda}=\frac{L_n(\frac{1}{2})}{{T_{\frac{1}{2}}}}[/tex]
     
  10. May 11, 2008 #9
    Right, so does e^(ln(1/2)/T1/2) = e^ln(1/2)/e^T1/2?
     
  11. May 11, 2008 #10

    malty

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    Well yes, if

    [tex] e^{ab}=e^a*e^b [/tex]

    then [tex] e^{-ab}=e^a*e^{-b}=\frac{e^a}{e^b}[/tex]
     
  12. May 11, 2008 #11
    Wait, doesn't e^a*e^b = e^(a+b) not e^ab? e^ab would be (e^a)^b, right?
     
  13. May 11, 2008 #12

    malty

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    Lol yeah your right is does, I have absolutely no idea what made me think it was well that...

    :redface:
     
    Last edited: May 11, 2008
  14. May 11, 2008 #13

    malty

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    Ahh now I remember the equation I used was

    [tex] {-\lambda}=\frac{L_n(\frac{1}{2})}{{T_{\frac{1}{2}}} }[/tex]

    I took the exponent of this:




    [tex] e^{-\lambda}=\frac{e^{L_n(\frac{1}{2})}}{e^{T_{\frac{1 }{2}}}}= \frac{1}{2}e^{-T_{\frac{1}{2}}}[/tex]

    And that is still correct, I believe, my whole escapade with the awful exponent rules was well becasue...I'm tired (and dreaming) :p

    :smile:
     
  15. May 11, 2008 #14
    Hmm I don't know if I'm just tired too or what but we seem to be back at square one. Didn't we conclude that e^(ln(1/2)/T1/2) does not equal e^ln(1/2)/e^T1/2?
     
  16. May 11, 2008 #15

    malty

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    No this is actually true because

    [tex]=e^{-\lambda}=e^{- \frac{ ln\frac{1}{2}}{T_{\frac{1}{2}}}}=e^{(ln\frac{1}{2})^{-T_{\frac{1}{2}}}}=\frac{1}{2}e^{-T_{\frac{1}{2}}}}[/tex]

    Least I think it, *is brain dead*
     
  17. May 11, 2008 #16
    Ok, I think I figured it out but I don't think the equation you just wrote is correct. e^(ln1/2)/T1/2 = 1/2^(T1/2^-1). Now, subbing this into our original equation, we get N = N0(1/2)^t/t1/2. Thanks so much for your help anyways, it got my brain going.
     
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