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Half life

  1. Feb 3, 2006 #1
    (a) If galactic cosmic rays produce 75 gm of [itex] ^{14} C [/itex] per year in Earth’s atmosphere, the [tex]^{14} C [/itex] resides in the atmosphere as 14CO2 and the radioactive decay half life of [itex] ^{14} C [/itex] is 5730 years, calculate the total mass of [itex] ^{14} [/itex]CO2 in the atmosphere.

    ok the half life equation is quite simple... quantity after time, t is
    [tex] q(t) = 2^{-\frac{t}{5760}} [/tex]
    would this form something of a differential equation?
    [tex] \frac{dq}{dt} = 75g/yr - \frac{2^{-\frac{t}{5760}}}{5760} [/tex]
    the equilibrium of this sytem would be when the out rate is equal tothe in rate... from this i can solve for t... which i then sub into the equation for q(t) ?

    Is that good?

    (b) If the cosmic ray production rate suddenly increased to 150 gm of 14C per year which of the following would be true about the total mass of 14CO2 in the atmosphere 11 years after the increase in the cosmic ray production rate (explain your answer):
    (i) The amount of 14CO2 in the atmosphere would have doubled.
    (ii) The amount of 14CO2 in the atmosphere would have increased by (11/5730) x 100%
    (iii) The amount of 14CO2 in the atmosphere would have increased only very slightly.


    i guess i could answer this one if i knew if 1 was a) was done correctly?
     
    Last edited: Feb 4, 2006
  2. jcsd
  3. Feb 3, 2006 #2

    Tide

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    For (a), you want something more like

    [tex]\frac {dq}{dt} = r - \frac {q}{\tau}[/tex]

    where q is the amount of the isotope in the atmosphere, r is its rate of production and [itex]\tau = 5760/ ln(2) yr[/itex].
     
  4. Feb 4, 2006 #3
    so then the equilibrium is when dq/dt = 0? From thsi i can find a t (time) value. Using this in the equation for q(t) (for which i need to solve the diff eq) i can find the quantitiy in the atmosphere? Is that correct?
     
  5. Feb 4, 2006 #4

    Tide

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    That is correct! You can use the steady state value you obtained as the initial condition when the new production rate takes effect.
     
  6. Feb 5, 2006 #5
    a couple of doubts i need to clarify

    how did you get the ln(2) term for tau? Shouldnt it be 2^something?
    Also is dq/dt
    [tex] \frac{dq}{dt} = 75 - \frac{qt \ln(2)}{5760} [/tex]

    however... i am unable to solve the diff eq. I tried using mathematica and it gave me an answer that involves erfi...
    as a result i am unable to find q(t) from the diff eq

    so do i simply use this equation for q(t)?

    [tex] q(t) = 2^{-\frac{t}{5760}} [/tex]
    when the new production rate takes over then the dq/dt equation changes with replacing 75 with 150. But if i cant solve the diff eq then how can i solve for the q(11)?
     
  7. Feb 5, 2006 #6

    Tide

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    The time constant is related to the half-life by

    [tex]\frac {1}{2} = e^{-t_{1/2}/\tau}[/tex]

    Your DE has a factor of t (second term - right side) that does not belong. The solution will involve a simple exponential.
     
  8. Feb 5, 2006 #7
    Solving
    [tex] \frac{dq}{dt} = 75 - \frac{q \ln(2)}{5760} [/tex]

    yields [tex] q(t) = -\frac{5760}{\ln 2} C e^{t \ln2/5760} + \frac{75 * 5760}{\ln 2} [/tex]
    if i put dq/dt =0 get for q = 6.2 x 10^5 kg (a few tonnes, which agrees with my prof's answer range)
    using this value for q(0) in

    [tex] \frac{dq}{dt} = 150 - \frac{q \ln(2)}{5760} [/tex]
    i can solve the constant of integration
    and i can compute q(11)
    all good?
     
  9. Feb 5, 2006 #8

    Tide

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    Looks good, Stunner! Now take it the rest of the way. :)
     
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