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Radioactive Decay and Linear Graphing

  1. Jan 22, 2009 #1
    1. The problem statement, all variables and given/known data
    This particular exercise has no "problem statement" but I'll explain it in detail using all the information provided to me.

    The class was presented with a table of data containing values for time (in days) and activity (in cts/sec) for a radioactive isotope Iodine-131. We are to create a Linear graph using the data and fine Iodine-131's half life.

    Here's the data, Time in days first - then the activity in cts/sec.
    05 | 6523
    10 | 4191
    15 | 2736
    20 | 1722
    25 | 1114
    30 | 0722
    35 | 0507
    40 | 0315

    2. Relevant equations
    Here are the equations given to us on the table.
    A = Ao[tex]^{-\lambda t}[/tex]
    [tex]\lambda[/tex]=(ln2)/T[tex]_{1/2}[/tex]

    Ao=10,000 cts/sec

    3. The attempt at a solution
    I've fiddled with the equation and wasn't getting anywhere near a y=mx+b solution so I decided to look up first order exponential decay on the web which landed me this equation.

    ln(A) = -[tex]\lambda[/tex]*t + ln(Ao)

    So I made a separate table using the natural log of the Activity data and graphed it which comes out linear with an R2=1, the best fit is: f(x)=-0.09x+9.2

    I used the slope to determine T1/2 or half life of the sample. I calculated a value of 7.7 days - the accepted value is barely over 8 days. I believe that the equation I found for first order exponential decay is in fact correct for this case but I have no idea how to go from A = Ao[tex]^{-\lambda t}[/tex] to ln(A) = -[tex]\lambda[/tex]*t + ln(Ao) algebraically.

    In essence my question is: Is the ln(A) = -[tex]\lambda[/tex]*t + ln(Ao) correct for this particular case of atomic decay and how do I get to this equation from the equation given.

    Thanks alot!
     
  2. jcsd
  3. Jan 23, 2009 #2

    LowlyPion

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    I think the equations are equivalent.

    Recall that Bx = (elnB)x = exlnB

    When you take the ln of both sides of A = Ao-λt

    keeping in mind that B above is your Ao

    Then you get lnA = -λ*t + lnAo
     
  4. Jan 23, 2009 #3
    Thanks a lot but I have a little problem.
    Okay so if:
    A = Aox // x = -λ*t
    A = ex*ln(Ao)

    then

    ln(A) = x*ln(Ao)
    ln(A) = -λ*t*ln(Ao)

    Why is my version -λ*t*ln(Ao) as opposed to the correct version -λ*t+ln(Ao)? Why is it "+ln(Ao)"
     
  5. Jan 23, 2009 #4

    LowlyPion

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    This step is incorrect. Exponents add, not multiply.
     
  6. Jan 23, 2009 #5
    I'm a little confused, which step is incorrect?

    A power raised to a power is multiplication, which is how you get A = ex*ln(Ao) no?

    Since ln() is the inverse of e it shouldn't change the exponent of e correct? ln(A) = x*ln(Ao)
     
  7. Jan 23, 2009 #6

    LowlyPion

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    I think you should familiarize yourself with ln arithmetic.

    eln(x) = x

    ln(ex) = x

    ln(ea*b) = a + b
     
  8. Jan 23, 2009 #7
    If I substitute numbers in for a and b using my calculator I don't get a true statement.
    a = 2
    b = 5

    ln(e^(a*b)) = a + b
    ln(e^(10)) = 7
    10 = 7

    What am I doing wrong? :(
     
  9. Jan 23, 2009 #8

    LowlyPion

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    Don't despair. Maybe I'm not remembering my ln math.

    Sorry if I confused you at all. I'll have to look it up.
     
  10. Jan 23, 2009 #9

    LowlyPion

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    OK. I don't know what I was thinking, because of course it's equal to the product.

    I think however that your original equation is

    A = Ao*e-λt

    Taking the ln of both sides:

    ln(A) =ln(Ao) -λt

    That makes sense. My fault. I should have recognized it immediately.
     
  11. Jan 23, 2009 #10
    That certainly makes more sense than before! The worksheet must have a typo on the equation, I double checked and the handout and the -λt is the exponent of Ao and Euler's number is nowhere to be found.

    Thanks alot.
     
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