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Calculating half life from experimental data

  1. Jun 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Decay of Carbon-11 from given data of count rate at time 0 to count rate at time 14 minutes in 2 minute intervals.
    i6TTRgV.png
    2. Relevant equations
    I feel like I've gone wrong here because the answer I get of 19.6 minutes is actually negative if I go back and do the last equation again. I think I need to be using the gradient of the graph rather than the gradient I found from two points for my calculations; it ends up giving a positive result when subbed into the last equation but turns up completely wrong compared to secondary data. Can anyone see where I went wrong?

    3. The attempt at a solution
    3SQVmp9.png
     
  2. jcsd
  3. Jun 3, 2015 #2

    ehild

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    You have to use the gradient of the line fitted to all the measured points instead of just using two pairs of measurement. If you use all measured data, that decreases the error.
    You got negative result as you calculated ΔN on the wrong way: It is N(final)-N(initial).
    The other cause of the error is that the measurement time is too short. You should follow the count rate at least till the expected half-life.
    Anyway, about 4% deviation of your result from the real half life is not too big!
     
  4. Jun 3, 2015 #3
    So use -0.0375 as the decay constant in becquerels (represented by lambda). This ends up giving a result of 8.21 minutes. That's a very long way off the actual half life?
     
  5. Jun 3, 2015 #4

    ehild

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    How did you get that value? You measured the time t in minutes, so 0.0375 is in 1/minutes. The fitted line is ln(N)=m t + ln(No). The reciprocal of m is the time constant. You have to multiply it by ln2 to get the half life.
     
  6. Jun 4, 2015 #5
    Can you explain that in a bit more detail please? I don't really understand why I'm supposed to use the reciprocal of the gradient.

    t = [1/ -0.0357] * ln(2) ??? That also gives a wrong answer

    I don't understand :/
     
  7. Jun 4, 2015 #5

    ehild

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    The count number is proportional to the number of the radioactive atoms present, which is an exponentially decreasing function of time: N=No e-t/τ. τ is the time constant.
    If you take the logarithm of the equation, it becomes ln(N)=ln(No)-(1/τ) t. It is a straight line as function of t, and the gradient is m=-1/τ.
    The gradient of your line is m = -0.0375 min-1. The time constant is the negative reciprocal of that.
    The half life time is ln(2) times the time constant.
     
  8. Jun 4, 2015 #6
    Understood.

    That gives a correct answer which is good but...

    Not only do I want to understand what is going on here for my own benefit but I need to be able to understand what is happening here so I can reproduce it in the form of working. Sorry for my ignorance regarding logarithms; we haven't really done much with them in maths b or c yet. We only briefly went over them when we were doing this unit in class.

    Are you saying the formula I used in the picture above is wrong? Because my understanding was that the formula that I'm using is just
    Nt= No e-λ*t rearranged. It actually appears that we have different formulas? Mine came directly out of my class's notes which were written by the teacher but don't seem to provide the right answer... Nor does the first equation you mentioned?
     
  9. Jun 4, 2015 #7

    ehild

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    Nt= No e-λ*t is equivalent to Nt= No e-t/τ if λ=1/τ.
    τ is the time needed to decrease the count number to 1/e times of the initial value: If t=τ, t/τ = 1 and N=No e-1=No/e.
    You can not rearrange the exponential formula to get a linear function of time. You have to take the logarithm: it is log(N)=log(No)-λt.
    See the page about logarithm. https://www.mathsisfun.com/algebra/exponents-logarithms.html
    The half life is the time needed to decrease the count rate to half the initial value. No/2=No e-λt, that is 2=eλt. Taking the logarithm, ln2=λt, that is the half life is t1/2 =ln(2)(1/λ).
     
  10. Jun 4, 2015 #8
    So full working would be:

    Just wait; No/2 = No e-λt
    2/2= 2 * e-λt
    2 = 4 * e-λt (did u just miss the negative lambda exponent or am i missing something?)


    "ln2=λt"
    shouldn't it be this?
    ln(2) = ln(-λt)


    What is the equation I need to start with? Maybe from there I could get a better idea of what I'm supposed to be doing. Is it No/2 = No e-λt
    where No is 2?
     
  11. Jun 4, 2015 #9

    ehild

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    No. You can simplify the first equation with No. That gives 1/2 = e-λt. Taking the logarithm, ln(1/2) = -λt, but ln(1/2)= - ln(2), so ln2=λt.

    No is not 2. It is half the initial count number, so No/2.

    Start with the general equation N = No e-λt.

    If you need the half-time, substitute half of the initial count number for N.
     
  12. Jun 4, 2015 #10
    Here's what I have now:

    N=No e^-λt
    N/2=No e^-λt
    208.5= 417* e^0.0357t
    0.5= e^0.0357t
    ln(0.5)= ln(0.0357t)
    ln(0.5)= t * ln(0.0357)
    t= 0.208???

    :/
     
  13. Jun 4, 2015 #11

    ehild

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    You miss the minus sign in the exponent.
    ##208.5 = 417 e ^{-0.0357 t} ##
    Proceed.
     
  14. Jun 4, 2015 #12
    Oh yeah whoops, just a typo. How do I get the natural log of a negative number??? My calculator returns "nonreal answer"
     
  15. Jun 4, 2015 #13

    haruspex

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    Why are you trying to take the log of a negative number? Please post the working that leads to this.

    Edit: it's this working I guess:
    That step is wrong. You've taken the log of the right hand side twice over.
     
  16. Jun 5, 2015 #14
    N=No e^-λt
    N/2=No e^-λt
    208.5= 417* e^-0.0357t
    0.5= e^-0.0357t
    0.5= ln(-0.0357t)
    0.5= t * -ln(0.0357)
    0.15=t???
     
  17. Jun 5, 2015 #15

    haruspex

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    That's even worse than the previous mistake at this step!
    You are supposed to be taking the ln() of each side of the first of those two equations.
    What is the ln() of the left hand side?
    What is the ln() of the right hand side?
     
  18. Jun 5, 2015 #16
    N=No e^-λt
    N/2=No e^-λt
    208.5= 417* e^-0.0357t
    0.5= e^-0.0357t
    ln(0.5)= -0.0357t
    t= 19.416!!!!!!!!!!!!!!!!!!!

    YES!!!

    Thank you very much haruspex and ehild :)

    I thought when taking the natural log you had to do it to both sides.
     
  19. Jun 6, 2015 #17

    haruspex

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    You do. Any operation you do to an equation you should do to both sides. But ##\ln(e^x)=x##, not ##\ln(x)##.
     
  20. Jun 6, 2015 #18
    Ah right. I still did it right though didn't I?
     
  21. Jun 6, 2015 #19

    haruspex

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    Yes.
     
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