Using a graph to calculate the half life of proactinium? How do I do the gradient?

Discussion in 'Physics & Math' started by Cat_with_no_eyes, Nov 6, 2010.

  1. Cat_with_no_eyes

    Cat_with_no_eyes Registered Senior Member

    I had carried out an experiement to determine the half life of proactinium and had plotted the graphs for the decay counts.

    But now I must calculate from the graphs, to find out the half life. There are 2 graphs, one is corrected count rate against time and the other in Corrected count rate (log) against time.

    I have to use a gradient for one of them and I dont know which one?! And how do I work out the other one?! What method do I use?!
  2. AlphaNumeric

    AlphaNumeric Fully ionized

    By the decay counts I presume you mean something like number of clicks a Geiger counter gives off in some unit of time, like over a minute. If that's the case then the general principles are pretty straight forward to go through.

    The number of clicks per unit time is proportional to the rate of decay. In radioactive systems the rate of emission is proportional to the amount of the unstable particles. The amount of radioactive material left after a certain amount of time is an analytically computable quantity, [tex]N(t) = N_{0}e^{-\lambda t} = N_{0}e^{-t/\tau}[/tex]. This is perhaps somewhat circular reasoning in the way I've just outlined but the equation is the important thing. The [tex]\tau[/tex] quantity is related to the half life via [tex]t_{\frac{1}{2}} = \tau \ln 2[/tex].

    The count rate (call it [tex]\gamma[/tex]) will be proportional to the total and thus proportional to [tex]e^{-t/\tau}[/tex], ie [tex]\gamma = Ke^{-t/\tau}[/tex], so [tex]\ln \gamma = -\frac{t}{\tau} + \ln K[/tex]. This is a straight line function of the form [tex]y = mx + c[/tex] where y is the log of the count rate, the constant is [tex]\ln K[/tex] and the gradient is [tex]-\frac{1}{\tau}[/tex] and then you just work out [tex]\tau \ln 2 = t_{\frac{1}{2}}[/tex]. Job done :)

    I don't know why you'd want to do it from the non-log version, it'd just be more hassle.
  3. Billy T

    Billy T Please use Sugar Cane Alcohol Fuel

    I'm not sure of your post, but it seems to me you want the log graph and it slope (perhaps that is what you mean by "gradient") is your measure of half life. The log graph is the only one that will be a line and the half life is a constant.
  4. Cat_with_no_eyes

    Cat_with_no_eyes Registered Senior Member

    Thanks, what i meant by the decay count was actually recording the decay of the proactinium using a GM tube for every 30 seconds untill I reached 300 seconds. I then corrected this to the 10 second count.

    I plotted the results on the graph and I do not know how I actually calculate the results for that graph and the other one in which is used log on the corrected count rate?
  5. Cat_with_no_eyes

    Cat_with_no_eyes Registered Senior Member

    so do I just do a line of best fit on the non log graph? and then what about the log graph?
  6. AlphaNumeric

    AlphaNumeric Fully ionized

    You use the log graph, not the standard one. The log graph should be a straight line, or very close to one, while the other is going to be an exponential decay curve. Its possible to extract the results from the exponential decay graph but its less pleasant, stick with the log graph.

    As I explained previously, the gradient of the log graph is going to be equal to [tex]-\frac{1}{\tau}[/tex] and the half life is equal to [tex]\tau \ln 2[/tex]. Thus draw a straight line of best fit and work out its gradient [tex]m = -\frac{1}{\tau}[/tex]. Then you just work out the value of the expression [tex]-\frac{1}{m}\ln 2[/tex].

    For instance, suppose you draw a line with gradient -5 (it'll always be negative, as I'll explain in a moment in case you don't see why). Then the half life is [tex]-\frac{1}{-5}\ln 2 = \frac{1}{5}\ln 2 = 0.138\ldots[/tex], in units of whatever you work. Your method of counting has been 'activity per 10 seconds' so in such a case my example would be 0.138*10 = 1.38 seconds.

    I hope that helps.

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