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Derivations of the Decay Constant Equation.

  1. Feb 5, 2008 #1
    I have used this equation in the past but have no knowledge of how the equation is formed. I will probably learn about it next year, but that is a long time to wait for a burning question. :shy:

    Here is the equation I am talking about. I would like be sure that I know what everything stands for before I understand how it is formed.

    [tex]T^{\frac{1}{2}}[/tex] is the Half-Life

    [tex]ln(2)[/tex] I know this is a logarithm, but I am not sure why this is used in particular if anyone could explain why this is used I would be most grateful.

    [tex]\lambda[/tex] is the Decay Constant


    Here the equation is.

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


    Any help on how this equation is constructed would be great, It would be nice to see a step by step method in how you would get there. I'm not sure if there is really much to say on the construction, but even an explantion on how it is constructed would be nice.
     
  2. jcsd
  3. Feb 5, 2008 #2

    Hootenanny

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    From experimental observation we know that the decay rate is proportional to the number of radioactive nuclei (N) in a given sample. That is,

    [tex]\frac{dN}{dt} \propto N \Rightarrow \frac{dN}{dt} = -\lambda N[/tex]

    The negative sign indicates that it is a decay (i.e. decrease) rather than an increasing function. [itex]\lambda[/itex] is the constant of proportionality and is called the decay constant. Now, the differential equation is first order and separable. I wouldn't imagine that you've solved any ODE's before so I'll give a fair amount of detail.

    First we divide by N, assuming of course that [itex]N\neq0[/itex], which is a valid assumption since if N=0, there would be no radioactive nuclei and hence no decay,

    [tex]\frac{1}{N}\frac{dN}{dt} = -\lambda[/tex]

    Now we integrate with respect to t,

    [tex]\int\frac{1}{N}{\frac{dN}{dt}dt = -\lambda\int dt[/tex]

    Now, you can think of the next step as canceling the dt's on the RHS. This isn't rigorously correct, we actually make a change of variable, but it will do for illustrative purposes.

    [tex]\int \frac{dN}{N} = -\lambda\int dt[/tex]

    When we perform the integrals we obtain,

    [tex]\ln(N) = \lambda t + const.[/tex]

    We now exponentiate the equation,

    [tex]N(t) = Ae^{-\lambda t}[/tex]

    Where A is an arbitrary constant. We can fix this constant be saying that at t=0, we have a set number of radioactive nuclei, that is [itex]N(0) = N_0[/itex]. Hence, we obtain the radioactive decay law,

    [tex]N(t) = N_0e^{-\lambda t}[/tex]

    I'll address your additional questions in the next post, I just wanted to post this to make sure that I didn't lose it.
     
  4. Feb 5, 2008 #3

    Kurdt

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    The decay of a radioactive substance is proportional to the number of atoms in the substance.

    [tex] -\frac{dN}{dt} = \lambda N [/tex]

    The solution of the differential equation is:

    [tex] N=N_0 e^{-\lambda t}[/tex]

    The half life is defined to be the time it takes for half the nuclei to decay therefore:

    [tex] \frac{N}{N_0} = \frac{1}{2} = e^{-\lambda T_{1/2}} [/tex]

    Now to get the [itex] \lambda T_{1/2} [/itex] we have to take the inverse exponential function of both sides which is the natural logarithm, [itex] \ln [/tex].

    [tex] \ln \left(\frac{N}{N_0}\right) = -\lambda T_{1/2} [/tex]

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

    thus, [tex] T_{1/2} = \frac{\ln (2)}{\lambda} [/tex]

    The last part comes from a property of logarithms where:

    [tex] \ln \left(\frac{1}{a}\right) = - \ln (a) [/tex]

    EDIT: damn beaten to it! :tongue:
     
  5. Feb 5, 2008 #4

    Hootenanny

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    Now with respect to the half life, this is defined as the time taken for half the radioactive nuclei to decay, that is the time [itex]t = T_{1/2}[/itex] when [itex]N = N_0/2p[/itex]. So, we start from our exponential decay law we derived in the previous post;

    [tex]N(t) = N_0e^{-\lambda t}[/tex]

    And we set [itex]t=T_{1/2}[/itex] and [itex]N\left(T_{1/2}\right) = \frac{1}{2}N_0[/itex],

    [tex]N\left(T_{1/2}\right) := \frac{1}{2}N_0 = N_0e^{-\lambda T_{1/2}}[/tex]

    We divide through by [itex]N_0[/itex] and take the logarithm,

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

    And then solve for [itex]T_{1/2}[/itex],

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

    By the laws of logarithms we can take the "-1" up as an exponent of the logarithm to obtain the result you quoted in your OP,

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

    I hope that answered all your questions.

    Edit: But you beat me to the final result Kurdt :tongue2:
     
    Last edited: Feb 5, 2008
  6. Feb 5, 2008 #5

    Kurdt

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    Good to have a couple of different perspectives though.
     
  7. Feb 5, 2008 #6

    Hootenanny

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    I agree, especially when I can't seem to get LaTeX working today :grumpy:
     
  8. Feb 5, 2008 #7
    Thanks you two, I've read over that a few times and it is all slowly but surely slipping in, if I marked this thread as solved would it eventually be deleted as I may want to come back to it later for further reference.
     
  9. Feb 5, 2008 #8

    Hootenanny

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    No problem. I don't think that it would be deleted, I imagine it would be moved to the Archives if anything, but I'm not 100% sure so it may be worth a question in the Feedback forum.
     
  10. Feb 5, 2008 #9

    I understand that N and A are proportional but I think of that as [tex]N \propto A[/tex] what do all those other letters symbolise?
     
  11. Feb 5, 2008 #10

    Hootenanny

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    A is the activity, which is the rate of decay of radioactive nuclei. In other words the activity is a measure of how fast the radioactive nuclei decay (the rate of change of N with time). Therefore, we can write,

    [tex]A = -\frac{dN}{dt}[/tex]

    This is the time derivative of the number of radio active nuclei and is the mathematical way to describe the rate of radioactive decay. However, as you correctly said [itex]A\propto N[/itex], therefore we can write,

    [tex]A = -\frac{dN}{dt} \propto N[/tex]

    or adding a constant of proportionality,

    [tex]A = -\frac{dN}{dt} = \lambda N[/tex]

    Which is the same as the expression we started with before.

    Do you follow? I apologise if I seam to be patronising you, but I'm not sure how much mathematics you know.
     
    Last edited: Feb 5, 2008
  12. Feb 5, 2008 #11

    Kurdt

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    Marking as solved doesn't delete the thread. We like to keep them so other people can reference them. It just helps HH's in the homework forums recognise which threads have been dealt with.
     
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