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Using the derivative of the formula of the number of nuclei

  1. Apr 19, 2015 #1
    Hi all,

    I have a question concerning the derivative of the formula of the number of nuclei. I hope I've posted this in the right section, I'm new here :P. Anyway, in the question, the given values are:
    At a certain time t, there is an amount of radioactive Br-82. The activity A is 7.4*1014 Bq, the number of nuclei N is 9.6*1018.
    I'm meant to calculate t1/2, the time it takes before the number of nuclei has halved.


    2. Relevant equations
    N = N0 * (½)t/t1/2
    The derivative of the formula above (given by the coursebook, so it's correct): A = ( ln(2) * N) / t1/2.

    3. The attempt at a solution
    What seemed like a plausible solution to me was to use the derivative to calculate t1/2.
    So: t1/2 = ( ln(2) * N) / A.
    That gave me: t1/2 = ( ln(2) * 9.6*1018 ) / ( 7.4*1014) = 8992 seconds. This derivative always uses t1/2 in seconds.
    This is equivalent to 8992/3600 = 2.50 hours. So t1/2 = 2.50 hours.
    The correct answer, however, is 35.3 hours.

    Can anybody explain what mistake I have made, or why I can't use the derivative in this situation?
    Thank you very much in advance :)
     
  2. jcsd
  3. Apr 19, 2015 #2

    Fredrik

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    When you write ##t_\frac{1}{2}=\frac{N\ln 2}{A}##, you haven't actually solved for ##t_\frac{1}{2}##, since N depends on ##t_\frac{1}{2}##.
     
  4. Apr 19, 2015 #3
    By this, I assume you mean I cannot use the starting values of A and N in that formula? Because in a different question where I non-starting values for A and N, my method did actually work.
     
  5. Apr 19, 2015 #4
     
  6. Apr 19, 2015 #5

    Fredrik

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    I'm going to write T instead of ##t_\frac{1}{2}## to reduce the amount of typing.

    I've had a closer look at the problem now. At first I thought you needed to solve the equation ##A(t)=\frac{N_0\cdot 2^{-t/T}\cdot\ln 2}{T}## for ##T##, but you don't. The problem gives you numerical values for A(t) and N(t) at one specific (but unspecified) time t. At that time, we have
    $$A(t)=\frac{N(t)\ln 2}{T}.$$ So I see nothing wrong with the result
    $$T=\frac{N(t)\ln 2}{A(t)}.$$ I will think about this some more, but right now I see nothing wrong with what you did. The result will be in seconds, because N(t) is given as a number of nuclei, and A(t) is given as a number of nuclei per second.
     
    Last edited: Apr 19, 2015
  7. Apr 19, 2015 #6
    Did you find out yet? :)
     
  8. Apr 19, 2015 #7

    Fredrik

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    I still don't see anything wrong. The problem is saying that there's a real number t such that ##N(t)=9.6\cdot 10^{18}## and ##A(t)=7.4\cdot 10^{14}##. Since
    $$A(t)=-N'(t)=-\frac{d}{dt}\left(N_0\cdot 2^{-t/T}\right) =-N_0\cdot 2^{-t/T}(\ln 2)\left(-\frac{1}{T}\right) =\frac{N_0\cdot 2^{-t/T}\cdot \ln 2}{T} =\frac{N(t)\ln 2}{T},$$ this means that
    $$T=\frac{N(t)\ln 2}{A(t)}=\frac{9.6\cdot 10^{18}\cdot \ln 2}{7.4\cdot 10^{14}} =\frac{9.6\ln 2}{7.4}\cdot 10^4\approx 8992.$$
     
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