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Math Help - Explanation for a series of numbers (radioactive Decay)

  1. #1
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    Explanation for a series of numbers (radioactive Decay)

    Hello Again!
    Thank you to all who helped with my last problem! i sincerely appreciate it!
    I have done all the maths to calculate the radioactive decay of the three substances (a,b & c), but i have been asked by my tutor to describe the results over time (paying particular attention to limits). Can anybody help explain what he might mean by this or perhaps share some opinions?

    thanking you all in advance!
    -corey
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  2. #2
    MHF Contributor chisigma's Avatar
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    That is a typical nuclear beta-decay reaction of the form...

    A \rightarrow B \rightarrow C

    If we indicate with m_{a} (t), m_{b} (t) and m_{c}(t) the mass of the elements as a function of time and t_{a}, t_{b} and t_{c} their 'half life time' and consider that C is 'stable' [i.e. t_{c} = \infty...] , we obtain...

    m_{a} (t) = m_{a} (0) \cdot 2^{-\frac{t}{t_{a}}}

    m_{b} (t) = \{m_{b} (0) + m_{a} (0) (1-2^ {-\frac{t}{t_{a}}}) \} \cdot 2^ {-\frac{t}{t_{b}}}

    m_{c} (t) = m_{c} (0) + \{m_{b} (0) + m_{a} (0) (1-2^ {-\frac{t}{t_{a}}}) \} \cdot (1- 2^ {-\frac{t}{t_{b}}})

    From the diagrams it seems to be...

    m_{a} (0)= 800 g

    m_{b} (0) = 100 g

    m_{c} (0) = 0 g

    The numerical value of t_{a} and t_{b} can be extrapolated from the diagrams...

    A very interesting problem!...

    Kind regards

    \chi \sigma
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  3. #3
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    thanks alot! I think i am missing something however, how does this relate to limits? thats the piece i am struggling with

    (+rep to you btw)
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  4. #4
    MHF Contributor chisigma's Avatar
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    Very easy!... at the time t=0 the global mass is m_{a} (0) + m_{b} (0) + m_{c} (0) = 900 g. At the time t=\infty the global mass is the mass of C alone and for the law of canservation must be m_{c} (\infty) = 900 g...

    Kind regards

    \chi \sigma
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