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Tricky mean life question (potassium)

  1. Feb 7, 2005 #1
    Natural potassiu, has an atomic weight of 39.089 and contains 0.0118 atomic percent of the isotope 19 K 40, which has 2 decay modes:

    19 K 40 -> 20 Ca 40 + beta particle + neutrino (nu bar)

    19 K 40 + e- -> 18 Ar 40 * + beta particle + neutrino (nu no bar)

    where 18 Ar 40 * means an excited state of 18 Ar 40. In this case, this excited state decays to the fround state by emitting a single gamma ray. The total intensity of beta particles emitted is 2.7*10^4 kg^-1 . s^-1 of natural potassium and on average there are 12 gamma rays emitted to every 100 beta particles emitted. Estimate the mean life of 19 K 40.

    If someone can set up the question for me, I will take it from there. I am very confused right :cry: :confused: now as to how the intensity can be used to compute the lifetime. Can someone please explain this using equations?
    James
     
  2. jcsd
  3. Feb 7, 2005 #2
    Please guys..any equations you can write to help me would be greatly appreciated.

    James
     
  4. Feb 8, 2005 #3
    Come on guys...someone msut be able to help me...please :cry:

    JAmes
     
  5. Feb 8, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I'm no expert on this but you are told how many, on average, beta particles are emitted per second. Since each Argon atom has to emit a beta particle when it disintegrates, that gives you the the average number of Argon atoms that disintegrate per second (atoms/second) which should be the reciprocal of the average lifetime (seconds/atom).

    Any suggestions, Doc Al?
     
  6. Feb 8, 2005 #5
    I agree with HallsofIvy. Problem can be solved with some mathematics. However, since there are two modes of decay, which one do we use to find the mean life? For the first mode of decay, one needs to consider the number of beta particles emitted per second.(mean life is the reciprocal of that as HallsofIvy said)


    regards.
     
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