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Rate of decay

  1. Dec 19, 2004 #1
    Rate of decay....

    "The rate of decay is durectly proprtional to the number of atoms present, following an exponential law, the rate of decay decreasing with time" - but why is this the case?

    Thanks in advance. :smile:
  2. jcsd
  3. Dec 19, 2004 #2


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    As you said, the rate is proportional to the number present. This decreases with time (by decaying!). The usual measure of the decay rate is half-life, i.e. the time for half the atoms to have decayed.
  4. Dec 21, 2004 #3
    But what does the statement actually mean in terms of physics?
  5. Dec 21, 2004 #4


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    What part of the original statement are you having trouble with?
  6. Dec 21, 2004 #5


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    Basically in nature, it has been observed that radionuclides decay primarily by beta-emission or alpha-emission, and sometimes by gamma-emission. The process is very random, in the sense that one cannot predict precisely when a given unstable atom will decay. Instead, one can take a population of the particular atom and observe that decays do occur according to a very simple first order differential equation.

    dN/dt = -[tex]\lambda[/tex]N, where N is the number of particles at any given time (e.g. N= N(t)) and [tex]\lambda[/tex] is the decay constant, which is unique to that nuclide.

    The decay constant [tex]\lambda[/tex] = (ln 2)/t1/2, where t1/2 is the half-life, which is the period after which approximately one-half the radioactive atoms have decayed.

    Here are some useful references:

    Radioactive decay - http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli.html#c1

    Half-life - http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli2.html#c1

    Radioactivity - http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radact.html#c1



    Please read these and if you still have questions, we will address them.
  7. Dec 25, 2004 #6
    But why is the first equation you mention ( dN= -lambdaNdT) true? ie - Why does the rate of decay depend on the number of nuclei present?

    Thanks. :smile:
  8. Dec 25, 2004 #7


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    Whether or not any particle decays is independent of what other particles are doing. Therefore the decay rate (per unit time) is directly proportional to the number of particles still around.
  9. Dec 26, 2004 #8
    But why is the rate of decay directly proportional to the number of nuclei present? What does the number of nuclei have to do with the rate if the decay of any particular nucleus is completely random?

    Thanks. :smile:
  10. Dec 26, 2004 #9


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    It is a statistical fact. Specifically, the probability that any particle decays during a very small unit of time is fixed. Call it p. Then on average, np particles will decay (where n is the number of particles at that time) during this small interval of time. Taking the limit as the interval of time goes to 0, we end up with the simple differential equation as shown above (Astronuc).
  11. Dec 26, 2004 #10
    Look, I'll try and give it my best shot.
    radioactive decay is a quantummechanical effect, Fermi's Golden Rule (which was found by Dirac, really) is an expression that gives the probability for a transition from undecayed to decayed nucleus. Apparently the probability for decay per unit time is independant of the time, a radioactive nucleus is equally likely to decay in say 5 minutes now than it is likely to decay in 15 minutes when it hasn't decayed 10 minutes from now.
    When you have 2 kazillion nuclei, we can expect 50% (1 kazillion) of them to have decayed when one halflife has expired, if you have 4 kazillion nuclei, virtually split that up in 2x2 kazillion nuclei. Both sets of 2 kazillion will have 50% (1 kazillion) decayed nuclei, yielding 2 kazillion decayed nuclei in total, which is double that of when you started out with 2 kazillion nuclei, ater the same time of 1 halflife. Hence the direct proportionality.
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