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Radio active decay

  1. Oct 30, 2011 #1
    I need some clarifications on the following concepts:

    1.When an α-particle is released by a radio active sample, it loses a charge of +2. Similarly, when it loses a β particle it gains one positive charge. But there is no change in the number of electrons in the outer orbits of the radio active atoms. It means each atom after releasing an α or β particle becomes either a negative or positive ion. Does it mean that any given radio sample consists of a mixture of neutral atoms and ions? Also does it mean that the sample is in a charged state unlike a non radio active samples which are only neutral? If that is true, what is the physical significance or consequence of such a state?

    2.Also, when a radio active sample disintegrates, it continuously undergoes transformations into new elements due to α and β decay. For example, if a Uranium sample emits two successive α-particles it becomes Th and Ra and so on. Does it mean that at any given time a radio active sample will contain different proportions of almost all the other possible radio active elements?

    I will be thankful to any body who enlightens me over these points.
     
  2. jcsd
  3. Oct 30, 2011 #2
    1. You have the radioactive part correct. Electrons in most materials are very mobile so rapidly move to neutralise charges so the radioactive sample is not very different from any other matter. However the ions do have a range of effects on the chemistry of the sample. All manner of chemical reactions can occur which otherwise would not – look up 'radiation chemistry'
    2. No. An 'old' sample of what was originally a pure element will contain only the elements in its 'decay chain' and in very definite proportions depending on the age. This is the basis of many forms of 'radioactive dating'.
     
  4. Oct 30, 2011 #3
    1. Don't know for certain, but I would have thought the atom that emits an α/β would become ionised for a short time, just as αs eventually capture two electrons and become He atoms. In a longer timeframe, it would attract or repel the requisite electrons to become neutral again. This would be modified if the atom is already part of a larger molecule.

    2. It's a long time since I studied this properly, but I've a recollection there is a mathematical proof that shows that, over a period of time, the proportions of elements within a radioactive decay series converge to a fixed set of numbers that is determined by the respective half lives of the isotopes in question. In those proportions, the rate of decay of each isotope minus the rate of its creation from the isotope that precedes it in the decay chain give the same overall decay rate for each isotope.

    The following is in no way rigorous, but it's the best I can muster these days and I hope gives some insight.

    Suppose we have a simple chain A → B → C, fractions x of A and y of B decay in a small time interval δt, and 1 > y > x > 0. Then, between t and t+δt, a.x Bs are created and b.y of them decay, where a(t) and b(t) are the amounts of A and B respectively at time t. At time t+δt, we then have (approximately)

    a(t+δt) = a(t).(1-x) [1], and
    b(t+δt) = b(t).(1-y) + a(t).x [2]​

    (the above ignore nuclei which undergo both decays during the interval δt).

    Now consider the case when
    a(t) = b(t) (y-x) / x [3]​

    Then

    b(t+δt) = b(t) [1 - y + x (y-x) / x]
    = b(t) (1-x) [4]​

    Therefore (dividing [4] by [1])

    b(t+δt)/a(t+δt) = b(t)/a(t)​

    so the overall proportions the two isotopes remain the same.

    Note that I had to stipulate that y > x, and that a/b goes negative in [3] if y < x. In this case, A has a shorter half-life so that (effectively) all of it eventually decays before the resulting Bs can.

    BTW I haven't done anything like this for ages so please could someone correct me if I've made any glaring mistakes!
     
  5. Oct 30, 2011 #4
    I agree with your point 1. It is the short term ionisation which produces the various chemical effects, even if the atom is in a molecule.

    As to your point two I would refer you to an excellent source
    www.ruf.rice.edu/~ctlee/Chapter3A.pdf
    This starts with the very basics and section 3.3 covers the material you are interested in.

    Hope this helps

    Regards

    Sam
     
  6. Nov 1, 2011 #5
    The two forms of equilibrium are secular equilibrium and transient equilibrium. The equilibrium type depends on relative half life of each isotope.
     
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