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Half life question

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  1. Mar 20, 2017 #1
    1. The problem statement, all variables and given/known data
    A radioactive source emits alpha particles at a constant rate 3.5x10^6 s^-1. The particles are collected for a period of 40 days.
    BY reference to the half life of the source, suggest why it may be assumed that rate of emission of alpha particles remain constant?


    3. The attempt at a solution
    Since the decay constant is very small, ##A_o-A \approx A_o## where ##A_o## is initial activity and ##A## is a decrease in activity over a significant period of time.

    Since Activity remains constant, the rate of emission remains constant.

    I cannot, however, find a way to incorporate the half-life into the answer which the question specifically mentions. I am aware of the relation between decay constant and half-life, but for this situation, I cannot develop a clear logic which involves the half-life of the source. More specifically how does rate of emission remains constant if a half-life is high?

    Secondly, how do I know half life is high in this question?

    Thirdly, what is the significance of the provided period of 40 days?
     
    Last edited: Mar 20, 2017
  2. jcsd
  3. Mar 20, 2017 #2
    Good questions. To which I would add: What is the mass of the source and its atomic weight, or equivalently the number of gram-moles it contains? A flux of 3.5x10^6 particles per second is one thing if it's coming from a 5 kg source, and something else if it's produced by 1 mg.

    Reference https://www.physicsforums.com/threads/half-life-question.908387/

    Is this problem related to an earlier one where more information is provided?

    I'd take the decay equation, expand the exponential and consider what might be a good approximation.

    Edit: Is it possible you're being asked to say something in terms of a general (long) half-life?
     
  4. Mar 20, 2017 #3
    Nothing else is provided. This is the whole question
     
  5. Mar 20, 2017 #4
    Well the answer of this question given was

    either the half life of the source very long
    or decay constant is very small
    or half life >> 40 days
    or decay constant << 0.02 day ^-1
     
  6. Mar 20, 2017 #5
    Btw I get what you are saying,
    For the equation, ##A = λN## to work, I should know the value of N. Earlier, I drew my conclusions by prematurely assuming that N = 6.02*10^23.
     
  7. Mar 20, 2017 #6
    I think I'd approach it this way. Assume the source is reasonably pure and macroscopic--say a few grams. Pick a credible mass number, and estimate the number of alpha emitters in the source. Compare that number to the number of decays, at the specified rate, in one second, an hour, a day, 40 days . . . (Note that whatever the half life, the rate can only go down; so assuming a constant rate gives an upper bound on the actual number of decays in each period.)

    Edit With luck you'll find that the details of your assumptions (mass, mass number, purity) don't matter much.
    Edit:
    Each of these pairs says the same thing twice. The half life is inversely proportional to the decay constant, so if one is large, the other must be small.
     
    Last edited: Mar 20, 2017
  8. Mar 20, 2017 #7
  9. Mar 20, 2017 #8

    haruspex

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    The wording of this question is atrocious. Are you sure you have quoted it exactly? Is it a translation?
    The time for which the particles are collected provides no information in itself. I can only suppose it intends to convey that the collection rate appeared constant for 40 days.
     
  10. Mar 21, 2017 #9
    Okay but lets assume the half life is of the order of 10000 years.

    Now how does this half life constitutes a constant rate of emission?
     
  11. Mar 21, 2017 #10

    jbriggs444

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    Would you agree that it would result in an approximately constant rate of emission over a span of 40 days? Can you calculate how much the rate would have reduced between start and end of that span?
     
  12. Mar 21, 2017 #11
    Let me see,
    ##A = A_oe^{-\frac{ln2}{t_{1/2}}t}##
    ##A = A_o*e^{-\frac{ln2}{10000*365}40}##
    ##A=0.99999A_o##

    Got your point.
     
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