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Radioactivity and decay constant

  1. May 26, 2009 #1
    A piece of radioactive substance gives a received count rate of 6000 counts per minute in a detector whose efficiency is known to be 5%. If the sample contains 10^10 atoms, what is the decay constant ( λ ) of this radioactive substance ?

    No idea how to solve this problem. Any help would be really appreciated.

    Edit: I am a medical student studying abroad, and the physics teacher doesn't speak English properly. Please believe that I have tried to understand, but after 1 day of looking for information on the internet, I still have no idea how to solve this problem.
     
    Last edited: May 26, 2009
  2. jcsd
  3. May 26, 2009 #2
    Hi there,

    To answer your question, there are many assumptions to be made:
    1. I assume 6000 counts/minute in the first minute.
    2. [tex]10^{10}[/tex] atoms is supposed to be be the initial excited/radioactive atoms in the sample
    3. 6000 counts/minute assumes that all radiation emitted passes through the detector, with a 5% efficiency.

    If all these assumptions are true, then you can simply apply the radioactive decay equation to your problem and the answer is solved in two lines.
     
  4. May 26, 2009 #3
    Here are my calculations :

    N(t)=No*exp(-λt)

    10^10 - 6000 = (10^10) * exp (-λ*60)

    ( 10^10 -6000) / (10^10) = exp (-λ*60)

    ln ( ( 10^10 -6000) / (10^10) ) = -λ*60

    λ= ( - ln ( ( 10^10 -6000) / (10^10) ) ) / 60

    λ = 10*(-8) per second
     
    Last edited: May 26, 2009
  5. May 26, 2009 #4
    The answer is supposed to be 2*10^(-7) though :(

    I guess it has something to do with the detector's efficiency ?
     
    Last edited: May 26, 2009
  6. May 26, 2009 #5
    Hi there,

    You are forgetting the efficiency of the detector in your calculations. Don't forget that only 5% of the particles are detected.
     
  7. May 26, 2009 #6
    "only 5% of the particles are detected" This sentence suddenly made everything clear to me. I guess I couldn't understand the meaning of "efficiency". Anyway, I replaced 6000 by 120 000 (100%) in my calculations, and I find the correct answer.

    Thank you very much :)
     
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