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Another question regarding Radioactive

  1. May 20, 2005 #1
    I have my doubt , solution and the question in the attachment that followed.
    Thanks for anybody that spend some time on this question.
     

    Attached Files:

  2. jcsd
  3. May 20, 2005 #2

    learningphysics

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    Homework Helper

    I think you made some careless errors.

    Your answer for a is right.

    I'd like to write out the equations...
    [tex]\frac{-dn}{dt} = 5000(\frac{1}{2})^{t/8}[/tex] where t is in hours

    Just plug in t=1 into the right side and you get your answer for a. 4585

    If we integrate both sides we get

    [tex]n(t) = -\frac{5000}{ln(1/2)}(\frac{1}{2})^{t/8}[/tex] or

    [tex]n(t) = \frac{5000}{ln(2)} (\frac{1}{2})^{t/8} [/tex]

    Now since the time in the equations are hours... there's no need to convert to seconds.

    n(0) = 5000/ln(2) = 7213
    n(1) = (5000/ln(2)) (1/2)^(1/8) = 6615

    So for b) n(0) - n(1) = 598

    c) She comes back 24 hours later. So in the equation we'll need to plug in t=25 (24 hours after t=1) to see how much of the radioactive stuff is left.
    n(25) = (5000/ln(2)) (1/2)^(25/8) = 827

    To get the same treatment she needs to get 598 atoms just like part b). So the total needs to come down from 827 to 827-598=229 after the second treatment is finished

    We solve n(t) = 229, I get t=40 hours.

    So the second treatment starts at t=25 hours, and finished t=40 hours to get 15 hours of treatment.

    Check the work yourself. I might have made some careless errors.
     
    Last edited: May 20, 2005
  4. May 21, 2005 #3
    Thanks for your explaination and comment to me for my unneccasary confusing way of calculation.I think I have understand what the question want and I will try to find out the answer.Thanks again,learningphysics.
     
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