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Half life

  1. Aug 10, 2007 #1
    If the rate of an isotope in 18 days has dropped by one eight of its initial value. what is the half life of the isotope?

    My answer
    In 18 days the rate has dropped by 1/8 so therefore the half life is 4/8 which is 18*4 = 72days
     
  2. jcsd
  3. Aug 10, 2007 #2

    Dick

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    That's wrong. What's the formula for the rate as a function of initial rate, elapsed time and half life?
     
  4. Aug 10, 2007 #3

    mgb_phys

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    Or even if you haven't studied the formula.
    After one half-life what is the rate, after two, after three ....?
     
  5. Aug 10, 2007 #4
    T1/2 = ln(2)/lambda
    Thats the formula i think will work.
     
  6. Aug 10, 2007 #5
    1/2*1/2*1/2 = 1/8
    so there will be three half lives after 18 days. So therefore one half life is 6 days?????
     
  7. Aug 10, 2007 #6

    Dick

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    Here's a better one:

    R(t)=R(0)*(2^(-t/th)). Where R(t) is the rate at time t, and th is the half life time.
     
  8. Aug 10, 2007 #7

    Dick

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    Yes, if you mean the final rate is 1/8 of the initial rate. I thought by saying 'dropped by one eighth' you meant that the final rate was 7/8 of the initial rate.
     
  9. Aug 10, 2007 #8
    Yes that is what it means, sorry about the mistake.
    Thank you very much for the clarification.
     
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