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Radioactive decay, need to find half life

  1. Mar 31, 2015 #1
    1. The problem statement, all variables and given/known data

    1. The number of radioactive nuclei in a particular sample decreases over a period of 26 days to one-fifteenth of the original number. What is the half-life of the radioactive material, in days?

    2. Relevant equations
    R=DN/Dt=(lambda)(N)=(N)(ln2/T1/2)

    3. The attempt at a solution
    DN= N - (1/15)N
    DT=26 days
    Plugged these in and solved for T(1/2), but CAPA says my answer is wrong.
     
  2. jcsd
  3. Mar 31, 2015 #2

    SammyS

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    What did you get for half-life ?
     
  4. Mar 31, 2015 #3
    19.3 days
     
  5. Mar 31, 2015 #4

    SammyS

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    Please explain what your variables represent.

    Where did you get this equation?
    R=DN/Dt=(lambda)(N)=(N)(ln2/T1/2)
     
  6. Mar 31, 2015 #5
    R=rate of decay
    DN= change in number of nuclei
    DN=change in time
    lambda=decay constant
    N=number of nuclei
    T1/2= half life
    I got the equation from a slide my prof has for our lectures.
     
  7. Apr 1, 2015 #6

    SammyS

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    Starting with the following (which is equivalent to ##\displaystyle\ N=N_0\,e^{-\ln(2)\,t/t_{1/2}}## )
    ##\displaystyle\ N=N_0\, e^{-\lambda t}\ ##, where N0 is the number of nuclei at time, t = 0 , (the start),​
    and taking the derivative w.r.t. time, we get:
    ##\displaystyle\ \frac{dN}{dt}=-\lambda N_0e^{-\lambda t}\ =-\lambda N\ .##
    Also, ##\displaystyle\ -\lambda N=-\ln(2)/t_{1/2}\ .##​
    This is the decay rate, similar to yours, except for a sign. However, it's the instantaneous decay rate, not the same as ΔN/Δt .

    By the way, 19.3 days is way off for the half-life. At that rate it would take 38.6 days (2 half-lives) to get to 1/4 the original number of nuclei.

    Getting to 1/16 in 28 days would give a half-life of 7 days.
     
  8. Apr 1, 2015 #7
    Okay, so I'm still not really sure how to solve it..plugging in the numbers just keeps on giving me 19.3..
     
  9. Apr 1, 2015 #8

    SammyS

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    Solve ## \displaystyle\ N=N_0\, e^{-\lambda t}\ ## for λ , if t = 26, and N = N0/15 .

    Then use your equation to find the corresponding half-life.
     
  10. Apr 1, 2015 #9

    BvU

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    Dear PM,

    In physics it's always good to make an estimate.
    You know that after one half-life time, the activity (hence also the number of active nuclei) drops to one half the original number.
    After two to one fourth
    After three to one eighth
    And after four to one sixteenth. So your answer should be close to 26 days / 4 = 6.5 days, and a bit lower.

    From the above: Activity = $$A(t) = A(0) / 2^{\;(t/\tau_{1/2})}$$ or$$A(0)/A(t) = 2^{\;(t/\tau_{1/2})}$$ so what you want to solve is ## 2^{\;(26 {\rm \; days}/\tau_{1/2})} = 15## . That sound good ?
     
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