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Half life with calculus

  1. Aug 17, 2011 #1
    1. The problem statement, all variables and given/known data
    This is the equation
    y = y0e-kt


    2. Relevant equations

    b) Show using your expression for λ that if at time t1 the amount is y1, then at time t1 + λ it will be y1/2, no matter what t1 is.

    3. The attempt at a solution

    y1=y0ekt1

    that part I got. Then the answer key goes into the next step saying:λ = (−ln2/k)y0ek(t1+λ)

    That is where I got lost. What is the step to go from the y1 equation into the lambda equation? I know that for half life, y1=y0/2 so that makes t=ln2/k . The time it takes for an element to decompose to half its mass is t=ln2/k . How do I make the connection to the very last lambda equation?
     
  2. jcsd
  3. Aug 17, 2011 #2

    ehild

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    The statement "at time t1 the amount is t1" is the same as y1=y0e-kt1. At any time, the amount is y=y0e-kt.
    At time t=t1+λ, the amount is y=y0e-k(t1+λ)=y0e-kt1-kλ. You know the identity ax+y=ax ay, don't you? And λ is the half-life, so y0e-kλ=y0/2.
    Can you proceed from here?


    ehild
     
  4. Aug 17, 2011 #3
    using your method and identity, I got: y=y0e-kt1*e-k[itex]\lambda[/itex]
    so
    y=y1*e-k[itex]\lambda[/itex]

    and then I am stuck again. lambda = half life, but what does that mean for me to get to y0e-kλ=y0/2.
     
  5. Aug 17, 2011 #4

    ehild

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    y0e-kλ=y0/2, that is e-kλ=1/2.

    ehild
     
  6. Aug 17, 2011 #5
    I still do not see it. is lambda in this equation being treated as a variable or constant? is lambda the same as time?
     
  7. Aug 17, 2011 #6

    SammyS

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    λ is the time it takes for a sample of an element to decompose so that the mass of the original element in the resulting sample is half of the original mass. Thus, λ, must be the half-life.
     
  8. Aug 18, 2011 #7
    Ok I think I got it. This is my understanding, tell me if I am correct.

    e-k[itex]\lambda[/itex] is the multiplier it takes to make an element exactly 1/2 of itself therefore we can set e-k[itex]\lambda[/itex]=1/2
    So when I come to y=y0e-kt1*e-kλ That is the same thing as y=y0e-kt1*(1/2) and knowing that y0e-kt1=y1 we can say y=y1*(1/2)

    am I right?
     
  9. Aug 18, 2011 #8

    ehild

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    Yes, You are right, well done!

    ehild
     
  10. Aug 18, 2011 #9
    THank you all those who helped.
     
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