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Half-Life of a Radionuclide

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data
    The activity (decays per second) of a certain radionuclide is observed to decrease by 87.5% in 30 hours.

    What is the half-life of the radionuclide?


    2. Relevant equations
    (1) A(t) = A[itex]_{0}[/itex]e[itex]^{-λt}[/itex]

    (2) T(1/2) = 0.693/λ


    3. The attempt at a solution
    I'm just stuck on which approach to use, do I need to use the above equations or would I simply be able to cross-multiply and divide like this:

    [itex]\frac{0.875}{30 hrs}[/itex] = [itex]\frac{0.50}{?}[/itex] in which I find the half life to be 17.14 hours.

    With the other approach I used equation 1 and solved for λ using a hypothetical A[itex]_{0}[/itex] value of 100 so that A(t)=12.5 and found λ to be 0.0693 with t=30 hrs. Then I used the second equation to obtain T(1/2)=9.998 hours.

    I'm just not sure which is the correct approach.
     
  2. jcsd
  3. Nov 18, 2013 #2

    gneill

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    Staff: Mentor

    Your first method is assuming a linear relationship between rate of decay and time. This is not correct, since the rate of decay (number of decays per second) depends on the total number of atoms.

    The second approach looks better, but could be improved. Have you studied calculus yet? If so, then you should know that the rate of decay (decays per second) will be given by the derivative w.r.t. time of your equation (1).
     
  4. Nov 18, 2013 #3
    Oh! Thank you! This makes more sense now :)
     
  5. Nov 20, 2013 #4
    Hey so in regards to this how will taking the derivative help you? I also considered both her methods and realized the one assumed a linear relationship. Would you take the derivative and set it = 1/2? and even so you have more than one unknown variable...
     
  6. Nov 20, 2013 #5

    gneill

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    Staff: Mentor

    The derivative gives you the rate of decay, decays per second, otherwise known as the activity. You're given a time span over which the activity changes by a given percentage. Set up a ratio.
     
  7. Nov 21, 2013 #6
    Ok sure, so the derivative of A(t)=A0e(-λt) is

    A'(t)= -λA0e(-λt) correct?

    and you propose to set this equal to the 0.875/30 hours? Such that:
    0.875/30hrs= -λA0e(-λt)
    However you would have three unknowns still? I think im just missing somthing small. But i just don't see how you can solve without (as mentioned above) finding the theoretical value for λ? I'm really not one for hand-holding im just missing somthing fundamental or having a brain fart I think.
     
  8. Nov 21, 2013 #7

    gneill

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    Staff: Mentor

    Okay, 0.875/30hrs makes no sense as a ratio; they're not the same things. One is time and the other a percentage.

    The given percentage tells you how the rates are related at two different times. What are those times? Write the rate expression for both of them.
     
  9. Nov 21, 2013 #8
    Yes you're right that doesn't make sence... So for the rates as you said before i must use the derivative of the function
    d/dt A(t)= -λA0e(-λt)

    So:

    A'(30hr)= -λA0e(-30hrλ)=0.125

    and the other? Is it just
    A'(t0)= -λA0e(-λt0)=1 ?

    You'll have to forgive me if i dont immedietly see this. And if so do we consider t0=0 ?

    Then take the ratio of the equations?
     
  10. Nov 21, 2013 #9

    gneill

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    Staff: Mentor

    ??? Where does 0.125 come from?

    You don't know (yet) what the rate is at t = 30hr. You only know that it's 87.5% of the initial rate (which you also don't know). So no numbers yet! Work with the symbols.

    Yes. Counting from an initial arbitrary instant always begins at zero.

    Yes, because you know what that ratio should work out to be (that's where your percentage comes in).
     
  11. Nov 21, 2013 #10
    hahah oh wow, so you still have to solve for your lambda first. I thought there was going to be some way to directly solve for t1/2 without first solving for lambda haha.


    if thats the case i could of just as easily set up: 0.125A0=A0e(-λt)
    at t=30hrs haha this yields the same result giving λ=6.93x10-2. Although i guess i just used intuition...and taking the ratio of the derivative is fundamentally what i did...wierd haha. But thanks alot for your help!Cheers!

    This question just threw me through a loop becuase they ask for t1/2 initially and then λ as part b)...jeez.
     
  12. Nov 21, 2013 #11
    And just incase anyone else looks at this or needs help the correct equations are:

    A'(30)= -λA0e(-30λ)
    A'(t0)= A'(0) = -λA0e(0)= -λA0

    (Because e0=1 of course)

    And we know that A'(30)/ A'(0) = 0.125


    Thus the ratio of A'(30)/ A'(0) = e(-30λ)= 0.125

    Taking ln of both sides and dividing by -30 yields λ= ln(0.125)/-30= 0.0692 (approx)
    Using this lambda value you can set up your equation again (that is) A(t)= A0e(-λt) just setting A(t)= 1/2A0. Now solve for your new time (half-life time).
     
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