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Half life and percentage remaining

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data
    137Cs has a half life t1/2 = 30.07 years. After 50 years, what fraction of 137Cs will remain?

    2. Relevant equations
    maybe...
    dN/dλ = λN
    λ = ln2/t1/2, half life in seconds
    N = Total mass /(137(1.67x10-17kg))

    3. The attempt at a solution
    i thought maybe i needed to figure out a decay rate at first but i dont have a total mass. unless i just use the mass of the cesium. so my N will simply equal 1/(1.67x10-17)

    but that gives me a huge number. 43 million decays per second.

    so then i thought if i lose 50% in 30 years. then in 20 years, which is 66% the amount of time needed for another half life to pass, then i should lose 66% of the 50% which is only 33%. so in 50 years i lose 33% of the 50% lost in the first 30 years. but then i get lost in my logic and dont know how to proceed.

    edit: if i take .50 - .33 = .17. then add the .17 to the .50 i get .67. this makes sense to me as after two half lives have passed then .75 of the Cs will be left over.
     
    Last edited: Apr 11, 2015
  2. jcsd
  3. Apr 11, 2015 #2
    This equation is incorrect. It should read:

    $$\frac{dN}{dt}=-\lambda N$$

    where
    λ = ln2/t1/2, half life in seconds

    If t1/2 = 30.07 years, from this equation, what is the value of λ?

    If N = N0 at time t = 0, do you know how to solve the above differential equation for the value of N at arbitrary time t (in terms of N0, λ, and t)?

    Chet
     
  4. Apr 11, 2015 #3
    i do not know how to solve that equation.

    but what i did was 50/30.07 = 1.66. and if 2-1 is .5. then 2-1.66 is .68. which leave 32% of the cesium remaining.
     
  5. Apr 12, 2015 #4

    haruspex

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    If you do not know how to solve that equation, and are not expected to be able to solve it, then you must have been given another relevant equation. Maybe one involving e?
     
  6. Apr 12, 2015 #5
    Yes. I agree with haruspex. On what mathematical basis did you choose this expression over the infinite number of other possibilities available?

    Chet
     
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