Calculus question 15 points!!

Polonium-210 is a radioactive element with a half-life of 140 days.

Assume that 10mg of the element are placed in a lead container.

1)How man mg of the material be left after 10 weeks?

2)How long will it take for 70% of the original material to decay?

Just to FYI, i have learned separable Differential Equations, Linear Differential Equations, and homogeneous differential equation.

I know that it's about the separable differential equations, but I don't know how to start...

2 個解答

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  • 最愛解答

    By the fact that the rate of decay is proportional to the no. of radioactive nuclei present, we have:

    dN/dt = - kN

    where k is a constant.

    So we have:

    dN/N = - kdt

    Integrating both sides:

    ln N = - kt + C where C is a constant.

    With the fact that when t = 0, N = N0 (starting no. of radioactive nuclei):

    ln N0 = C

    Hence

    ln N = - kt + ln N0

    N = N0 e-kt

    For the sample in this Q, its half-life is 140 days and so:

    e-140k = 1/2

    - 140 k = - ln 2

    k = (ln 2)/140

    Thus, for (a): t = 70

    N = N0 e-(ln 2)/2 = 0.707 N0

    So mass of material left after 10 weeks is 7.07 mg

    For (b), e-kt = 0.3 since 70% of the original material has decayed.

    - kt = ln 0.3

    t = 243.2 days

    2010-12-30 10:45:10 補充:

    N is the no. of radioactive nuclei and should not be mixed up with the mass.

    However, even the mass is still obeying the equation M = M0e^(-kt) since it is also an exponential decay

    資料來源: 原創答案
  • 10 年前

    RE:翻雷滾天 風卷殘雲

    Since there's 10mg of substance in the container,

    Can I say when t=0, N=10 so that I can find the value of C?

    Because the N0 is kind of confusing to me..

    2011-01-01 09:29:03 補充:

    Thank you so much;)

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