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...
- 翻雷滾天 風卷殘雲Lv 79 年 前最佳解答
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
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
- 9 年 前
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;)