# 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

• 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
• 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;)