anonymous 5 years ago The half-life of cesium-137 is 30 years. Suppose we have a 120 mg sample. (a) Find the mass that remains after t years.

1. anonymous

These systems die off in proportion to themselves. So the rate of decay is given by $\frac{dx}{dt}=-kx$ where k is positive. The negative sign ensures the rate is a decaying one. This is separable, so$\frac{dx}{x}=-k dt \rightarrow \int\limits_{}{}\frac{dx}{x}=\int\limits_{}{}-kdt=\log x =-$

2. anonymous

log x = -kt +c

3. anonymous

k= (ln 2) / 30

4. anonymous

Exponentiate both sides to get,$x(t)=x_0e^{-kt}$

5. anonymous

Now, at time t=0, you have 120mg, so$x(0)=120mg=x_0e^0=x_0$

6. anonymous

x(t) = 120 x e^((ln2)/30)

7. anonymous

x(t) = 120 x e^-((ln2)/30)

8. anonymous

Also, after 30 years, 1/2 the mass remains, so$x(30)=60=120 e^{-30k} \rightarrow \frac{1}{2}=e^{-30k} \rightarrow k=-\frac{1}{30}\log \frac{1}{2}$

9. anonymous

i.e. $k=\frac{\log 2}{30}$So,$x(t)=120e^{-\frac{\log 2}{30}t}$

10. anonymous

Thanks suzi20 and lokisan

11. anonymous

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