Question

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.

Answer 1

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\[\frac{dx}{x}=-k dt \rightarrow \int\limits_{}{}\frac{dx}{x}=\int\limits_{}{}-kdt=\log x =-\]

Answer 2

log x = -kt +c

Answer 3

k= (ln 2) / 30

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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Answer 11

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