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.

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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.

Mathematics
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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\[\frac{dx}{x}=-k dt \rightarrow \int\limits_{}{}\frac{dx}{x}=\int\limits_{}{}-kdt=\log x =-\]
log x = -kt +c
k= (ln 2) / 30

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Exponentiate both sides to get,\[x(t)=x_0e^{-kt}\]
Now, at time t=0, you have 120mg, so\[x(0)=120mg=x_0e^0=x_0\]
x(t) = 120 x e^((ln2)/30)
x(t) = 120 x e^-((ln2)/30)
Also, after 30 years, 1/2 the mass remains, so\[x(30)=60=120 e^{-30k} \rightarrow \frac{1}{2}=e^{-30k} \rightarrow k=-\frac{1}{30}\log \frac{1}{2}\]
i.e. \[k=\frac{\log 2}{30}\]So,\[x(t)=120e^{-\frac{\log 2}{30}t}\]
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