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anonymous
 5 years ago
The halflife of cesium137 is 30 years. Suppose we have a 120 mg sample.
(a) Find the mass that remains after t years.
anonymous
 5 years ago
The halflife of cesium137 is 30 years. Suppose we have a 120 mg sample. (a) Find the mass that remains after t years.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0These 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\[\frac{dx}{x}=k dt \rightarrow \int\limits_{}{}\frac{dx}{x}=\int\limits_{}{}kdt=\log x =\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exponentiate both sides to get,\[x(t)=x_0e^{kt}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, at time t=0, you have 120mg, so\[x(0)=120mg=x_0e^0=x_0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x(t) = 120 x e^((ln2)/30)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x(t) = 120 x e^((ln2)/30)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Also, after 30 years, 1/2 the mass remains, so\[x(30)=60=120 e^{30k} \rightarrow \frac{1}{2}=e^{30k} \rightarrow k=\frac{1}{30}\log \frac{1}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i.e. \[k=\frac{\log 2}{30}\]So,\[x(t)=120e^{\frac{\log 2}{30}t}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks suzi20 and lokisan
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