# The half - life of a certain radioactive material is 85 days. An initial amount of the material has a mass of 801 kg Write an exponential function that models the decay of this material. Find how much radioactive material remains after 10 days. Round your answer to the nearest thousandth.

Question
Exponential models
The half - life of a certain radioactive material is 85 days. An initial amount of the material has a mass of 801 kg Write an exponential function that models the decay of this material. Find how much radioactive material remains after 10 days. Round your answer to the nearest thousandth.

2021-01-03
Given initial mass:
$$\displaystyle{m}_{{0}}={801}$$kg at t=0
Let
m(t)=Mass at time t
The exponential function,
$$\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{{k}{t}}}$$, where k =constant
Also given:
half life=85days
$$\displaystyle\Rightarrow{m}{\left({85}\right)}={\frac{{{m}_{{0}}}}{{{2}}}}$$
Now,
When t=85days:
$$\displaystyle{m}{\left({85}\right)}={m}_{{0}}{e}^{{{k}\times{85}}}$$
$$\displaystyle\Rightarrow{\frac{{{m}_{{0}}}}{{{2}}}}={m}_{{0}}{e}^{{{85}{k}}}$$
$$\displaystyle\Rightarrow{e}^{{{85}{k}}}={\frac{{{1}}}{{{2}}}}$$
$$\displaystyle\Rightarrow{e}^{{k}}={\left({\frac{{{1}}}{{{2}}}}\right)}^{{{\frac{{{1}}}{{{85}}}}}}$$
$$\displaystyle\Rightarrow{k}={{\ln{{\left({\frac{{{1}}}{{{2}}}}\right)}}}^{{{\frac{{{1}}}{{{85}}}}}}}$$
$$\displaystyle\Rightarrow{k}=-{\frac{{{1}}}{{{85}}}}{\ln{{2}}}$$
Therefore,
When t=10days:
$$\displaystyle{m}{\left({10}\right)}={801}{e}^{{{k}\times{10}}}$$
$$\displaystyle={801}{\left[{e}^{{{10}\times{\left(-{\frac{{{1}}}{{{85}}}}{\ln{{2}}}\right\rbrace}}}\right]}$$
$$\displaystyle={738.273}$$kg
Hence,
The radioactive material remains after 10 days =738.273kg

### Relevant Questions

The half - life of a certain radioactive material is 85 days. An initial amount of the material has a mass of 801 kg Write an exponential function that models the decay of this material. Find how much radioactive material remains after 10 days. Round your answer to the nearest thousandth.
Scientists are working with a sample of cobalt-56 in their laboratory. They begin with a sample that has 60 mg of cobalt-56, and they measure that after 31 days, the mass of cobalt-56 sample is 45.43 mg. Recall that the differential equation which models exponential decay is $$\frac{dm}{dt}=-km$$ and the solution of that differential equation if $$m(t)=m_0e^{-kt}$$, where $$m_0$$ is the initial mass and k is the relative decay rate.
a) Use the information provided to compute the relative decay rate k. Show your calculation (do not just cit a formula).
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a) Use the information provided to compute the relative decay rate k. Show your calculation (do not just cit a formula).
b) Use the information provided to determine the half-life of cobalt-56. Give your answer in days and round to the second decimal place. Show your calculation (do not just cite a formula).
c) To the nearest day, how many days will it take for the initial sample of 60mg of cobalt-56 to decay to just 10mg of cobalt-56?
d) What will be the rate at which the mass is decaying when the sample has 50mg of cobalt-56? Make sure to indicate the appropriate units and round your answer to three decimal places.
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N(t)=?
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(b) (Logistic Growth) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}$$ where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{l}}}$$ can be estimated from $$\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}$$ for $$\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.$$
(c) Assume that $$\displaystyle{N}{\left({0}\right)}={1}$$ and $$\displaystyle{N}{\left({10}\right)}={1000}.$$ Estimate $$\displaystyle{r}_{{{e}}}$$ and $$\displaystyle{r}_{{{l}}}$$ for both $$\displaystyle{K}={1001}$$ and $$\displaystyle{K}={10000}.$$
(d) Use your answer in (c) to explain the following quote from Stanley (1979): There must be a general tendency for calculated values of $$\displaystyle{\left[{r}\right]}$$ to represent underestimates of exponential rates,because some radiation will have followed distinctly sigmoid paths during the interval evaluated.
(e) Explain why the exponential model is a good approximation to the logistic model when $$\displaystyle\frac{{N}}{{K}}$$ is small compared with 1.