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). 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 init

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). 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 init

Question
Exponential models
asked 2020-11-22
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 \(\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}=-{k}{m}\) and the solution of that differential equation if \(\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{-{k}{t}}}\), where \(\displaystyle{m}_{{0}}\) 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).
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.

Answers (1)

2020-11-23
Since you have posted a question with multiple subparts, we will answer the first three subparts (a,b,c). If you want the remaining subpart to be answered, repost the question and mention the subpart you want us to answer in your message.
Given:
Scientists with a sample that has 60 mg of Cobalt-56, and after 31 days, the mass becomes 45.43mg.
The differential equation which models expontial decay is \(\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}=-{k}{m}\) and the solution or differential equation is \(\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{-{k}{t}}}\), where \(\displaystyle{m}_{{0}}\) is the initial mass and k is the relative decay rate.
Then, \(\displaystyle{m}_{{0}}={60}\)mg and m(31)=45.43 mg
a) Compute the relative decay rate k using the given information as follows.
\(\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{-{k}{t}}}\)
\(\displaystyle{m}{\left({31}\right)}={m}_{{0}}{e}^{{-{k}{\left({31}\right)}}}\)
\(\displaystyle{45.43}={60}{e}^{{-{31}{k}}}\)
Solve for k as shown below.
\(\displaystyle{e}^{{-{31}{k}}}={\frac{{{45.43}}}{{{60}}}}\)
=0.75716667
Take natural logarithm on both sides,
\(\displaystyle{\ln{{\left({e}^{{-{31}{k}}}\right)}}}={\ln{{\left({0.75716667}\right)}}}\)
\(\displaystyle-{31}{k}={\ln{{\left({0.75716667}\right)}}}\)
\(\displaystyle{k}=-{\frac{{{\ln{{\left({0.75716667}\right)}}}}}{{{31}}}}\)
\(\displaystyle\approx{0.00897}\) mg\day
Therefore, the relative decay rate is approximately 0.00897 mg\day
b) The half life of Cobalt-56 is the time after which only half of the original material remains.
The original amount taken is 60 mg. Then half of it will be 30 mg.
Now obtain the half-life of Cobalt-56 as shown below.
\(\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{-{k}{t}}}\)
\(\displaystyle{30}={60}{e}^{{-{k}{t}}}\)
\(\displaystyle{30}={60}{e}^{{-{0.00897}{t}}}\)
Solve for t as follows.
\(\displaystyle{e}^{{-{0.00897}{t}}}={\frac{{{30}}}{{{60}}}}\)
\(\displaystyle{\ln{{\left({e}^{{-{0.00897}{t}}}\right)}}}={\ln{{\left({\frac{{{1}}}{{{2}}}}\right)}}}\)
\(\displaystyle-{0.00897}{t}=-{\ln{{\left({2}\right)}}}\)
\(\displaystyle{t}={\frac{{{\ln{{\left({2}\right)}}}}}{{{0.00897}}}}\)
\(\displaystyle\approx{77.27}\) days
Therefore, the half-life of Cobalt-56 is 77.27 days.
c) Obtain the number of days taken for the initial sample of 60 mg of Cobalt-56 to decay to just 10 mg as follows.
\(\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{-{k}{t}}}\)
\(\displaystyle{10}={60}{e}^{{-{k}{t}}}\)
\(\displaystyle{10}={60}{e}^{{-{0.00897}{t}}}\)
Solve for t as follows
\(\displaystyle{e}^{{-{0.00897}{t}}}={\frac{{{10}}}{{{60}}}}\)
\(\displaystyle{\ln{{\left({e}^{{-{0.00897}{t}}}\right)}}}={\ln{{\left({\frac{{{1}}}{{{6}}}}\right)}}}\)
\(\displaystyle-{0.00897}{t}=-{\ln{{\left({6}\right)}}}\)
\(\displaystyle{t}={\frac{{{\ln{{\left({6}\right)}}}}}{{{0.00897}}}}\)
\(\displaystyle={199.750219}\)
\(\displaystyle\approx{200}\) days
0

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