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 aft

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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 ${\displaystyle \frac{dm}{dt}=-km}$ and the solution or differential equation is ${\displaystyle m\left(t\right)=m_0{e}^{-kt}}$, 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.
${\displaystyle m\left(t\right)=m_0{e}^{-kt}}$
${\displaystyle m\left(31\right)=m_0{e}^{-k\left(31\right)}}$
${\displaystyle 45.43=60e^{-31k}}$
Solve for k as shown below.
${\displaystyle e^{-31k}=\frac{45.43}{60}}$
=0.75716667
Take natural logarithm on both sides,
${\displaystyle \ln \left(e^{-31k}\right)=\ln \left(0.75716667\right)}$
${\displaystyle -31k=\ln \left(0.75716667\right)}$
${\displaystyle k=-\frac{\ln \left(0.75716667\right)}{31}}$
${\displaystyle \approx 0.00897}$ 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.
${\displaystyle m\left(t\right)=m_0{e}^{-kt}}$
${\displaystyle 30=60e^{-kt}}$
${\displaystyle 30=60e^{-0.00897t}}$
Solve for t as follows.
${\displaystyle e^{-0.00897t}=\frac{30}{60}}$
${\displaystyle \ln \left(e^{-0.00897t}\right)=\ln \left(\frac{1}{2}\right)}$
${\displaystyle -0.00897t=-\ln \left(2\right)}$
${\displaystyle t=\frac{\ln \left(2\right)}{0.00897}}$
${\displaystyle \approx 77.27}$ 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.
${\displaystyle m\left(t\right)=m_0{e}^{-kt}}$
${\displaystyle 10=60e^{-kt}}$
${\displaystyle 10=60e^{-0.00897t}}$
Solve for t as follows
${\displaystyle e^{-0.00897t}=\frac{10}{60}}$
${\displaystyle \ln \left(e^{-0.00897t}\right)=\ln \left(\frac{1}{6}\right)}$
${\displaystyle -0.00897t=-\ln \left(6\right)}$
${\displaystyle t=\frac{\ln \left(6\right)}{0.00897}}$
${\displaystyle =199.750219}$
${\displaystyle \approx 200}$ days