Answer:
The expected mass of the sample in the year 2009 is approximately 60 miligrams.
Step-by-step explanation:
The mass of radioactive isotopes decays exponentially in time, the equation that represents this phenomenon is presented below:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex] (1)
Where:
[tex]m(t)[/tex] - Current mass of the radioactive isotope, in miligrams.
[tex]m_{o}[/tex] - Initial mass of the radioactive isotope, in miligrams.
[tex]t[/tex] - Time, in years.
[tex]\tau[/tex] - Time constant, in years.
First, we must find the time constant associated to the decay of the radioactive isotope by clearing the respective term in (1):
[tex]-\frac{t}{\tau} = \ln \frac{m(t)}{m_{o}}[/tex]
[tex]\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }[/tex]
If we know that [tex]t = 5\,y[/tex], [tex]m(t) = 190\,mg[/tex] and [tex]m_{o} = 800\,mg[/tex], then the time constant is:
[tex]\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }[/tex]
[tex]\tau = -\frac{5\,y}{\ln \frac{190\,mg}{800\,mg} }[/tex]
[tex]\tau \approx 3.478\,y[/tex]
Now, we calculate the mass of the radioactive isotope in 2009 ([tex]t = 9\,y[/tex]). If we know that [tex]t = 9\,y[/tex], [tex]\tau \approx 3.478\,y[/tex] and [tex]m_{o} = 800\,mg[/tex], then the mass of the radioactive isotope is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
[tex]m(9\,y) = (800\,mg)\cdot e^{-\frac{9\,y}{3.478\,y} }[/tex]
[tex]m(9\,y) \approx 60.155\,mg[/tex]
The expected mass of the sample in the year 2009 is approximately 60 miligrams.