Answer:
[tex]Q(t) = 50e^{-0.0527t}[/tex]
Step-by-step explanation:
A certain radioactive material is known to decay at a rate proportional to the amount present.
This means that the situation can be described by the following differential equation:
[tex]\frac{dQ}{dt} = -rQ[/tex]
In which r is the decay rate.
Solving the differential equation by separation of variables, we have that:
[tex]\frac{dQ}{Q} = -r dt[/tex]
Integrating both sides:
[tex]\ln{Q} = -rt + K[/tex]
In which K is the integrative constant.
Applying the exponential to both sides, to remove the ln, we get:
[tex]Q(t) = Ke^{-rt}[/tex]
In which K is the initial amount present.
Initially there is 50 mg/L
This means that [tex]K = 50[/tex].
After 2 hours it is observed that the material has lost 10 percent of its original concentration
This means that [tex]Q(2) = 0.9K[/tex]. So
[tex]0.9K = Ke^{-2r}[/tex]
[tex]e^{-2r} = 0.9[/tex]
[tex]\ln{e^{-2r}} = \ln{0.9}[/tex]
[tex]-2r = \ln{0.9}[/tex]
[tex]r = -\frac{\ln{0.9}}{2}[/tex]
[tex]r = 0.0527[/tex]
So
[tex]Q(t) = Ke^{-rt}[/tex]
[tex]Q(t) = 50e^{-0.0527t}[/tex]
