Question

A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 50 mg/L of the material present and after 2 hours it is observed that the material has lost 10 percent of its original concentration, find (a) an expression for the concentration of the material remaining at any time.

Answer 1

Answer:

$Q(t) = 50e^{-0.0527t}$

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:

$\frac{dQ}{dt} = -rQ$

In which r is the decay rate.

Solving the differential equation by separation of variables, we have that:

$\frac{dQ}{Q} = -r dt$

Integrating both sides:

$\ln{Q} = -rt + K$

In which K is the integrative constant.

Applying the exponential to both sides, to remove the ln, we get:

$Q(t) = Ke^{-rt}$

In which K is the initial amount present.

Initially there is 50 mg/L

This means that $K = 50$.

After 2 hours it is observed that the material has lost 10 percent of its original concentration

This means that $Q(2) = 0.9K$. So

$0.9K = Ke^{-2r}$

$e^{-2r} = 0.9$

$\ln{e^{-2r}} = \ln{0.9}$

$-2r = \ln{0.9}$

$r = -\frac{\ln{0.9}}{2}$

$r = 0.0527$

So

$Q(t) = Ke^{-rt}$

$Q(t) = 50e^{-0.0527t}$