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

`Q(t) = 50e^(-0.0527t)`

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:

`(dQ)/(dt) = -rQ`

In which r is the decay rate.

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

`(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 = -(ln(0.9))/(2)`

`r = 0.0527`

So

`Q(t) = Ke^(-rt)`

`Q(t) = 50e^(-0.0527t)`