Chapter 0.5, Problem 46E

(a.)

To determine

The amount of substance remaining as a function of time $t$ .

Answer to Problem 46E

It has been determined that the amount of substance remaining as a function of time $t$ , is $N\left(t\right)=8e^{-\frac{\ln \left(2\right)}{12}t}$ .

Explanation of Solution

Given:

The half-life of a certain radioactive substance is $12$ hours. There are $8$ grams present initially.

Concept used:

The differential equation that models radioactive decay is given as $\frac{dN}{dt}=kN$ , where $N$ is the amount of the radioactive substance, $t$ is the time in hours, and $k$ is the decay constant.

Calculation:

As discussed, the equation for radioactive decay is,

  $\frac{dN}{dt}=kN$

Simplifying,

  $\frac{dN}{N}=kdt$

Now, it is given that there are $8$ grams present initially.

Then, $N=8$ when $t=0$ .

It is also given that the half-life of the radioactive substance is $12$ hours.

Then, $N=\frac{8}{2}=4$ when $t=12$ .

Integrating both sides of $\frac{dN}{N}=kdt$ with appropriate integration limits,

  ${\displaystyle {\int }_8^4\frac{dN}{N}}=k{\displaystyle {\int }_0^{12}dt}$

Solving,

  ${\left[\ln \left(N\right)\right]}_8^4=k{\left[t\right]}_0^{12}$

On further solving,

  $\ln \left(4\right)-\ln \left(8\right)=k\left(12-0\right)$

Simplifying,

  $\ln \left(8\right)-\ln \left(4\right)=-12k$

On further simplifying,

  $\ln \left(\frac{8}{4}\right)=-12k$

Continuing simplification,

  $\ln \left(2\right)=-12k$

Finally,

  $k=-\frac{\ln \left(2\right)}{12}$

Put $k=-\frac{\ln \left(2\right)}{12}$ in $\frac{dN}{N}=kdt$ to get,

  $\frac{dN}{N}=-\frac{\ln \left(2\right)}{12}dt$

Integrating both sides with appropriate limits, assuming that $N=N\left(t\right)$ at time $t$ ,

  ${\displaystyle {\int }_8^{N\left(t\right)}\frac{dN}{N}}=-\frac{\ln \left(2\right)}{12}{\displaystyle {\int }_0^{t}dt}$

Solving,

  ${\left[\ln \left(N\right)\right]}_8^{N\left(t\right)}=-\frac{\ln \left(2\right)}{12}{\left[t\right]}_0^t$

On further solving,

  $\ln \left(N\left(t\right)\right)-\ln \left(8\right)=-\frac{\ln \left(2\right)}{12}\left(t-0\right)$

Simplifying,

  $\ln \left(\frac{N\left(t\right)}{8}\right)=-\frac{\ln \left(2\right)}{12}t$

On further simplifying,

  $\frac{N\left(t\right)}{8}=e^{-\frac{\ln \left(2\right)}{12}t}$

Finally,

  $N\left(t\right)=8e^{-\frac{\ln \left(2\right)}{12}t}$

This is the required expression for the amount of substance remaining as a function of time $t$ .

Conclusion:

It has been determined that the amount of substance remaining as a function of time $t$ , is $N\left(t\right)=8e^{-\frac{\ln \left(2\right)}{12}t}$ .

(b.)

To determine

The time when there will be $1$ gram remaining.

Answer to Problem 46E

It has been determined that there will be $1$ gram of the substance remaining after $36$ hours.

Explanation of Solution

Given:

The half-life of a certain radioactive substance is $12$ hours. There are $8$ grams present initially.

Concept used:

The differential equation that models radioactive decay is given as $\frac{dN}{dt}=kN$ , where $N$ is the amount of the radioactive substance, $t$ is the time in hours, and $k$ is the decay constant.

Calculation:

As determined previously, the amount of substance remaining as a function of time $t$ , is $N\left(t\right)=8e^{-\frac{\ln \left(2\right)}{12}t}$ .

Note that, here, the unit of $N$ is assumed to be grams when the unit of $t$ is hours.

Now, according to the problem, $N=1$ .

Put $N=1$ in $N\left(t\right)=8e^{-\frac{\ln \left(2\right)}{12}t}$ to get,

  $8e^{-\frac{\ln \left(2\right)}{12}t}=1$

Simplifying,

  $\begin{array}{l}8e^{-\frac{\ln \left(2\right)}{12}t}=1\\ e^{-\frac{\ln \left(2\right)}{12}t}=\frac{1}{8}\\ -\frac{\ln \left(2\right)}{12}t=\ln \left(\frac{1}{8}\right)\\ \frac{\ln \left(2\right)}{12}t=-\ln \left(\frac{1}{8}\right)\end{array}$

On further simplification,

  $\begin{array}{l}\frac{\ln \left(2\right)}{12}t=\ln \left(8\right)\\ t=\frac{12}{\ln \left(2\right)}\ln \left(8\right)\\ t=\frac{12}{\ln \left(2\right)}\ln \left(2^3\right)\\ t=\frac{12}{\ln \left(2\right)}\times 3\ln \left(2\right)\end{array}$

Continuing simplification,

  $\begin{array}{l}t=12\times 3\\ t=36\end{array}$

This implies that there will be $1$ gram of the substance remaining after $36$ hours.

Conclusion:

It has been determined that there will be $1$ gram of the substance remaining after $36$ hours.