Chapter 8.8, Problem 4PPS

a.

To determine

To find : the value of $k$ and the equation of decay for Rubidium-87

Answer to Problem 4PPS

The value of $k\approx 1.42\times {10}^{-11}$

Explanation of Solution

Given information : The question is

The half-life of Rubidium-87 is about $48.8$ billion years

Calculation :

Exponential decay can be modeled by the function $f\left(x\right)=a{e}^{-kt}$ , where $a$ is the initial value, $t$ is the time in years, $k$ is a constant representing the rate of continuous decay

Substitute $t=4.8\times {10}^9$ and $f\left(x\right)=0.5a$

  $\begin{array}{l}f\left(x\right)=a{e}^{-kt}\\ 0.5a=a{e}^{-k\left(4.8\times {10}^9\right)}\\ 0.5=e^{-k\left(4.8\times {10}^9\right)}\\ \ln 0.5=-k\left(4.8\times {10}^9\right)\\ k=-\frac{\ln 0.5}{48.8\times {10}^9}\\ k\approx 1.42\times {10}^{-11}\end{array}$

b.

To determine

To calculate : how long will it take the specimen to decay to only $18$ milligrams of Rubidium-87

Answer to Problem 4PPS

It will take $71,947,270,950years$

Explanation of Solution

Given information : The question is;

The specimen currently contain $50$ milligrams of Rubidium-87

Calculation :

To find how long will it take the specimen to decay to only $18$ milligrams of Rubidium-87, we have to substitute $a=50,k=1.42\times {10}^{-11}\&f\left(x\right)=18$

  $\begin{array}{l}f\left(x\right)=a{e}^{-kt}\\ 18=50e^{-\left(1.42\times {10}^{-11}\right)t}\\ \frac{18}{50}=e^{-\left(1.42\times {10}^{-11}\right)t}\\ \ln \frac{18}{50}=\ln -1.42\times {10}^{-11}t\\ t=\frac{\ln 18-\ln 50}{-1.42\times {10}^{-11}}\\ t=71,947,270,950years\end{array}$

c.

To determine

To calculate : how many milligrams of Rubidium-87 will be left after $800$ million years

Answer to Problem 4PPS

About $49.4$ milligrams of Rubidium-87 will be left after $800$ million years

Explanation of Solution

Given information : The question is

The half-life of Rubidium-87 is about $48.8$ billion years

Calculation :

To find how many milligrams of Rubidium-87 will be left after $800$ million years, we have to substitute $a=50,k=1.42\times {10}^{-11}\&t=800,000,000$

  $\begin{array}{l}f\left(x\right)=a{e}^{-kt}\\ f\left(x\right)=50e^{-\left(1.42\times {10}^{-11}\right)800,000,000}\\ f\left(x\right)=50e^{-0.01136}\\ f\left(x\right)\approx 49.4\end{array}$

d.

To determine

To calculate : how long will it take Rubidium-87 to decay to one-sixteenth to its original amount

Answer to Problem 4PPS

It will take $195,252,726,918years$

Explanation of Solution

Given information : The question is

The half-life of Rubidium-87 is about $48.8$ billion years

Calculation :

To find out how long will it take Rubidium-87 to decay to one-sixteenth to its original amount, we have to substitute $a=50,k=1.42\times {10}^{-11}\&f\left(x\right)=\frac{50}{16}$

  $\begin{array}{l}f\left(x\right)=a{e}^{-kt}\\ \frac{50}{16}=50e^{-\left(1.42\times {10}^{-11}\right)t}\\ \frac{1}{16}=e^{-\left(1.42\times {10}^{-11}\right)t}\\ \ln \frac{1}{16}=-\left(1.42\times {10}^{-11}\right)t\\ t=\frac{\ln 1-\ln 16}{-1.42\times {10}^{-11}}\\ t=195,252,726,918years\end{array}$