Chapter 8.8, Problem 1CYU

(a)

To determine

To Find: The value of $k$ and the equation of decay for potassium -40.

Answer to Problem 1CYU

The value of $k=5.545\times {10}^{-10}$ and the equation is $y=a{e}^{-5.545\times {10}^{-10}t}$ .

Explanation of Solution

Given:

The half-life of potassium 40 is about 1.25 billion years.

Calculation:

Consider the equation for the decay is,

  $y=a{e}^{-kt}$

Then,

  $\begin{array}{l}y=a{e}^{-kt}\\ \frac{a}{2}=a{e}^{-kt}\\ 0.5=e^{-\left(1.25\right)\left({10}^9\right)k}\\ \ln 0.5=-1.25\times {10}^{9}k\end{array}$

Then,

  $\begin{array}{c}k=\frac{\ln 0.5}{-1.25\times {10}^9}\\ =5.545\times {10}^{-10}\end{array}$

Then, the expression is,

  $y=a{e}^{-5.545\times {10}^{-10}t}$

The value of $k=5.545\times {10}^{-10}$ and the equation is $y=a{e}^{-5.545\times {10}^{-10}t}$ .

(b)

To determine

To Find: The time taken by the specimen to decay to only 15 milligrams to only 15 milligrams of the potassium 40.

Answer to Problem 1CYU

The time decay to 15 milligrams is $t=1.57727\times {10}^9$ years.

Explanation of Solution

Given:

The half-life of potassium 40 is about 1.25 billion years.

The specimen contains 36 milligrams of potassium -40.

Calculation:

Consider the equation is,

  $y=a{e}^{-5.545\times {10}^{-10}t}$

Then,

  $\begin{array}{l}15=\left(36\right)e^{-5.5\times {10}^{-10}t}\\ e^{-5.5\times {10}^{-10}t}=0.42\\ \ln e^{-5.5\times {10}^{-10}t}=\ln 0.42\\ t=\frac{\ln 0.42}{-5.5\times {10}^{-10}}\end{array}$

The time decay to 15 milligrams is $t=1.57727\times {10}^9$ years.

(c)

To determine

To Find: The amount of milligrams of potassium 40 that will left after 300 million years.

Answer to Problem 1CYU

The remaining potassium 40 after 300 million year is about 30.52 mg.

Explanation of Solution

Given:

The half-life of potassium 40 is about 1.25 billion years.

Calculation:

Consider the equation is,

  $\begin{array}{c}y=36e^{-5.545\times {10}^{-10}t}\\ =30.52\end{array}$

The remaining potassium 40 after 300 million year is about 30.52 mg.

(d)

To determine

To Find: The time taken to take potassium 40 to decay to one eighth of its original amount.

Answer to Problem 1CYU

The potassium 40 will decay to one eight of its original amount in $3.7808\times {10}^9$ years.

Explanation of Solution

Given:

The half-life of potassium 40 is about 1.25 billion years.

Calculation:

Consider the general equation is,

  $y=a{e}^{-5.545\times {10}^{-10}t}$

Then for $y=\frac{a}{8}$ is,

  $\begin{array}{l}0.125=a{e}^{-5.545\times {10}^{-10}t}\\ \ln 0.125=-0.55\times {10}^{-9}t\\ t=\frac{\ln 0.125}{-0.55\times {10}^{-9}}\\ t=3.7808\times {10}^9\end{array}$

Thus, the potassium 40 will decay to one eight of its original amount in $3.7808\times {10}^9$ years.