Question

1. how long will it take 600 grams of plutonium - 239 (half - life = 24400 years) to decay to 18.75 grams?
2. a sample of wood that originally contained 100 grams of c - 14 (half - life 5568 years) now only contains 25 grams of c - 14. approximately how many years ago was this sample part of a living tree?
3. a rock initially contained 0.01 g of a radioactive substance when it first formed. how much remains after 4 half - lives?

Explanation:

Step1: Recall radioactive - decay formula

The amount of a radioactive substance $A$ after $n$ half - lives is given by $A = A_0\times(\frac{1}{2})^n$, where $A_0$ is the initial amount.

Step2: Solve for the first question

For Plutonium - 239, $A_0 = 600$ grams and $A = 18.75$ grams. We need to find $n$.
Set up the equation $18.75=600\times(\frac{1}{2})^n$.
First, divide both sides by 600: $\frac{18.75}{600}=(\frac{1}{2})^n$.
$\frac{18.75}{600}=\frac{1875}{60000}=\frac{1}{32}$.
So, $(\frac{1}{2})^n=\frac{1}{32}$, and since $\frac{1}{32}=(\frac{1}{2})^5$, $n = 5$.
The half - life $T_{1/2}=24400$ years. The time $t=n\times T_{1/2}=5\times24400 = 122000$ years.

Step3: Solve for the second question

For C - 14, $A_0 = 100$ grams and $A = 25$ grams.
Set up the equation $25 = 100\times(\frac{1}{2})^n$.
Divide both sides by 100: $\frac{25}{100}=(\frac{1}{2})^n$, so $(\frac{1}{2})^n=\frac{1}{4}$.
Since $\frac{1}{4}=(\frac{1}{2})^2$, $n = 2$.
The half - life $T_{1/2}=5568$ years. The time $t=n\times T_{1/2}=2\times5568 = 11136$ years.

Step4: Solve for the third question

For a radioactive substance with $A_0 = 0.01$ g and $n = 4$ half - lives.
Using the formula $A = A_0\times(\frac{1}{2})^n$, substitute $A_0 = 0.01$ and $n = 4$.
$A=0.01\times(\frac{1}{2})^4=0.01\times\frac{1}{16}=0.000625$ g.

Answer:

1. 122000 years
2. 11136 years
3. 0.000625 g