Question

Use the information in the table to calculate the ages of the meteorites. The half-life of potassium-40 is 1.3 billion years.

Meteorite Initial
Amount
(g) Final
Amount
(g)
1 30 7.5
2 80 5
3 100 12.5
Use your calculations to answer the questions.

How old is meteorite 1?

How old is meteorite 2?

How old is meteorite 3?

Answers

Answered by GPT-4o mini
To calculate the age of each meteorite, we can use the decay of potassium-40. The half-life of potassium-40 is given as 1.3 billion years.

### Steps to Calculate the Age of Each Meteorite:

1. **Determine the number of half-lives that have passed:**
The formula for the remaining amount after a certain number of half-lives is:

\[
\text{Final Amount} = \text{Initial Amount} \times \left(\frac{1}{2}\right)^n
\]

where \( n \) is the number of half-lives that have passed.

2. **Rearranging the formula to find \( n \):**

\[
n = \log_2\left(\frac{\text{Initial Amount}}{\text{Final Amount}}\right)
\]

3. **Calculate the age of each meteorite using:**

\[
\text{Age} = n \times \text{Half-Life}
\]

### Calculations for Each Meteorite:

#### Meteorite 1
- **Initial Amount:** 30 g
- **Final Amount:** 7.5 g

Calculate \( n \):

\[
n = \log_2\left(\frac{30}{7.5}\right) = \log_2(4) = 2
\]

Calculate the age:

\[
\text{Age} = n \times 1.3 \, \text{billion years} = 2 \times 1.3 = 2.6 \, \text{billion years}
\]

#### Meteorite 2
- **Initial Amount:** 80 g
- **Final Amount:** 5 g

Calculate \( n \):

\[
n = \log_2\left(\frac{80}{5}\right) = \log_2(16) = 4
\]

Calculate the age:

\[
\text{Age} = n \times 1.3 \, \text{billion years} = 4 \times 1.3 = 5.2 \, \text{billion years}
\]

#### Meteorite 3
- **Initial Amount:** 100 g
- **Final Amount:** 12.5 g

Calculate \( n \):

\[
n = \log_2\left(\frac{100}{12.5}\right) = \log_2(8) = 3
\]

Calculate the age:

\[
\text{Age} = n \times 1.3 \, \text{billion years} = 3 \times 1.3 = 3.9 \, \text{billion years}
\]

### Summary of Ages:
- **Meteorite 1 age:** 2.6 billion years
- **Meteorite 2 age:** 5.2 billion years
- **Meteorite 3 age:** 3.9 billion years

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