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?
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
### 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