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.
use your calculations to answer the questions.
how old is meteorite 1?
how old is meteorite 2?
how old is meteorite 3?

Explanation:

Step1: Calculate the number of half - lives for meteorite 1

The formula for the number of half - lives \(n\) is \(N = N_0\times(\frac{1}{2})^n\), where \(N_0\) is the initial amount and \(N\) is the final amount.
For meteorite 1: \(N_0 = 30\), \(N = 7.5\)
\(7.5=30\times(\frac{1}{2})^n\)
\(\frac{7.5}{30}=(\frac{1}{2})^n\)
\(\frac{1}{4}=(\frac{1}{2})^n\)
Since \(\frac{1}{4}=(\frac{1}{2})^2\), \(n = 2\)
The age \(t=n\times T\), where \(T = 1.3\) billion years
\(t=2\times1.3\)

Step2: Calculate the number of half - lives for meteorite 2

For meteorite 2: \(N_0 = 80\), \(N = 5\)
\(5 = 80\times(\frac{1}{2})^n\)
\(\frac{5}{80}=(\frac{1}{2})^n\)
\(\frac{1}{16}=(\frac{1}{2})^n\)
Since \(\frac{1}{16}=(\frac{1}{2})^4\), \(n = 4\)
The age \(t=n\times T\)
\(t=4\times1.3\)

Step3: Calculate the number of half - lives for meteorite 3

For meteorite 3: \(N_0 = 100\), \(N = 12.5\)
\(12.5=100\times(\frac{1}{2})^n\)
\(\frac{12.5}{100}=(\frac{1}{2})^n\)
\(\frac{1}{8}=(\frac{1}{2})^n\)
Since \(\frac{1}{8}=(\frac{1}{2})^3\), \(n = 3\)
The age \(t=n\times T\)
\(t=3\times1.3\)

Answer:

Meteorite 1: \(2.6\) billion years
Meteorite 2: \(5.2\) billion years
Meteorite 3: \(3.9\) billion years