Question

what is the age of a meteorite if potassium - 40 decayed from 80 g to 10 g? the half - life of potassium - 40 is 1.3 billion years.
1.3 billion years
2.6 billion years
3.9 billion years
5.2 billion years

Explanation:

Step1: Calculate the number of half - lives

The formula for the amount of a radioactive substance remaining is \(N = N_0\times(\frac{1}{2})^n\), where \(N\) is the final amount, \(N_0\) is the initial amount, and \(n\) is the number of half - lives.
Given \(N_0 = 80\space g\), \(N = 10\space g\).
Substitute into the formula: \(10=80\times(\frac{1}{2})^n\).
Divide both sides by 80: \(\frac{10}{80}=(\frac{1}{2})^n\), so \(\frac{1}{8}=(\frac{1}{2})^n\).
Since \(\frac{1}{8}=\frac{1}{2^3}\), then \(n = 3\).

Step2: Calculate the age of the meteorite

The age \(t\) of the meteorite is given by \(t=n\times T\), where \(T\) is the half - life.
Given \(T = 1.3\) billion years and \(n = 3\).
Then \(t=3\times1.3\) billion years.
\(t = 3.9\) billion years.

Answer:

3.9 billion years