what is the age of a meteorite if potassium - 40 decayed from 80 g to 10 g? the half - life of potassium…

Explanation:

Step1: Determine number of half - lives

We know the formula for radioactive decay $N = N_0\times(\frac{1}{2})^n$, where $N$ is the final amount, $N_0$ is the initial amount, and $n$ is the number of half - lives. Given $N_0 = 80$ g and $N = 10$ g. So, $10=80\times(\frac{1}{2})^n$. Then $\frac{10}{80}=(\frac{1}{2})^n$, and $\frac{1}{8}=(\frac{1}{2})^n$. Since $\frac{1}{8}=(\frac{1}{2})^3$, $n = 3$.

Step2: Calculate age of meteorite

The half - life of potassium - 40, $t_{1/2}=1.3$ billion years. The age of the meteorite $t=n\times t_{1/2}$. Substituting $n = 3$ and $t_{1/2}=1.3$ billion years, we get $t=3\times1.3$ billion years.

Answer:

3.9 billion years