# What is the rock's age? A rock contains one-fourth of its original amount of potassium-40 The half-life of potassium-40 is 1.3 billion years?

Feb 14, 2015

The rock's age is approximately $\text{2.6 billion}$ years.

There are essentially two ways of solving nuclear half-life problems. One way is by applying the half-life formula, which is

$A \left(t\right) = {A}_{0} \left(t\right) \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\frac{1}{2}\right)}$ , where

$A \left(t\right)$ - the quantity that remains and has not yet decayed after a time t;
${A}_{0} \left(t\right)$ - the initial quantity of the substance that will decay;
${t}_{\frac{1}{2}}$ - the half-life of the decaying quantity;

In this case, the rock contains $\text{1/4th}$ of the orignal amount of potassium-40, which means $A \left(t\right)$ will be equal to $\frac{{A}_{0} \left(t\right)}{4}$. Plug this into the equation above and you'll get

$\frac{{A}_{0} \left(t\right)}{4} = {A}_{0} \left(t\right) \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\frac{1}{2}\right)}$, or $\frac{1}{4} = {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\frac{1}{2}\right)}$

This means that $\frac{t}{t} _ \left(\frac{1}{2}\right) = 2$, since $\frac{1}{4} = {\left(\frac{1}{2}\right)}^{2}$.

Therefore,

$t = 2 \cdot {t}_{\frac{1}{2}} = 2 \cdot \text{1.3 = 2.6 billion years}$

A quicker way to solve this problem is by recognizing that the initial amount of the substance you have is halved with the passing of each half-life, or ${t}_{\frac{1}{2}}$.

This means that you'll get

$A = \frac{{A}_{0}}{2}$ after the first 1.3 billion years

$A = \frac{{A}_{0}}{4}$ after another 1.3 billion years, or $2 \cdot \text{1.3 billion}$

$A = \frac{{A}_{0}}{8}$ after another 1.3 billion years, or $2 \cdot \left(2 \cdot \text{1.3 billion}\right)$

and so on...