You find a rock that has the element Lokuim and duaghter of Lokium (DOL) in it. You take a sample and find that there is 75% DOL and 25% lokium in the sample. You know that the half life of the element lokium, is 3500 years. Using this information explain how you find the age of the rock. (4 points)

Answer 1

To determine the age of the rock containing Lokuim and its daughter isotope DOL, you can follow these steps:

1. Understand the Relationship Between Parent and Daughter Isotope: The element Lokuim (parent) decays into DOL (daughter) over time. In this scenario, you have a sample that contains 75% DOL and 25% Lokuim, meaning that for every 1 part of Lokuim, there are 3 parts of DOL in the sample.

2. Determine the Ratio of Parent to Daughter Isotope: Since the sample has 25% Lokuim remaining, this indicates that 75% of the original Lokuim has decayed into DOL. This gives a ratio of Lokuim to DOL that can be expressed as: $$\text{Ratio}=\frac{\text{Amount of Lokuim}}{\text{Amount of DOL}}=\frac{25}{75}=\frac{1}{3}$$

3. Use the Decay Formula: The relationship between the parent isotope and the daughter isotope can be described using the decay equation: $$N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_{1/2}}}$$ where $N$ is the remaining quantity of the parent isotope (Lokuim), $N_0$ is the initial quantity, $t$ is the time elapsed (age of the rock), and $t_{1/2}$ is the half-life of Lokuim (3500 years).

4. Calculate the Age: Rearranging the decay equation to find time: $$\text{Given that }\frac{N_0}{N}=4\text{(from Lokuim to DOL ratio)}$$ This indicates that the initial amount of Lokuim was 4 times the current amount (since it has decayed to 1/4 of its original amount). Applying the equation: $$\frac{N_0}{N}=4={\left(\frac{1}{2}\right)}^{\frac{t}{3500}}$$ Taking the log base 2 of both sides: $$4=2^2\Rightarrow \frac{t}{3500}=2\Rightarrow t=2\times 3500=7000\text{ years}$$

Thus, the age of the rock is 7000 years.