Question

The age of an igneous rock that has solidified from magma can be found by analyzing the amount of \({ }^{40} \mathrm{~K}\) and \({ }^{40}\) Ar. Potassium- 40 emits a positron to produce argon- 40 and the half-life of \({ }^{40} \mathrm{~K}\) is \(1.25 \times 10^{9}\) years. $$ { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{18}^{40} \mathrm{Ar}+{ }_{1}^{0} $$ If the rock contains \(3.35\) mmol of \({ }^{40} K\) and \(0.25 \mathrm{mmol}\) of \({ }^{40} \mathrm{Ar}\), how long ago did the rock cool?

Short answer

The rock cooled approximately 128 million years ago.

Step-by-step solution

Understand the decay mechanism

The decay of \( {}^{40} \mathrm{K} \) to \( {}^{40} \mathrm{Ar} \) occurs via a radioactive decay process where \( {}^{40} \mathrm{K} \) emits a positron to transform into \( {}^{40} \mathrm{Ar} \). The half-life of \( {}^{40} \mathrm{K} \) is \( 1.25 \times 10^9 \) years.

Set up the decay equation

The decay equation is used to express the relation between the amount of initial isotope, current isotope, and time. The general formula is:\[N_t = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\]where \( N_t \) is the amount of isotope remaining, \( N_0 \) the initial amount, \( t \) is the time passed, and \( T_{1/2} \) is the half-life.

Calculate initial amount of \( {}^{40} \mathrm{K} \)

Since \( {}^{40} \mathrm{K} \) decays to \( {}^{40} \mathrm{Ar} \), the initial \( {}^{40} \mathrm{K} \) is the sum of current \( {}^{40} \mathrm{K} \) and \( {}^{40} \mathrm{Ar} \):\[N_0 = 3.35 + 0.25 = 3.60 \text{ mmol}\]

Calculate fraction remaining

Find the fraction of \( {}^{40} \mathrm{K} \) remaining by dividing the current amount with the initial amount:\[\frac{N_t}{N_0} = \frac{3.35}{3.60}\]This simplifies to \( 0.9306 \).

Solve for \( t \) using the decay equation

Use the decay equation to solve for \( t \):\[0.9306 = \left(\frac{1}{2}\right)^{\frac{t}{1.25 \times 10^9}}\]Take logarithms on both sides:\[\ln(0.9306) = \frac{t}{1.25 \times 10^9} \cdot \ln\left(\frac{1}{2}\right)\]Solving for \( t \) gives:\[t = \frac{\ln(0.9306)}{\ln(0.5)} \times 1.25 \times 10^9\]

Calculate \( t \)

Compute the time \( t \):\[\ln(0.9306) \approx -0.071156\quad \text{and}\quad \ln(0.5) \approx -0.6931\]\[t \approx \frac{-0.071156}{-0.6931} \times 1.25 \times 10^9 \approx 1.282 \times 10^8 \text{ years}\]

Key concepts

Potassium-40 Decay

Potassium-40 (\(^ {40} \text{K}\)) is an isotope that undergoes radioactive decay to transform into Argon-40 (\(^ {40} \text{Ar}\)). This process involves the emission of a positron, which is a particle similar to an electron but with a positive charge. During the decay, one of the protons in the \(^ {40} \text{K}\) nucleus is converted into a neutron, resulting in the transformation into \(^ {40} \text{Ar}\).

• This decay process is crucial for dating rocks and minerals.
• By measuring the amounts of Potassium-40 and Argon-40 in a sample, we can determine the elapsed time since the rock solidified.
• The decay equation provides a mathematical way to relate these amounts over time.

Understanding Potassium-40 decay is essential in fields like geology and archaeology, where dating ancient materials accurately is necessary for reconstructing Earth’s history and past life events.

Half-Life Calculation

The half-life of \(^ {40} \text{K}\) is an essential parameter, defined as the time required for half of the radioactive atoms in a sample to decay. For Potassium-40, this period is \(1.25 \times 10^9\) years.
A half-life calculation is key to understanding how a radioactive substance changes over time. The principle can be summed up using the formula:
\[N_t = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\]This equation shows how the number of remaining atoms (\(N_t\)) is linked to the initial amount (\(N_0\)), the elapsed time (\(t\)), and the half-life (\(T_{1/2}\)).

• The half-life enables precise calculations regarding the age of geological samples.
• It helps in determining how long rocks have been solidified based on current measurements of decay products.
• By using logs, we can rearrange the decay formula to solve for the time that has passed since decay began.

Half-life calculations are a foundational principle for dating various natural occurrences and for scientific analysis of ancient materials.

Argon-40 Production

Argon-40 (\(^ {40} \text{Ar}\)) production is an outcome of the decay of Potassium-40 (\(^ {40} \text{K}\)). As Potassium-40 decays, it loses a positron and forms Argon-40 as a stable end product.
This production is critical in dating because the amount of Argon-40 generated relates directly to the time elapsed since the rock containing Potassium-40 solidified.

• The more \(^ {40} \text{Ar}\) present in a sample, the longer the elapsed time since the formation of the rock.
• The calculation of Argon-40 produced over time helps scientists determine the geological age of a specimen.
• Argon-40 does not escape from solid rock easily, which makes it a reliable marker for dating.

Thus, Argon-40 production serves as a potent tool for scientists unraveling the timeline of Earth's history and understanding when key geological events took place.