Question

Potassium-40 decays to argon-40 with a half-life of \(1.27 \times 10^{9} \mathrm{yr}\). What is the age of a rock in which the mass ratio of \({ }^{40} \mathrm{Ar}\) to \({ }^{40} \mathrm{~K}\) is \(4.2 ?\)

Short answer

The age of the rock is approximately \(2.25 \times 10^9\) years.

Step-by-step solution

1. Write the radioactive decay formula

The radioactive decay formula is given by the equation: \[N(t) = N_0 \times 2^{-\frac{t}{t_{1/2}}} \] Where \(N(t)\) is the number of radioactive particles at time \(t\), \(N_0\) is the initial number of radioactive particles, \(t_{1/2}\) is the half-life of the radioactive element, and \(t\) is the time elapsed.

2. Find the remaining fraction of potassium-40

Since the ratio of \(^{40}\mathrm{Ar}\) to \(^{40}\mathrm{K}\) is 4.2, we can set up a relationship between the amount of the two isotopes: \[\frac{^{40}\mathrm{Ar}}{^{40}\mathrm{K}} = \frac{(N_0 - N(t))}{N(t)} = 4.2 \] We are interested in the remaining fraction of \(^{40}\mathrm{K}\): \[\frac{N(t)}{N_0} = \frac{1}{1 + 4.2} = \frac{1}{5.2} \]

3. Substitute the remaining fraction into the decay formula

Using the radioactive decay formula from step 1, substitute the remaining fraction: \[\frac{N(t)}{N_0} = 2^{-\frac{t}{t_{1/2}}} \] Substitute the remaining fraction of potassium-40: \[\frac{1}{5.2} = 2^{-\frac{t}{t_{1/2}}} \]

4. Solve for time t

To solve for time t, first, take the natural logarithm of both sides: \[ \ln{\left(\frac{1}{5.2}\right)} = -\frac{t}{t_{1/2}}\times \ln{2} \] Now, divide by -\(\ln{2}\) and multiply by the given half-life of potassium-40, which is \(1.27 \times 10^9\) years: \[t = \left(\frac{\ln{\left(\frac{1}{5.2}\right)}}{-\ln{2}}\right)\times \left(1.27 \times 10^9\mathrm{yr}\right) \]

5. Calculate the age of the rock

Using the formula derived in the previous step, plug in the given half-life and calculate the age of the rock: \[t = \left(\frac{\ln{\left(\frac{1}{5.2}\right)}}{-\ln{2}}\right)\times \left(1.27 \times 10^9\mathrm{yr}\right) \approx 2.25 \times 10^9\mathrm{yr}\] So, the age of the rock is approximately \(2.25 \times 10^9\) years.

Key concepts

Potassium-40 Decay

Potassium-40 ( ^{40} K) is a naturally occurring isotope that undergoes radioactive decay. This process involves the transformation of the unstable ^{40} K isotope into a more stable form, such as argon-40 ( ^{40} Ar). Radioactive decay is spontaneous and happens at a predictable rate, characterized by a property called the half-life. The half-life of ^{40} K is known to be approximately 1.27 billion years. It is the time it takes for half of the initial amount of ^{40} K to convert into ^{40} Ar or other decay products.
The significance of potassium-40 decay lies in its application in dating geological formations, as it can provide insights into the age of rocks and contribute to our understanding of Earth's history. By observing the amounts of ^{40} K and its decay products in a given sample, scientists can determine how much time has passed since the rock solidified.

Argon-40

Argon-40 ( ^{40} Ar) is the product of the radioactive decay of potassium-40 ( ^{40} K). Once formed, ^{40} Ar being a noble gas, does not chemically bond or interact easily with other elements. Hence, when trapped inside a mineral structure, it remains stable over immense geological timescales. This makes ^{40} Ar exceptionally useful in dating geological specimens through the method known as "potassium-argon dating."
In potassium-argon dating, scientists measure the ratio of ^{40} K to ^{40} Ar within a rock sample. The increase in the concentration of ^{40} Ar alongside the gradual decrease of ^{40} K enables the calculation of the sample's age. This method is particularly useful for dating ancient volcanic rocks and is key to constructing the geological time scale.

Half-Life Calculation

The half-life of a radioactive isotope is the time it takes for half of the radioactive substance to decay. In the context of potassium-40 decay, the half-life is about 1.27 billion years. Calculating the age of a sample that contains potassium-40 involves using the decay formula:\[ N(t) = N_0 \times 2^{-\frac{t}{t_{1/2}}} \]Here, \(N(t)\) is the amount of ^{40}K remaining at time \(t\), and \(N_0\) is the initial quantity. To determine the elapsed time \(t\), we rearrange the formula taking into account the ratio between remaining ^{40}K and original ^{40}K:\[ \frac{N(t)}{N_0} = 2^{-\frac{t}{t_{1/2}}} \]Given this ratio, one can solve for \(t\) by taking the logarithm of both sides of the equation. This calculation is fundamental in geological dating, helping to derive the age of ancient rock samples with considerable accuracy.

Isotope Ratio

The ratio of isotopes within a sample can reveal a lot about its age and history. For potassium-40 decay, the isotope ratio of ^{40} Ar to ^{40} K is a key factor in determining rock age. The ratio reflects the degree of decay from ^{40} K to ^{40} Ar.

• An isotope ratio of greater than 1, such as 4.2 in the exercise, indicates that a substantial amount of ^{40} K has transformed into ^{40} Ar.
• This transformation rate helps us calculate geological time lapses, as the original ^{40} K decreases while ^{40} Ar increases.
• By measuring this ratio accurately, scientists calculate the time that has passed since the rock's formation, leading to precise age determinations.

Understanding isotope ratios is crucial not only in geological studies but also in several other scientific fields where accurate dating is required.