Problem 84 - Chemistry

Question

Rubidium- 87 decays by \(\beta\) -particle production to strontium- 87 with a half-life of \(4.7 \times 10^{10}\) years. What is the age of a rock sample that contains 109.7 \mug of \(^{87} \mathrm{Rb}\) and 3.1\(\mu \mathrm{g}\) of \(^{87} \mathrm{Sr} ?\) Assume that no \(^{87}\) Sr was present when the rock was formed. The atomic masses for \(^{87}\mathrm{Rb}\) and \(^{87} \mathrm{Sr}\) are 86.90919 \(\mathrm{u}\) and 86.90888 u, respectively.

Step-by-Step Solution

Answer

The age of the rock sample is approximately \(3.03 \times 10^9\) years.

Step 1: Identify the given information

We are given: 1. Half-life of Rubidium-87 (\(T_{1/2}\)) = \(4.7 \times 10^{10}\) years 2. Final amount of Rubidium-87 (\(N_t\)) = 109.7 µg 3. Final amount of Strontium-87 (\(N_{Sr}\)) = 3.1 µg 4. Initial amount of Rubidium-87 (\(N_0\)) = \(N_t + N_{Sr}\) (since no \(^{87} \mathrm{Sr}\) was present initially)

Step 2: Calculate the initial amount of Rubidium-87

To find the initial amount of Rubidium-87, we add the final amounts of Rubidium-87 and Strontium-87. \(N_0 = N_t + N_{Sr} = 109.7 \, \mathrm{µg} + 3.1 \, \mathrm{µg} = 112.8 \, \mathrm{µg}\)

Step 3: Find the decay constant

We are given half-life of Rubidium-87, so we can find the decay constant λ using: \(T_{1/2} = \frac{\ln{2}}{\lambda}\) Now, solve for \(\lambda\): \(\lambda = \frac{\ln{2}}{T_{1/2}} = \frac{\ln{2}}{4.7 \times 10^{10} \, \mathrm{years}}\)

Step 4: Apply the decay formula to find the age of the rock sample

The decay formula is given by: \(N_t = N_0 \cdot e^{-\lambda t}\) We need to solve for t (time in years). We can rearrange the formula for t: \(t = \frac{\ln{\frac{N_0}{N_t}}}{\lambda}\) Plug in the values: \(t = \frac{\ln{\frac{112.8 \, \mathrm{µg}}{109.7 \, \mathrm{µg}}}}{\frac{\ln{2}}{4.7 \times 10^{10} \, \mathrm{years}}} = \frac{\ln{1.028}}{\frac{\ln{2}}{4.7 \times 10^{10} \, \mathrm{years}}}\) Now, calculate t: \(t \approx 3.03 \times 10^9 \, \mathrm{years}\) So, the age of the rock sample is approximately \(3.03 \times 10^9\) years.

Key Concepts

Half-life

The concept of half-life is essential in understanding radioactive decay. Half-life, denoted by the symbol \(T_{1/2}\), is the time required for half of the radioactive nuclei in a sample to decay. In simpler terms, if you start with a certain amount of a radioactive substance, in one half-life, only half of it will remain, while the rest will have transformed into another element. This is a constant value unique to each radioactive isotope.
For example, Rubidium-87 has a half-life of \(4.7 \times 10^{10}\) years. This means it takes that amount of time for half of a sample of Rubidium-87 to decay into Strontium-87. By knowing the half-life, scientists can estimate the age of a sample by determining how many half-lives have passed based on the amount of remaining radioactive material versus the amount of decay product.

Decay Constant

The decay constant, denoted by \(\lambda\), is a critical factor in the mathematics of radioactivity. It provides insight into the rate at which a particular radioactive isotope decays. The decay constant is linked to the half-life by the formula:

• \(T_{1/2} = \frac{\ln{2}}{\lambda}\)

By rearranging this equation, one can find \(\lambda\) as:

• \(\lambda = \frac{\ln{2}}{T_{1/2}}\)

For Rubidium-87, this decay constant determines how quickly it turns into Strontium-87 over time. By calculating \(\lambda\), you gain a deeper understanding of the decay process kinetics—expressed in the number of decays per unit time. With the decay constant, you can also use the decay formula \(N_t = N_0 \cdot e^{-\lambda t}\) to estimate the time elapsed, understanding the historical timeline of a rock or mineral.

Rubidium-87

Rubidium-87 is a naturally occurring radioactive isotope notable for its long half-life, making it very stable compared to many other radioactive substances. It undergoes beta decay, a process where a neutron in the nucleus is transformed into a proton while emitting an electron (beta particle). As a result, Rubidium-87 is converted into Strontium-87, another stable element. Understanding Rubidium-87 is crucial in the field of geochronology—the science of determining the age of rocks. Because it decays so slowly, it is particularly valuable in dating very old geological samples. By measuring the ratio of Rubidium-87 to its decay product, Strontium-87, scientists can calculate the age of a rock with impressive precision. This technique, known as Rubidium-Strontium dating, provides vital insights into Earth's history and helps in piecing together the chronological sequence of geological events. When analyzing a rock sample, the amount of Rubidium-87 and Strontium-87 present allows researchers to pinpoint how much time has passed since the rock's formation.