Problem 73 - Chemistry The Molecular Science

Question

The oldest known fossil cells were found in South Africa. The fossil has been dated by the reaction $$ { }_{37}^{87} \mathrm{Rb} \longrightarrow{ }_{38}^{87} \mathrm{Sr}+{ }_{-1}^{0} \mathrm{e} \quad t_{1 / 2}=4.9 \times 10^{10} \text { years } $$ If the ratio of the present quantity of \({ }^{87} \mathrm{Rb}\) to the original quantity is \(0.951,\) calculate the age of the fossil cells.

Step-by-Step Solution

Answer

The age of the fossil cells is approximately 3.44 billion years.

Step 1: Understand the Decay Process

The decay process described is a radioactive beta decay where \({}_{37}^{87} \mathrm{Rb}\) decays into \({}_{38}^{87} \mathrm{Sr} + {}_{-1}^{0} \mathrm{e}\). The half-life (\(t_{1/2}\)) of \({}_{37}^{87} \mathrm{Rb}\) is given as \(4.9 \times 10^{10}\) years. The given ratio is the remaining \({}^{87} \mathrm{Rb}\) to the original amount, which is 0.951.

Step 2: Use the Decay Formula

The formula to find the age of the sample based on radioactive decay is given by \(N = N_0 \times 0.5^{t/t_{1/2}}\), where \(N\) is the remaining quantity, \(N_0\) is the initial quantity, and \(t\) is the time. Rearrange it to \(N/N_0 = 0.5^{t/t_{1/2}}\).

Step 3: Plug in the Known Values

We know that \(N/N_0 = 0.951\) and \(t_{1/2} = 4.9 \times 10^{10}\) years. Substitute these into the equation: \(0.951 = 0.5^{t/(4.9 \times 10^{10})}\).

Step 4: Solve for Time \\(t\\)

Take the logarithm of both sides to solve for \(t\): \(\log(0.951) = \left(t \over 4.9 \times 10^{10}\right) \log(0.5)\). Rearrange to solve for \(t\): \(t = \frac{\log(0.951)}{\log(0.5)} \times 4.9 \times 10^{10}\).

Step 5: Calculate the Result

Compute \(t\) using a calculator: \(\log(0.951) \approx -0.02115\) and \(\log(0.5) \approx -0.301\). Thus, \(t = \frac{-0.02115}{-0.301} \times 4.9 \times 10^{10}\) which simplifies to \(t \approx 3.44 \times 10^9\) years.

Key Concepts

Half-life Calculation

Understanding half-life is key to grasping radioactive decay. Half-life is the time it takes for half of a radioactive substance to decay. Each element has a specific half-life. For example, Rubidium-87 has a half-life of \(4.9 \times 10^{10}\) years.

In our calculation, we find how much time has passed by using the initial and remaining amounts of Rubidium-87. We use the formula \[N/N_0 = 0.5^{t/t_{1/2}} \] where:

• \(N\) is the remaining quantity
• \(N_0\) is the initial quantity
• \(t\) is the time elapsed
• \(t_{1/2}\) is the half-life

This rearranges to \[N/N_0 = 0.5^{t/t_{1/2}} \] showing that the quantity decreases exponentially over time. By comparing the known ratio, we solve for time \(t\). This tells us how long ago the radioactive decay started, in this instance, giving us the age of the fossil.

Rubidium-Strontium Dating

Rubidium-Strontium dating is a radiometric dating technique used to determine the age of rocks and fossils. It relies on the decay of Rubidium-87 into Strontium-87. This method is particularly useful for geological samples millions or even billions of years old.

Rubidium-87 is a beta-decaying isotope, meaning it releases an electron during decay, transforming into Strontium-87. By measuring the ratio of Rubidium-87 to Strontium-87, scientists can calculate the age of the sample. This technique is powerful because it can detect even small changes in these ratios over geological time scales.

In this exercise, the given ratio reflects how much Rubidium-87 remains compared to its original amount. The age calculated tells us how long it took for Rubidium-87 in the sample to decrease to this ratio, offering insights into the history of the Earth's earliest life forms.

Logarithmic Calculation

The use of logarithms is crucial in solving exponential decay problems, such as those involving radioactive decay. Logarithms help us convert multiplicative relationships into additive ones, making it easier to solve for unknown variables.

Let's consider the formula: \[0.951 = 0.5^{t/(4.9 \times 10^{10})}\]Taking the logarithm of both sides allows us to bring down the exponent:\[\log(0.951) = \left(\frac{t}{4.9 \times 10^{10}}\right) \log(0.5)\]This makes it simpler to isolate \(t\) by rearranging:\[t = \frac{\log(0.951)}{\log(0.5)} \times 4.9 \times 10^{10}\]Using logarithmic tables or calculators, we find:

• \(\log(0.951) \approx -0.02115\)
• \(\log(0.5) \approx -0.301\)

Plug these into the equation to calculate \(t\), which tells us how long the decay process has been occurring. This illustrates not only the power of logarithms in simplifying complex equations but also their utility in real-world scientific applications.