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 \(109.7 \mu \mathrm{g}\) of \({ }^{87} \mathrm{Rb}\) and \(3.1 \mu \mathrm{g}\) of \({ }^{87} \mathrm{Sr}\) ? Assume that no \({ }^{87} \mathrm{Sr}\) 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.

Short answer

The age of the rock sample is approximately \(1.69 \times 10^{10}\) years.

Step-by-step solution

Calculate the initial amount of Rubidium-87

Initially, all the Strontium-87 present in the sample would have been Rubidium-87 as it's assumed that no Strontium-87 was present when the rock was formed. So, we can add the current amounts of Rubidium-87 and Strontium-87 in the rock to find the initial amount of Rubidium-87. Initial Rubidium-87 = Current Rubidium-87 + Current Strontium-87 Initial Rubidium-87 = \(109.7\;\mu𝑔\) + \(3.1\;\mu𝑔\) Initial Rubidium-87 = \(112.8\;\mu𝑔\)

Calculate the decay constant (λ)

The decay constant (λ) can be calculated using the half-life formula: Half-life (t) = \(\frac{0.693}{\lambda}\) Rearranging to solve for λ: \(\lambda = \frac{0.693}{t}\) Using the given half-life of Rubidium-87, \(4.7 \times 10^{10}\) years: \(\lambda = \frac{0.693}{4.7 \times 10^{10}}\) \(\lambda \approx 1.474\times 10^{-11}\) years\(^{-1}\)

Calculate the age of the rock

Now we can use the decay rate formula to calculate the age of the rock: \(N_t = N_0e^{-\lambda t}\) Rearranging to solve for t: \(t = \frac{-\ln(\frac{N_t}{N_0})}{\lambda}\) Where \(N_0\) is the initial amount of Rubidium-87, \(N_t\) is the current amount of Rubidium-87, and \(\lambda\) is the decay constant. Plugging in the values: \(t = \frac{-\ln(\frac{109.7}{112.8})}{1.474\times 10^{-11}}\) \(t \approx 1.690 \times 10^{10}\) years So the age of the rock sample is approximately \(1.69 \times 10^{10}\) years.

Key concepts

Rubidium-87 Decay

Rubidium-87 is a naturally occurring isotope that undergoes radioactive decay through a process known as beta-particle production. In this process, a neutron in the nucleus of a Rubidium-87 atom is transformed into a proton while emitting a beta particle (an electron) and an antineutrino. This transformation results in the creation of a new element: Strontium-87. Over time, traces of Rubidium-87 in minerals convert steadily to Strontium-87, which can be measured to determine the age of geological samples, such as rocks.
To understand how much Rubidium-87 has decayed, it is important to consider the starting and current amounts in the sample. The original amount of Rubidium can be calculated by summing the present amounts of Rubidium-87 and the Strontium-87 decay product, assuming no Strontium-87 was initially present in the rock:
\[\text{Initial Rb-87} = \text{Current Rb-87} + \text{Current Sr-87}\]This equation is crucial in radiometric dating as it establishes a basis for subsequent age calculations.

Half-life Calculation

Half-life is the time required for half of a radioactive substance to decay into its daughter product. For Rubidium-87, the half-life is particularly long, about \(4.7 \times 10^{10}\) years. This prolonged half-life makes Rubidium-87 an excellent candidate for dating ancient geological events, as it allows for the determination of ages of rocks on a grand scale of billions of years.

The half-life can be mathematically connected to the decay constant using the formula:
\[ t_{1/2} = \frac{0.693}{\lambda} \]
The decay constant \(\lambda\) is a measure of the probability of decay per unit time. Rearranging the expression, we can solve for \(\lambda\) using Rubidium-87's known half-life:

• \(\lambda = \frac{0.693}{4.7 \times 10^{10}}\)
• This results in a decay constant approximately equal to \(1.474 \times 10^{-11} \text{ years}^{-1}\).

This relationship is pivotal for transforming the physical half-life into a numerical rate which can then be applied widely in age calculations.

Decay Constant

The decay constant \(\lambda\) is a fundamental part of the radioactive decay equation. It indicates how quickly a particular isotope will decay. For Rubidium-87, with a decay constant around \(1.474 \times 10^{-11} \text{ years}^{-1}\), this value dictates the rate at which it slowly converts to Strontium-87 over time.
The decay constant is used in the exponential decay formula:

• \(N_t = N_0 e^{-\lambda t}\)

Here, \(N_t\) represents the amount of the radionuclide present after time \(t\), \(N_0\) is the initial quantity, and \(e\) is the base of natural logarithms. By rearranging this formula, researchers can solve for \(t\), the age of the sample.
The derived formula, \(t = \frac{-\ln(\frac{N_t}{N_0})}{\lambda}\), allows scientists to calculate the time elapsed since the formation of the sample. This method is critical in determining geological time scales, effectively turning Rubidium-87 decay into a precise natural clock.