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One method of dating rocks is based on their \(^{87} \mathrm{Sr} /^{87} \mathrm{Rb}\) ratio. \(^{87} \mathrm{Rb}\) is a \(\beta^{-}\) emitter with a half- life of \(5 \times 10^{11}\) years. A certain rock has a mass ratio \(^{87} \mathrm{Sr} /^{87} \mathrm{Rb}\) of \(0.004 / 1.00 .\) What is the age of the rock?

Short Answer

Expert verified
The age of the rock is calculated by substituting the given values into the derived formula. This will give the final age in years.

Step by step solution

01

Identify given information

From the exercise, the half-life of \(^{87}\mathrm{Rb}\) is \(5 \times 10^{11}\) years. Also, the mass ratio \(^{87}\mathrm{Sr} /^{87}\mathrm{Rb}\) of the rock is 0.004/1.00.
02

Use the Decay Equation to find decay constant

The decay constant (\(λ\)) is given by \(\frac{ln 2}{t_{1/2}}\), where \(t_{1/2}\) is the half-life. So, \(λ = \frac{ln 2}{5 \times 10^{11}}\).
03

Use the ratio to compute time

In radioactive decay, the ratio of Sr-87 to Rb-87 is related to time (t) by the equation \(\frac{Nf}{Ni} = e^{-λt}\), where \(Nf\) is the final quantity (Sr-87 in this case), \(Ni\) is the initial quantity (Rb-87 in this case), and \(e\) is Euler's number. Rearranging for t gives \(t = -\frac{ln (\frac{Nf}{Ni})}{λ}\).
04

Substitute values and solve for t

Substitute \(Nf/Ni = 0.004/1.00\) and our previously calculated \(λ\) into the equation to find t. \(t = -\frac{ln (0.004)}{\frac{ln 2}{5 \times 10^{11}}}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rb-Sr Dating Method
The Rb-Sr dating method is a popular technique used to date ancient rocks and minerals. It relies on the decay of the radioisotope Rubidium-87 (\(^{87}\text{Rb}\)) into Strontium-87 (\(^{87}\text{Sr}\)). This process occurs over vast periods, making it ideal for dating Earth’s oldest formations.
A key factor in this method is the measurable ratio of \(^{87}\text{Sr}\) to \(^{87}\text{Rb}\) in a sample. When a rock forms, it initially contains a certain amount of \(^{87}\text{Rb}\) and no \(^{87}\text{Sr}\). Over time, the \(^{87}\text{Rb}\) decays into \(^{87}\text{Sr}\), thus altering the ratio.
To determine the age, scientists measure the \(^{87}\text{Sr}/^{87}\text{Rb}\) ratio and use the decay equation. This kind of dating is crucial for understanding geological processes and the history of the Earth.
Half-Life Calculation
The concept of half-life is central to understanding radioisotope dating methods. Half-life is the time required for half of the radioactive isotopes in a sample to decay.
For Rubidium-87, the half-life is an incredibly long \(5 \times 10^{11}\) years. This means that if you start with a certain amount of \(^{87}\text{Rb}\), half of it will transform into \(^{87}\text{Sr}\) after that time period.
Calculating the decay constant \(\lambda\) from the half-life is essential in dating calculations. The formula \(\lambda = \frac{\ln 2}{t_{1/2}}\) allows you to find the rate of decay. This constant is then used in various equations to ascertain the age of rocks or fossils.
Radioactive Decay Equation
The radioactive decay equation provides a mathematical way to determine the age of a rock. For the Rb-Sr method, the equation is expressed as:
\[ \frac{N_f}{N_i} = e^{-\lambda t} \]
Here, \(N_f\) represents the amount of \(^{87}\text{Sr}\) now present, while \(N_i\) is the initial amount of \(^{87}\text{Rb}\). The equation involves natural exponential functions, with \(e\) as Euler's number, which is approximately 2.718.
This equation can be rearranged to solve for time \(t\), giving: \[ t = -\frac{\ln(\frac{N_f}{N_i})}{\lambda} \]
By substituting the measured \(^{87}\text{Sr}/^{87}\text{Rb}\) ratio and the decay constant into the equation, scientists can determine the age of a rock with remarkable precision, offering insight into geological history and the evolution of the Earth.

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Most popular questions from this chapter

A sample of radioactive \(\frac{35}{16} \mathrm{S}\) disintegrates at a rate of \(1.00 \times 10^{3}\) atoms \(\min ^{-1} .\) The half-life of \(_{16}^{35} \mathrm{S}\) is \(87.9 \mathrm{d}\) How long will it take for the activity of this sample to decrease to the point of producing (a) \(253 ;\) (b) \(104 ;\) and (c) 52 dis \(\min ^{-1} ?\)

What nucleus is obtained in each process? (a) \(\frac{234}{94}\) Pu decays by \(\alpha\) emission. \(\text { (b) } \begin{array}{l}248 \\\97\end{array}\)Bk decays by \(\beta^{-}\) emission. \(\text { (c) } \begin{array}{r}196 \\\82\end{array}\) Pb goes through two successive EC processes.

Which member of the following pairs of nuclides would you expect to be most abundant in natural sources: (a) \(_{10}^{20} \mathrm{Ne}\) or \(_{10}^{22} \mathrm{Ne} ;\) (b) \(_{8}^{17} \mathrm{O}\) or \(_{8}^{18} \mathrm{O} ;\) (c) \(_{3}^{6} \mathrm{Li}\) or \(_{3}^{7} \mathrm{Li} ?\) Explain your reasoning.

5\. In your own words, define the following symbols: (a) \(\alpha ;\) (b) \(\beta^{-} ;\) (c) \(\beta^{+} ;\) (d) \(\gamma ;\) (e) \(t_{1 / 2}\).

The packing fraction of a nuclide is related to the fraction of the total mass of a nuclide that is converted to nuclear binding energy. It is defined as the fraction \((M-A) / A,\) where \(M\) is the actual nuclidic mass and \(A\) is the mass number. Use data from a handbook (such as the Handbook of Chemistry and Physics, published by the CRC Press) to determine the packing fractions of some representative nuclides. Plot a graph of packing fraction versus mass number, and compare it with Figure \(25-6 .\) Explain the relationship between the two.

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