Determine the age of a rock containing \(0.065 \mathrm{~g}\) of uranium-238 \(\left(t_{1 / 2}=4.5 \times 10^{9} \mathrm{yr}\right)\) and \(0.023 \mathrm{~g}\) of lead-206. (Assume that all the lead- 206 came from \({ }^{238} \mathrm{U}\) decay.

Short Answer

Expert verified
The age of the rock is approximately 2.2 billion years.

Step by step solution

01

- Understand the decay process

Uranium-238 \text{(}^{238}\text{U}\text{)} decays into Lead-206 \text{(}^{206}\text{Pb}\text{)} through a series of steps. Given the half-life, we can use the decay formula: \[ N(t) = N_0 e^{-\frac{t}{\tau}} \] where \text{N(t)} is the quantity of \text{Uranium-238} at time \text{t}, \text{N}_0 is the original quantity, and \[ \tau = \frac{ t_{1/2}}{\text{ln}(2)} \] The number of moles will help us determine the original quantity.
02

- Convert amounts to moles

The molar mass of Uranium-238 is approximately 238 g/mol and that of Lead-206 is approximately 206 g/mol. Convert the given masses to moles: \(n_{U} = \frac{ 0.065~ \text{g}}{238~ \text{g/mol}} \approx 2.73 \times 10^{-4} \text{mol} \) \(n_{Pb} = \frac{0.023~ \text{g}}{206~ \text{g/mol}} \approx 1.12 \times 10^{-4} \text{mol} \)
03

- Determine initial amount of Uranium-238

Since all the Lead-206 \text{(}^{206}\text{Pb}\text{)} came from the Uranium-238 \text{(}^{238}\text{U}\text{)} decay, the initial amount of Uranium-238 can be determined by: \(n_{0} = n_{Pb} + n_{U} \approx 1.12 \times 10^{-4}\text{ mol} + 2.73 \times 10^{-4} \text{mol} \approx 3.85 \times 10^{-4} \text{mol} \)
04

- Use the decay formula to find time

Using the decay formula and solving for time: \( N(t) = N_{0} e^{-\frac{t}{\tau}} \implies \frac{n_{U}}{n_{0}} = e^{-\frac{t}{\tau}}\) \(\frac{2.73 \times 10^{-4}}{3.85 \times 10^{-4}} = e^{-\frac{t}{(4.5 \times 10^9)/\text{ln}(2)}}\)
05

- Solving for time

Isolate \text{t}: \(\text{ln} \frac{2.73 \times 10^{-4}}{3.85 \times 10^{-4}} = -\frac{t}{ \tau}\) \(\text{ln}(0.709)= -\frac{t}{(4.5 \times 10^9)/ \text{ln} 2} \approx -0.344\) \(t \approx -0.344 \times -6.49 \times 10^9 \approx 2.2 \times 10^{9} \text{ years}\)

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Key Concepts

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

Radioactive Decay
Understanding radioactive decay is essential for solving problems related to Uranium-Lead dating. Radioactive decay is a random process in which an unstable atomic nucleus loses energy by emitting radiation. In the context of this problem, Uranium-238 (\(^{238}\text{U}\)) decays into Lead-206 (\(^{206}\text{Pb}\)) through a series of steps.
This decay process can be expressed with the formula: \[ N(t) = N_0 e^{-\frac{t}{\tau}} \] Here, \( N(t) \)- is the quantity of Uranium-238 at time \( t \), \( N_0 \)- is the original quantity of Uranium-238, and \( \tau = \frac{t_{1/2}}{ \text{ln}(2)} \)- is the mean lifetime of Uranium-238. Understanding this concept helps in calculating how much of the original material has decayed over a given period.
  • Radioactive isotopes transform into stable isotopes through decay.
  • This transformation rate is predictable and is expressed through mathematical formulas.
Radioactive decay is not dependent on external factors like temperature and pressure, making it a reliable means of dating geological samples.
Half-Life Calculation
Half-life is an important concept in radioactive decay. It is the time required for half of the radioactive nuclei in a sample to decay. For Uranium-238, the half-life (\( t_{1/2} \)) is 4.5 billion years.
The relationship between the half-life and decay constant is given by: \[ \tau = \frac{ t_{1/2}}{\text{ln}(2)} \] Where \ \text{ln} \ represents the natural logarithm.In this exercise:
  • We will use the half-life of Uranium-238 to determine the age of the rock sample.
  • By comparing the remaining amount of Uranium-238 to the initial quantity, we can estimate the time elapsed.
The decay formula is then used to solve for time, which helps in calculating the age of the rock. This calculation is crucial for understanding geological events and the age of different formations.
Isotope Geochemistry
Isotope geochemistry deals with the isotopic composition of elements within the Earth's crust. Uranium-238 decays to Lead-206, and this transformation serves as the basis for Uranium-Lead dating.
Isotope geochemistry provides insights into geological processes by examining isotopic ratios. In the given exercise:
  • We start by determining the amount of Lead-206 and Uranium-238 in the rock.
  • The initial amount of Uranium-238 can be inferred by adding the present Uranium-238 and Lead-206 amounts, acknowledging that all Lead-206 came from Uranium-238 decay.
Understanding isotope geochemistry allows us to connect isotopic data with geological ages, making it a powerful tool for studying Earth's history.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is used to convert between the mass of a substance and the amount of substance in moles. In our exercise, we needed to convert the given masses of Uranium-238 and Lead-206 to moles.
For Uranium-238, the molar mass is 238 g/mol, and for Lead-206, it is 206 g/mol. Using these values, we can calculate the number of moles (\(n\)):\[ n_{U} = \frac{0.065\text{ g}}{238\text{ g/mol}} \approx 2.73 \times 10^{-4} \text{ mol} \]\[ n_{Pb} = \frac{ 0.023\text{ g}}{206\text{ g/mol}} \approx 1.12 \times 10^{-4} \text{ mol} \]This conversion is essential to determine the initial amount of Uranium-238 accurately.
  • It allows calculation of chemical quantities from physical measurements.
  • Essential for understanding reactions and transformations in geochemical processes.
Mastering the concept of molar mass is key for performing precise calculations in chemistry and geosciences.

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Most popular questions from this chapter

The oxidation of methanol to formaldehyde can be accomplished by reaction with chromic acid: $$ \begin{array}{r} 6 \mathrm{H}^{+}(a q)+3 \mathrm{CH}_{3} \mathrm{OH}(a q)+2 \mathrm{H}_{2} \mathrm{CrO}_{4}(a q) \longrightarrow \\ 3 \mathrm{CH}_{2} \mathrm{O}(a q)+2 \mathrm{Cr}^{3+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(I) \end{array} $$ The reaction can be studied with the stable isotope tracer \({ }^{18} \mathrm{O}\) and mass spectrometry. When a small amount of \(\mathrm{CH}_{3}{ }^{18} \mathrm{OH}\) is present in the alcohol reactant, \(\mathrm{CH}_{2}{ }^{18} \mathrm{O}\) forms. When a small amount of \(\mathrm{H}_{2} \mathrm{Cr}^{18} \mathrm{O}_{4}\) is present, \(\mathrm{H}_{2}{ }^{18} \mathrm{O}\) forms. Does chromic acid or methanol supply the \(\mathrm{O}\) atom to the aldehyde? Explain.

How do chemical and nuclear reactions differ in (a) Magnitude of the energy change? (b) Effect on rate of increasing temperature? (c) Effect on rate of higher reactant concentration? (d) Effect on yield of higher reactant concentration?

A metastable (excited) form of \({ }^{50} \mathrm{Sc}\) changes to its stable form by emitting \(\gamma\) radiation with a wavelength of \(8.73 \mathrm{pm}\). What is the change in mass of \(1 \mathrm{~mol}\) of the isotope when it undergoes this change?

A sample of cobalt- \(60\left(t_{1 / 2}=5.27 \mathrm{yr}\right),\) a powerful \(\gamma\) emitter used to treat cancer, was purchased by a hospital on March 1 ,2012. The sample must be replaced when its activity reaches \(70 . \%\) of the original value. On what date must it be replaced?

The fraction of a radioactive isotope remaining at time \(t\) is \(\left(\frac{1}{2}\right)^{t / t_{12}},\) where \(t_{1 / 2}\) is the half-life. If the half-life of carbon-14 is 5730 yr, what fraction of carbon- 14 in a piece of charcoal remains after (a) \(10.0 \mathrm{yr} ;\) (b) \(10.0 \times 10^{3} \mathrm{yr} ;\) (c) \(10.0 \times 10^{4} \mathrm{yr} ?\) (d) Why is radiocarbon dating more reliable for the fraction remaining in part (b) than that in part (a) or in part (c)?

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