Uranium-238 decays through a series of \(\alpha\) and \(\beta\) emissions to lead-206, with an overall half-life for the entire process of \(4.5 \mathrm{Ga}\). How old is a uranium-bearing ore that is found to have a \({ }^{238} \mathrm{U} /{ }^{206} \mathrm{~Pb}\) ratio of (a) \(1.00\) and (b) \(1.25\) ?

Short Answer

Expert verified
The uranium-bearing ore is 4.5 Ga old for the case of a 1.00 ratio and older than 4.5 Ga for the case of a 1.25 ratio (exact age requires further calculation).

Step by step solution

01

Understanding the Decay Process

Uranium-238 decays to lead-206 through a series of alpha and beta emissions. The half-life of this decay process is the time it takes for half of the Uranium-238 to decay into lead-206. This half-life is given as 4.5 billion years (Ga).
02

Apply the Half-Life Decay Formula

We use the formula for exponential decay, which relates the half-life of a substance to the remaining amount after a certain time has passed: \[ N(t) = N_0 (1/2)^{t/T_{1/2}} \] where: \( N(t) \) = remaining amount of Uranium-238 at time \( t \),\( N_0 \) = initial amount of Uranium-238,\( t \) = time elapsed,\( T_{1/2} \) = half-life of the substance.
03

Relate the Uranium-Lead Ratio to the Decay Formula

For a given Uranium-238 to Lead-206 ratio, we can express the initial amount of Uranium-238 (\( N_0 \)) as the sum of the remaining Uranium-238 (\( N(t) \)) and the amount that has decayed into Lead-206.
04

Solve for the Time Elapsed - Case (a) Ratio 1.00

With a ratio of 1, this means that the amount of Uranium-238 is equal to the amount of Lead-206. Since we start with pure Uranium-238, for every atom of Lead-206, we have lost an atom of Uranium-238. This correlates to exactly one half-life passed. Therefore, the age is simply the half-life of Uranium-238. \[ t = T_{1/2} \]
05

Solve for the Time Elapsed - Case (b) Ratio 1.25

If the ratio is 1.25, this implies that for every 5 Lead-206 atoms, there are 6 Uranium-238 atoms left. Use the decay formula to find the time \( t \) that gives us this ratio: \[ \frac{N(t)}{N_0 - N(t)} = \frac{1.25}{1} \]We need to solve for \( t \) using this equation.
06

Calculation for Time Elapsed - Case (b)

Start by expressing the Uranium-238 to Lead-206 ratio in terms of the decay formula: \[ (1/2)^{t/T_{1/2}} = \frac{N(t)}{N_0} \]Now insert the ratio of Uranium-238 to Lead: \[ (1/2)^{t/T_{1/2}} = \frac{6}{6+5} \]Solve for \( t \) from this equation.
07

Express the Time as an Exponent Base of 1/2

To find \( t \), we use the equation from the previous step and solve for \( t \): \[ \log_{1/2}(\frac{6}{11}) = \frac{t}{T_{1/2}} \]Now, solve for \( t \) by multiplying both sides of the equation by the half-life \( T_{1/2} \).
08

Calculate the Numerical Value of Time Elapsed - Case (b)

Finally, calculate the value of \( t \) using the known half-life of 4.5 Ga: \[ t = \log_{1/2}(\frac{6}{11}) \times 4.5 \text{ Ga} \]

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Key Concepts

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

Understanding Radioactive Decay
Radioactive decay is a process by which an unstable atomic nucleus loses energy by emitting radiation, such as an alpha particle or a beta particle, which are represented by \( \alpha \) and \( \beta \) respectively in scientific notation. This process transmutes the element into another at a rate determined by the substance's half-life.

For example, Uranium-238 decays to lead-206 via a chain of these \( \alpha \) and \( \beta \) emitters. Each element in this decay chain has its own unique half-life, but the overall transformation from Uranium-238 to lead-206 occurs with a half-life of approximately 4.5 billion years. That's how long it takes for half of a sample of Uranium-238 to convert into lead-206.

Understanding this process is crucial for estimating the age of uranium-bearing ores and thus has significant applications in fields such as geology and archeology for dating rocks and fossils.
Half-Life Calculation Fundamentals
The half-life of a radioactive substance is a measure of the time it takes for half of the radioactive atoms in a sample to decay. In simpler terms, after one half-life, the amount of the substance is reduced to half of its initial quantity.

To calculate the age of a substance using its half-life, you need to know the starting amount of the substance and the ratio of the undecayed atoms to the decay products. For instance, in the exercise involving Uranium-238, knowing the ratio of Uranium-238 to lead-206 enables scientists to use the half-life of Uranium-238 to estimate the age of the sample.

The half-life is crucial when dealing with radioactive materials, as it helps determine the longevity of their radioactivity and can be used to assess potential risks or to date ancient objects, aiding researchers in piecing together the history of our planet.
Exploring the Exponential Decay Formula
The exponential decay formula is a mathematical representation that describes the decrease of a quantity at a rate proportional to its current value. For radioactive decay, the formula is often expressed as:

\[ N(t) = N_0 (1/2)^{t/T_{1/2}} \]

where \( N(t) \) represents the number of undecayed atoms at time \( t \), \( N_0 \) is the initial number of atoms, \( t \) is the time that has elapsed, and \( T_{1/2} \) is the half-life of the decaying substance.

When working with decay problems, we often want to find out the elapsed time since the formation of the material. To do this, we can rearrange the above formula to isolate \( t \), usually involving logarithms due to the exponential nature of the equation. This is the method used to determine the age of the uranium-bearing ore in the given exercise, highlighting the power and utility of the exponential decay formula in understanding radioactive decay processes.

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Most popular questions from this chapter

A chemist is studying the mechanism of the following hydrolysis reaction of the organic ester methyl acetate: \(\mathrm{CH}_{3} \mathrm{COOCH} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{OH}\). The question arises whether the \(\mathrm{O}\) atom in the product methanol comes from the methyl acetate or from the water. Propose an experiment using isotopes that would allow the chemist to determine the origin of the oxygen atom.

The activity of a certain radioactive source is \(5.3 \times 10^{5} \mathrm{~Bq}\). Express this activity in curies.

Use the law of radioactive decay to determine the activity of (a) a \(1.0\)-mg sample of radium-226 \(\left(t_{1 / 2}=1.60 \mathrm{ka}\right)\); (b) a \(2.0-\mathrm{mg}\) sample of strontium-90 \(\left.t_{1 / 2}=28.1 \mathrm{a}\right) ;\) (c) a \(0.43-\mathrm{mg}\) sample of promethium- \(147\left(t_{1 / 2}=2.6\right.\) a). The mass of each nuclide as a multiple of the atomic mass constant \(\left(m_{\mathrm{u}}\right)\) is equal to its mass number, within two significant figures.

Actinium-225 decays by successive emission of three \(\alpha\) particles. (a) Write the nuclear equations for the three decay processes. (b) Compare the neutron-to-proton ratio of the final daughter product with that of actinium-225. Which is closer to the band of stability?

State whether the following statements are true or false. If false, explain why. (a) A subcritical mass of fissionable material is unstable and likely to explode. (b) In order for fusion to occur, the colliding particles must have high kinetic energy. (c) Highly reactive fission products are considered to be no longer dangerous after two half-lives. (d) The larger the binding energy per nucleon, the more stable is the nucleus.

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