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\({ }^{92} \mathrm{U}^{234}\) decays to \({ }^{90} \mathrm{Th}^{230}\) which in turn decays to \({ }^{88} \mathrm{Ra}^{226}\). The half life of this uranium isotope is \(24.7 \times 10^{4}\) years, and of the thorium isotope \(8 \times 10^{4}\) years. (a) How many grams of \(^{92} \mathrm{U}^{234}\) and (b) how many grams of \(^{90} \mathrm{Th}^{230}\) will be present after a \(20 \mathrm{~g}\) sample of pure \({ }^{92} \mathrm{U}^{234}\) has decayed for \(15 \times 10^{4}\) years?

Short Answer

Expert verified
The final step is to find the mass of remaining Uranium-234 and Thorium-230 after performing these calculations. This can be done by entering the respective decay constants and decay time into the decay equation for each element and solve for the mass that remains.

Step by step solution

01

Determine the decay of Uranium-234

A decay equation can be used to determine the final amount (m) after time t of a substance that has a decay constant λ. The decay equation is: m = m0 * e^(-λt), where m0 is the initial mass and t is the decay time. Firstly, use the half-life of Uranium-234 to calculate the decay constant (λ = ln(2)/(24.7 * 10^4 years). Next, use this λ and the given decay time (t = 15 * 10^4 years) in the decay equation to find the final mass m for Uranium-234 after the given time.
02

Determine the amount decayed

Now that we have the final mass of Uranium-234, we can calculate the mass of Uranium that has decayed during the given time. This can be done by subtracting the final mass from the initial mass (20g). This shows us how much Uranium-234 has decayed into Thorium-230 over the given time period.
03

Determine the decay of Thorium-230

Now that we have the mass of Thorium-230 that was produced, we can find out how much of this remains after decaying for the same amount of time (t = 15 * 10^4 years). To do this, calculate the decay constant for Thorium-230 (λ = ln(2) / (8 * 10^4 years)) and use it in the decay equation as mentioned in step 1. This will tell us how much Thorium-230 remains after this time period.

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Key Concepts

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

Half-Life
The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. It is a measure of the rate at which a substance decays.

In mathematical terms, the half-life (\( T_{1/2} \)) is related to the decay constant (\( \lambda \)) through the formula: \( T_{1/2} = \frac{\ln(2)}{\lambda} \). This relationship is crucial for solving problems related to radioactive decay, as it allows us to find the decay constant when the half-life is known.

To find the remaining quantity of a radioisotope after a certain time, the decay formula is used: \( m = m_0 \times e^{-\lambda t} \), where \( m_0 \) is the initial mass, \( \lambda \) is the decay constant, \( t \) is time, and \( m \) is the remaining mass.

Understanding half-life helps us predict how long it will take for a substance to reduce to a certain quantity, which is particularly important in fields such as archaeology for radiocarbon dating or in environmental science to assess nuclear waste management.
  • Half-life defines the rate of decay.
  • Used to calculate the remaining quantity of a substance.
  • Helps in various scientific fields like archaeology and environmental science.
Uranium Decay
Uranium-234 decays into Thorium-230 through the process of radioactive decay. This transformation occurs in steps, where the uranium loses particles and changes its structure over time.

The decay of \( ^{92} \mathrm{U}^{234} \) involves the emission of alpha particles, and its half-life is \( 24.7 \times 10^4 \) years. In this decay process, uranium reduces its atomic number and changes into another element, usually a daughter isotope like thorium in this case.
  • The decay reduces the atomic number by emitting alpha particles.
  • It involves a large timescale due to the long half-life.
  • Changes the element from uranium to thorium.
Calculating how much uranium remains after a certain period involves using the decay equation and understanding the proportional decay over time. Observing these transformations helps scientists trace the history of geological formations and assess the age of materials.
Thorium Decay
Like uranium, thorium undergoes radioactive decay, turning into another element over time. Thorium-230 decays into radium, further progressing the chain of decay.

The half-life of \( ^{90} \mathrm{Th}^{230} \) is \( 8 \times 10^4 \) years, significantly shorter than that of uranium. This means thorium transforms into its next state more quickly, affecting how long the element remains predominant in any sample.

The decay constant for thorium can be determined similarly using the half-life formula, enabling calculations for the remaining mass after a particular time.
  • Thorium follows uranium in the decay series.
  • Has a shorter half-life compared to uranium.
  • Continues to decay into other isotopes like radium.
This decay process of thorium is crucial in understanding natural radioactive series, which can influence geological and environmental studies. It provides a timeline of earth’s geological history and is key in dating ancient materials due to its known decay rates and behavior.

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Most popular questions from this chapter

What are the differences between single particle states and many particle states? How are they related? What about \(\gamma\)-decaying states?

Why is it impossible for two protons to fuse, as in the first step of the proton-proton cycle, without a \(\beta\) decay simultaneously taking place?

A series of Mössbauer experiments is performed with the same emitter and absorber but with the emitter placed in various host materials. The absorber is always in the same host. (a) Show that the chemical shift (the absorber velocity corresponding to the center of the spectrum) is a linear function of the electron probability density \(\rho\) at the site of the emitter and so is given by \(v=a \rho+b\), where \(a\) and \(b\) do not depend on the sample in which the emitter is placed. (b) The following data was recorded for four samples : \(v_{1}=\) \(1.42 \mathrm{~mm} / \mathrm{sec}, v_{2}=0.23 \mathrm{~mm} / \mathrm{sec}, v_{3}=0.37 \mathrm{~mm} / \mathrm{sec}\), and \(v_{4}=0.95 \mathrm{~mm} / \mathrm{sec} .\) For the first two samples \(\rho\) was found using other experimental data, with the results \(\rho_{1}=8.0248 \times\) \(10^{34} \mathrm{~m}^{-3}\) and \(\rho_{2}=8.0286 \times 10^{34} \mathrm{~m}^{-3}\), respectively. Find the values of \(a\) and \(b\), then find the electron probability densities for samples 3 and 4 .

In \(\gamma\) decays from states of excitation energy around \(1 \mathrm{MeV}\), or less, to ground states, electric dipole radiation is almost never observed. Use the shell model to explain this.

Explain exactly why the optical model potential which a nucleus exerts on a bombarding nucleon of energy \(50 \mathrm{MeV}\) is different from the shell model potential which it exerts on one of its own nucleons. What would you expect the optical model potential to be like for a bombarding nucleon of energy \(5 \mathrm{MeV}\) ?

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