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The common isotope of uranium, \(^2$$^3$$^8\)U, has a halflife of 4.47 \(\times\) 10\(^9\) years, decaying to \(^2$$^3$$^4\)Th by alpha emission. (a) What is the decay constant? (b) What mass of uranium is required for an activity of 1.00 curie? (c) How many alpha particles are emitted per second by 10.0 g of uranium?

Short Answer

Expert verified
(a) \(4.92 \times 10^{-18} \text{s}^{-1}\), (b) \(2.97 \times 10^{6} \text{g}\), (c) \(1.24 \times 10^{5} \text{ particles/second}\).

Step by step solution

01

Calculate the decay constant

The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by the equation \( \lambda = \frac{\ln(2)}{t_{1/2}} \). The half-life of \(^2^3^8\text{U}\) is \( 4.47 \times 10^9 \) years. First, convert the half-life to seconds:\[ t_{1/2} = 4.47 \times 10^9 \times 3.156 \times 10^7 \text{ seconds/year} = 1.411 \times 10^{17} \text{ seconds} \] Then, calculate the decay constant:\[ \lambda = \frac{\ln(2)}{1.411 \times 10^{17}} \approx 4.92 \times 10^{-18} \text{ s}^{-1} \]
02

Calculate mass of uranium required for an activity of 1.00 curie

Activity \( A \) is defined by \( A = \lambda N \), where \( N \) is the number of radioactive atoms. A curie (1 Ci) is defined as \( 3.7 \times 10^{10} \text{ disintegrations per second}\). So, rearrange the formula to find \( N \):\[ N = \frac{A}{\lambda} = \frac{3.7 \times 10^{10}}{4.92 \times 10^{-18}} \approx 7.52 \times 10^{27} \text{ atoms} \] To find the mass, convert atoms to moles using Avogadro's number (\( 6.022 \times 10^{23} \text{ atoms/mole} \)):\[ \text{Moles of } ^2^3^8\text{U} = \frac{7.52 \times 10^{27}}{6.022 \times 10^{23}} \approx 12493.7 \text{ moles} \] Finally, multiply by the molar mass of uranium (238 g/mol):\[ \text{Mass of uranium} = 12493.7 \times 238 \approx 2.97 \times 10^{6} \text{ g} \]
03

Calculate alpha emissions from 10 g of uranium per second

First, calculate the number of moles in 10 g of \(^2^3^8\text{U}\):\[ \text{Moles of } ^2^3^8\text{U} = \frac{10}{238} \approx 0.042 \text{ moles} \] This corresponds to \( N \) atoms using Avogadro's number:\[ N = 0.042 \times 6.022 \times 10^{23} \approx 2.53 \times 10^{22} \text{ atoms} \] The activity \( A \) or the rate of alpha emission is given by \( A = \lambda N \):\[ A = (4.92 \times 10^{-18}) \times (2.53 \times 10^{22}) \approx 1.24 \times 10^5 \text{ alpha particles per second} \]

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Key Concepts

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

Uranium-238
Uranium-238 is one of the most common forms of uranium found in nature. It is a radioactive isotope, which means it is unstable and will eventually decay into a different element. In this case, Uranium-238 decays into Thorium-234 through a process called alpha emission.

This isotope is important in various scientific fields, such as geology and nuclear physics, because of its long half-life. This property makes it useful for dating ancient rocks and for understanding the age of the Earth.
  • Characteristics: Uranium-238 has an atomic mass of 238 units, which mainly contributes to its heavy nature.
  • Natural Abundance: It accounts for over 99% of the uranium found in nature.
  • Applications: Apart from geological dating, Uranium-238 can be used in nuclear reactors, albeit not as a primary fuel, due to its non-fissile nature.
Decay Constant
The decay constant, symbolized by \( \lambda \), is a fundamental parameter in the study of radioactive decay. It represents the probability of decay of a radioactive atom per unit time. The decay constant is crucial because it helps us understand how quickly a radioactive isotope like Uranium-238 transforms into another element.
  • Relationship with Half-life: The decay constant is inversely proportional to the isotope's half-life. The equation \( \lambda = \frac{\ln(2)}{t_{1/2}} \) links them directly.
  • Determination: In the case of Uranium-238, its decay constant can be calculated if the half-life is known, giving insights into its decay rate.

Understanding the decay constant allows scientists to quantify the stability of an isotope. A small decay constant, such as that of Uranium-238, indicates slow decay, aligning with its long half-life.
Half-life
Half-life is a key concept in physics that describes the time taken for half of a sample of radioactive material to decay. For Uranium-238, the half-life is approximately 4.47 billion years, a staggering duration that reflects its stability relative to other radioactive materials.
  • Significance: The half-life can be used to date ancient geological formations because it provides a measure of time over which uranium has been decaying.
  • Calculation: Given the half-life, scientists can determine how long it will take for half of a uranium sample to change into thorium, making it a powerful tool in geochronology.
  • Measurement: The half-life is unaffected by external conditions such as temperature or pressure, making it a reliable constant for scientific calculations.

This constancy allows Uranium-238 to serve as a natural clock, helping us estimate the timing of events in Earth's history.
Alpha Emission
Alpha emission is a common type of radioactive decay found in heavy elements like Uranium-238. During alpha emission, the atom ejects an alpha particle, which is essentially a helium nucleus composed of two protons and two neutrons.

This process reduces the atomic mass and atomic number of the original element, leading to the formation of a new element.
  • Consequences for Uranium-238: As Uranium-238 undergoes alpha emission, it loses an alpha particle and is converted into Thorium-234.
  • Significance: Alpha particles are heavily charged and relatively massive, meaning they have low penetration power but can cause significant damage if they interact with other matter.

Understanding alpha emission is vital for various applications, including the safe handling of radioactive materials and the design of radiation shielding. The knowledge of emission rates also aids in determining the activity of uranium samples and their potential uses in scientific and industrial fields.

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