Question

The first step in the radioactive decay of \({ }^{238}{ }_{92} \mathrm{U}\) is \({ }^{238}{ }_{92} \mathrm{U}={ }^{234}{ }_{90} \mathrm{Th}+{ }_{2}^{4} \mathrm{He}\). Calculate the energy released in this reaction. The exact masses of \({ }^{238}{ }_{92} \mathrm{U},{ }^{234}{ }_{90} \mathrm{Th}\), and \({ }_{2}^{4} \mathrm{He}\) are \(238.0508,234.0437\) and \(4.0026 \mathrm{amu}\), respectively. \(1.0073 \mathrm{amu}=1.673 \times 10^{-24} \mathrm{~g}\)

Short answer

= 238.0508 amu - (234.0437 amu + 4.0026 amu) = 238.0508 amu - 238.0463 amu = 0.0045 amu #tag_title#Step 2: Convert mass defect to energy#tag_content#Now that we have the mass defect, we can convert it to energy by using Einstein's equation E=mc^2. First, we need to convert the mass defect from amu to grams: mass defect = 0.0045 amu * (1.673 * 10^{-24} g/amu) = 0.0045 * 1.673 * 10^{-24} g = 7.529 * 10^{-27} g Now, we can plug this into Einstein's equation. Remember that the speed of light c is approximately 3 * 10^8 m/s: E = (7.529 * 10^{-27} g) * (3 * 10^8 m/s)^2 = 7.529 * 10^{-27} g * 9 * 10^{16} m^2/s^2 = 6.7761 * 10^{-10} g m^2/s^2 Finally, we can convert the energy to more convenient units, such as MeV (mega-electron volts). To do this, we need to use the conversion factor: \(1 \mathrm{MeV} = 1.60219 \times 10^{-13} \mathrm{J}\) and \(1 \mathrm{J} = 1 \mathrm{g} \cdot \mathrm{m}^2\mathrm{s}^{-2}\). E = 6.7761 * 10^{-10} g m^2/s^2 * (1 J/g m^2/s^2) * (1 MeV/1.60219 * 10^{-13} J) = 6.7761 * 10^{-10} * 1/1.60219 * 10^3 MeV ≈ 4.23 MeV The energy released in this reaction is approximately 4.23 MeV.

Step-by-step solution

Calculate the mass defect

We'll first calculate the mass defect, which is the difference between the mass of uranium-238 and the sum of the masses of thorium-234 and helium-4. Let's note down the given masses of U, Th, and He: U: 238.0508 amu Th: 234.0437 amu He: 4.0026 amu Now, find the mass defect: mass defect = mass of U - (mass of Th + mass of He)

Key concepts

Mass Defect

Understanding the concept of 'mass defect' is critical in nuclear chemistry and is the foundation for explaining the energy released during radioactive decay. Imagine mass defect as the discrepancy between the total mass of an atomic nucleus and the sum of the masses of its constituent protons and neutrons. In simpler terms, when a nucleus forms, the mass of the nucleus is less than the sum of its parts. This missing mass has been converted into energy during the fusion process that created the nucleus, and can be recovered during nuclear fission.

This energy is the binding energy, which serves as a nuclear 'glue' that holds the nucleus' components together. The mass defect is expressed using atomic mass units (amu), with the relation that 1 amu equals about 1.67 x 10^-27 kilograms. The calculation of mass defect requires subtracting the total mass of the individual separated nucleons from the mass of the actual nucleus. When a nucleus undergoes radioactive decay, the initial nucleus and the decay products may have different mass defects, and the difference in their binding energies corresponds to the energy released or absorbed in the process.

Uranium-238 Decay

Uranium-238 decay is a complex sequence known as a decay chain, which ultimately leads to a stable isotope. In the process of decay, uranium-238 emits an alpha particle, transforming into thorium-234—the first step in the series. The emitted alpha particle is actually a helium-4 nucleus, consisting of two protons and two neutrons.

To calculate the energy released when uranium-238 decays into thorium-234 and an alpha particle, we use the mass defect concept previously detailed. By taking the original mass of uranium-238 and subtracting the sum of the masses of thorium-234 and helium-4, we can find this mass defect. Since mass and energy are equivalent by Einstein's theory of relativity, represented by his famous equation E=mc^2, this mass defect can then be converted into an amount of energy released. This energy is what we observe as the radioactive decay energy. The process gives insights into the stability of atomic nuclei and the forces at work within them.

Nuclear Chemistry

Nuclear chemistry is the field of chemistry that deals with the reactions and transformations of atomic nuclei. This includes studying radioactive decay, such as the one exhibited by uranium-238, fission, where heavy nuclei split into smaller fragments, and fusion, which involves the combining of light atomic nuclei. The energy calculations in nuclear processes are enormous compared to typical chemical reactions, primarily due to the mass differences at play being on the order of 1000 times smaller, magnifying the impact of Einstein's mass-energy equivalence principle.

In the context of learning nuclear chemistry, being aware of the applications of these reactions such as in nuclear power generation, medical imaging, cancer treatment, and even archaeological dating is beneficial. It's also fascinating to see how the principles governing the atomic level repercuss in the macro world with real-world implications. The logical steps involved in solving problems in nuclear chemistry, like calculating decay energy, not only enhance students' problem-solving skills but also deepen their understanding of the subatomic world.