Problem 42 - University Physics with Modern Physics

Question

The United States uses \(1.0 \times 10^{20} \mathrm{J}\) of electrical energy per year. If all this energy came from the fission of \(^{235} \mathrm{U},\) which releases 200 MeV per fission event, (a) how many kilograms of 235 \(\mathrm{U}\) would be used per year and (b) how many kilograms of uranium would have to be mined per year to provide that much \(^{235} \mathrm{U} ?\) (Recall that only 0.70\(\%\) of naturally occurring uranium is \(^{235} \mathrm{U} .\) )

Step-by-Step Solution

Answer

1218 kg of \(^{235}\mathrm{U}\) are used per year, and 174,000 kg of uranium must be mined annually.

Step 1: Convert Energy Per Year to MeV

First, convert the annual electrical energy usage from joules to MeV, as the energy from fission is given in MeV. The conversion factor is 1 J = 6.242 × 10¹² MeV.Thus, \[1.0 \times 10^{20} \text{ J/year} \times 6.242 \times 10^{12} \text{ MeV/J} = 6.242 \times 10^{32} \text{ MeV/year}\]

Step 2: Calculate Number of Fission Events

The energy per fission event is 200 MeV. We need to find out how many such events are required to produce the total annual energy. Divide the total energy in MeV by the energy per fission event:\[\text{Number of fissions} = \frac{6.242 \times 10^{32} \text{ MeV/year}}{200 \text{ MeV/fission}}\]This calculation gives:\[3.121 \times 10^{30} \text{ fissions/year}\]

Step 3: Calculate Mass of Uranium-235 Used

Each fission event uses one atom of \(^{235}\mathrm{U}\). Calculate the number of moles of \(^{235}\mathrm{U}\) used per year using Avogadro's number \(6.022 \times 10^{23}\).\[\text{Moles of } ^{235}\mathrm{U} = \frac{3.121 \times 10^{30}}{6.022 \times 10^{23}}\]This results in approximately 5183 moles per year. The molar mass of \(^{235}\text{U}\) is 235 g/mol:\[\text{Mass of } ^{235}\mathrm{U} = 5183 \text{ moles} \times 235 \text{ g/mole} = 1.218 \times 10^6 \text{ g}\]Convert grams to kilograms:\[1.218 \times 10^3 \text{ kg/year}\]

Step 4: Calculate Total Uranium Mined

Only 0.7% of naturally occurring uranium is \(^{235}\text{U}\). Calculate the total mass of uranium that needs to be mined by dividing the mass of \(^{235}\text{U}\) by 0.007:\[\text{Total uranium mined} = \frac{1.218 \times 10^3 \text{ kg/year}}{0.007}\]This results in approximately:\[1.74 \times 10^5 \text{ kg/year}\]

Step 5: Final Summary

The calculations show that the total mass of \(^{235}\mathrm{U}\) needed per year is approximately 1218 kg. To obtain this, a total of roughly 174,000 kg of natural uranium would need to be mined annually.

Key Concepts

Energy Conversion

Energy conversion is a key concept in many technologies, including nuclear fission. In physics, it involves changing one form of energy into another. In the context of nuclear fission of uranium, we convert the energy released from splitting uranium nuclei into usable electrical energy.
This process starts with the absorption of a neutron by a uranium-235 ( U^{235} ) nucleus, which becomes unstable and splits into smaller nuclei, releasing energy in the form of kinetic energy and gamma rays. Each fission event of uranium-235 releases approximately 200 MeV (million electron volts).

• Understanding energy conversion is critical to calculating how nuclear energy can satisfy energy needs.
• In a nuclear reactor, the energy produced from millions of such events is converted into thermal energy, which is then used to produce electricity.

Converting different units of energy, such as from electron volts (eV) to joules (J) as shown above, is essential in applying these concepts to real-world energy needs.

Uranium Mining

Uranium mining is the process of extracting uranium ore from the ground, which is then processed to fuel nuclear reactors. Since only a small portion of naturally occurring uranium is uranium-235, significant amounts of uranium need to be mined to obtain sufficient quantities of ^{235}U for energy production.
Naturally occurring uranium is primarily composed of uranium-238, with only about 0.7% being uranium-235, which is fissile. Therefore, to obtain enough uranium-235, we often need to mine and process larger quantities of ore.

• Mining involves extracting uranium ore and then processing it to concentrate the uranium-235.
• To meet the energy demand outlined in the exercise, a significant volume of uranium ore must be extracted and refined.
• Uranium mining is a crucial first step in the nuclear fuel cycle that supplies power plants with the fuel needed to produce electricity.

This step is critical as it directly impacts both the economic and environmental aspects of nuclear energy.

Moles and Molar Mass

In chemistry, understanding moles and molar mass is essential for quantifying substances. A mole is a fundamental unit in chemistry that denotes a specific number of atoms, ions, or molecules, analogous to using "dozen" for a group of twelve objects.
Molar mass refers to the mass of a given substance per mole of that substance, expressed in grams per mole (g/mol). For uranium-235, the molar mass is 235 g/mol. This is used to convert between the number of moles and the mass of substance.

• Knowing the molar mass allows you to calculate how many moles of uranium are used, which is essential for determining the amount consumed during fission.
• The concept helps bridge the microscopic world of atoms and molecules with macroscopic amounts of material, like kilograms used in manufacturing and industry.

By utilizing Avogadro's number and the molar mass, one can translate the number of fission events into a measurable quantity of uranium.

Avogadro's Number

Avogadro's number is a constant that denotes the number of elementary particles, typically atoms or molecules, in one mole of a substance. The value is approximately 6.022 imes 10^{23} particles per mole. This constant facilitates the connection between bulk material measurements and the number of individual atoms.
In nuclear fission, it's crucial for calculating how many uranium atoms undergo fission, which directly ties to how much uranium is consumed. By applying Avogadro's number, one can figure out the number of moles of a substance from a given number of particles, and subsequently, determine its mass given the molar mass.

• This concept is foundational for understanding chemical reactions and stoichiometry in physics and chemistry.
• It enables the conversion of the dimensionless count of atoms to practical measurements, central to solving the exercise above efficiently.

Understanding Avogadro's number aids in the quantification of material in both chemical equations and real-world applications like energy production from nuclear fission.