Problem 37 - College Physics

Question

The United States uses \(1.0 \times 10^{19} \mathrm{J}\) 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; (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

(a) 1.217 x 10^5 kg of \(^{235}U\) is needed per year; (b) 1.74 x 10^7 kg of uranium must be mined.

Step 1: Convert MeV to Joules

First, we need to understand the energy release per fission in joules. Given that 1 MeV is equivalent to \(1.602 \times 10^{-13}\) J, the energy released per fission event is \(200 \times 1.602 \times 10^{-13} = 3.204 \times 10^{-11}\) J per fission.

Step 2: Number of Fissions Needed

Next, calculate the number of fission events required to produce \(1.0 \times 10^{19}\) J of energy in a year. Divide the total energy needed by the energy per fission event: \[\frac{1.0 \times 10^{19} \text{ J}}{3.204 \times 10^{-11} \text{ J/fission}} = 3.12 \times 10^{29} \text{ fissions}.\]

Step 3: Mass of Uranium required

Knowing that each fission event involves one nucleus of \(^{235}U\), we convert the total number of fissions to moles using Avogadro's number \(6.022 \times 10^{23}\) fissions/mole. Thus, the moles of \(^{235}U\) needed are \(\frac{3.12 \times 10^{29}}{6.022 \times 10^{23}} = 5.18 \times 10^{5}\) moles. Since the molar mass of \(^{235}U\) is about 235 g/mol, the mass needed is \(5.18 \times 10^{5} \times 235 \text{ g} = 1.217 \times 10^{8} \text{ g} = 1.217 \times 10^{5} \text{ kg}.\)

Step 4: Adjust for Natural Uranium

Since only 0.70\(\%\) of naturally occurring uranium is \(^{235}U\), the total uranium mined must account for this. Calculate the total mass of uranium needed by dividing the mass of \(^{235}U\) by 0.007: \[\frac{1.217 \times 10^{5} \text{ kg}}{0.007} \approx 1.74 \times 10^{7} \text{ kg}.\]

Key Concepts

Uranium-235

Uranium-235 is a specific isotope of uranium which makes it highly significant for producing nuclear energy. Unlike the more abundant Uranium-238, U-235 is fissile, meaning it can sustain a nuclear chain reaction. This property is why it is extensively used in nuclear reactors and nuclear weapons. Although natural uranium contains only about 0.7% U-235, it is the key player in energy production.
Uranium is a dense metal with a high atomic number, 92, signifying the number of protons in its nucleus. What makes U-235 special is its ability to capture a neutron, become unstable, and split into smaller nuclei, releasing a significant amount of energy in the process. This "fission" of the U-235 nucleus is a key part of the nuclear energy production process.

Fission

Fission is a nuclear process where the nucleus of an atom splits into two or more smaller nuclei, along with the release of energy. In the case of Uranium-235, when it absorbs a neutron, it becomes unstable and undergoes fission. The fission process is what releases the energy stored in the nucleus of U-235, around 200 MeV per fission event.
During fission, additional neutrons are also released, which can then induce fission in other nearby U-235 nuclei, creating a chain reaction. This type of reaction is the fundamental principle behind the operation of nuclear reactors, where the process is controlled, and nuclear bombs, where the reaction runs uncontrolled. Managing this chain reaction is crucial since it determines the amount of energy produced. The energy released during fission is harnessed for electricity generation in nuclear power plants.

Joule Conversion

Understanding joule conversion is essential for calculating energy outputs in nuclear reactions. For instance, we start by converting the energy released by a single fission event from MeV to joules. One mega-electronvolt (MeV) is equivalent to approximately \(1.602 \times 10^{-13}\) joules.
By converting 200 MeV (energy released per fission of a U-235 atom) to joules, we get \(200 \times 1.602 \times 10^{-13} = 3.204 \times 10^{-11}\) joules per fission. This conversion is crucial when comparing the energy produced by nuclear reactions to the total energy consumed or required. In industrial-scale energy production and scientific calculations, using joules allows for a standard unit of measurement, making global energy assessments and comparisons feasible.

Energy Consumption

The tremendous energy consumption of modern societies underlines why nuclear energy is significant. For example, the U.S. annual energy consumption can be estimated to be around \(1.0 \times 10^{19}\) joules. Understanding how much uranium would need to be used for providing this vast amount of energy is crucial for planning and resource management.
This example illustrates the significance of nuclear energy in meeting high energy demands, as the fission of uranium-235 offers a potent source of energy. Efficiently using U-235 for energy helps reduce dependency on fossil fuels, thus impacting economic and environmental dynamics. Addressing energy consumption is integral to the wider conversation about energy policy and sustainability.

Avogadro's Number

Avogadro's Number, approximately \(6.022 \times 10^{23}\), is a fundamental constant that defines the number of constituent particles, usually atoms or molecules, in one mole of a substance. In the context of nuclear energy, it is used to calculate the number of atoms or molecules needed for a reaction.
When converting the number of fission events into moles of U-235, Avogadro's Number is essential. For instance, if we need \(3.12 \times 10^{29}\) fission events, we divide by Avogadro's Number to find out the number of moles of uranium required. This allows us to figure out the mass of uranium needed for producing energy on a commercial scale, linking atomic scale reactions to real-world energy requirements.