Question

The United States uses about 1.4 \(\times\) 10\(^1$$^9\) J of electrical energy per year. If all this energy came from the fission of \(^2$$^3$$^5\)U, which releases 200 MeV per fission event, (a) how many kilograms of \(^2$$^3$$^5\)U would be used per year, and (b) how many kilograms of uranium would have to be mined per year to provide that much \(^2$$^3$$^5\)U? (Recall that only 0.70% of naturally occurring uranium is \(^2$$^3$$^5\)U.)

Short answer

For part (a), 170.57 kg of \\(^{235}U\\) is used per year. For part (b), approximately 24367.1 kg of uranium would need to be mined each year.

Step-by-step solution

Convert Energy Released per Fission to Joules

First, we need to convert the energy released per fission from MeV to joules. Given that 200 MeV is released per fission, we use the conversion factor 1 MeV = 1.60218 × 10⁻¹³ J.\[200 \text{ MeV} \times 1.60218 \times 10^{-13} \text{ J/MeV} = 3.20436 \times 10^{-11} \text{ J}\]

Determine Total Fissions Needed

To find out how many fission events are needed to produce the annual electricity needs, divide the total energy requirement by the energy per fission event:\[\frac{1.4 \times 10^{19} \text{ J}}{3.20436 \times 10^{-11} \text{ J/fission}} \approx 4.37 \times 10^{29} \text{ fissions}\]

Calculate Mass of Uranium Used

Calculate the mass of \(^{235}\text{U}\) atoms needed given each fission uses one atom. The molar mass of \(^{235}\text{U}\) is approximately 235 g/mol. Use Avogadro's number for the conversion:\[\frac{4.37 \times 10^{29} \text{ fissions}}{6.022 \times 10^{23} \text{ atoms/mol}} \times 235 \text{ g/mol} \approx 1.7057 \times 10^5 \text{ g}\]Convert grams to kilograms:\[1.7057 \times 10^2 \text{ kg}\]

Calculate Total Uranium Mined

Since only 0.7% of natural uranium is \(^{235}\text{U}\), calculate the total mass of uranium mined. Use the percentage to find the total initial uranium:\[\frac{1.7057 \times 10^2 \text{ kg}}{0.007} \approx 24367.1 \text{ kg}\]

Key concepts

uranium-235

Uranium-235 is a radioactive isotope of uranium that is commonly used as fuel in nuclear reactors and weapons. It has 92 protons and 143 neutrons in its nucleus, making the total mass number 235. This isotope is special because it is one of the few materials that can undergo nuclear fission, the process by which an atomic nucleus splits into smaller parts, releasing a significant amount of energy.
Uranium-235 is not very abundant in nature. Only about 0.7% of uranium found in minerals is the 235 isotope, the rest being mostly uranium-238. This rarity makes it necessary to enrich natural uranium to increase the concentration of uranium-235 in nuclear fuel applications.
When uranium-235 absorbs a neutron, it becomes unstable and splits into two smaller nuclei along with a few additional neutrons and a large amount of energy in the form of kinetic energy. This reaction is what powers nuclear reactors and atomic bombs.

energy conversion

Energy conversion in nuclear fission involves transforming the energy stored in the atomic nucleus into usable forms like heat and electricity. Each time a uranium-235 nucleus undergoes fission, it releases around 200 MeV (million electron volts) of energy.
To make this energy comprehensible for practical uses, it's converted into joules — the standard energy unit. For instance, in our specific case, the 200 MeV released per fission event equates to approximately \(3.20436 \times 10^{-11}\) Joules.
When dealing with large amounts of energy like those required by a country each year, extensive quantities of fission events are necessary. As calculated, providing 1.4 × 10\(^{19}\) J of energy annually would require about 4.37 × 10\(^{29}\) fission reactions. This energy can then be used to produce electricity for millions of homes.

Avogadro's number

Avogadro's number is a fundamental constant in chemistry and physics, often used to calculate the number of atoms or molecules in a given amount of substance. It is defined as 6.022 × 10\(^{23}\) entities per mole. This number allows us to connect the microscopic world of atoms and molecules to the macroscopic world.
In nuclear fission calculations, Avogadro's number helps determine the mass of uranium-235 needed for a specific number of fission events. For instance, knowing how many fission reactions are required, we can use this constant to calculate how much uranium-235 is needed, ensuring a seamless link between atomic-scale reactions and large-scale energy production.
This is vital for understanding how much material we need on a practical level, such as in the mining and enrichment of uranium for nuclear fuel production.

molar mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For uranium-235, its molar mass is approximately 235 g/mol.
This concept is crucial in nuclear calculations as it allows us to convert between mole-based quantities and mass-based quantities, which are more practical for large-scale operations. By knowing the molar mass of uranium-235, we can determine how many grams or kilograms are needed to sustain a given number of fission reactions.
In solving the exercise, calculating the total mass of uranium-235 utilized required knowing that one mole contains approximately 235 grams. Using the number of fissions needed annually, we converted this to gram mass, finally resulting in the calculation of uranium required per year. Molar mass thus bridges the gap between chemical quantities and real-world mass units.