Question

The fission of \({ }_{92}^{235} \mathrm{U}\) releases \(2.1 \times 10^{13}\) joules per mole of \({ }_{92}^{235} \mathrm{U}\). Calculate the energy released per atom and per gram of \({ }_{92}^{235} \mathrm{U}\).

Short answer

The energy released per atom of \({ }_{92}^{235} \mathrm{U}\) is \(3.487 \times 10^{-11} \, \text{J/atom}\) and the energy released per gram of \({ }_{92}^{235} \mathrm{U}\) is \(8.936 \times 10^{10} \, \text{J/g}\).

Step-by-step solution

Identifying the given information

We are given that the energy released per mole of \({ }_{92}^{235} \mathrm{U}\) is \(2.1 \times 10^{13} \, \text{J/mol}\). We will use this along with Avogadro's number and the molar mass of \({ }_{92}^{235} \mathrm{U}\) to calculate the energy released per atom and per gram.

Finding the energy released per atom of \({ }_{92}^{235} \mathrm{U}\)

We will divide the energy released per mole of \({ }_{92}^{235} \mathrm{U}\) by Avogadro's number, which is \(6.022 \times 10^{23} \, \text{atoms/mol}\), to find the energy released per atom: \[ \frac{2.1 \times 10^{13} \, \text{J/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} = 3.487 \times 10^{-11} \, \text{J/atom} \] So the energy released per atom of \({ }_{92}^{235} \mathrm{U}\) is \(3.487 \times 10^{-11} \, \text{J/atom}\).

Finding the energy released per gram of \({ }_{92}^{235} \mathrm{U}\)

To find the energy released per gram, we will divide the energy released per mole by the molar mass of \({ }_{92}^{235} \mathrm{U}\), which is \(235 \, \text{g/mol}\): \[ \frac{2.1 \times 10^{13} \, \text{J/mol}}{235 \, \text{g/mol}} = 8.936 \times 10^{10} \, \text{J/g} \] So the energy released per gram of \({ }_{92}^{235} \mathrm{U}\) is \(8.936 \times 10^{10} \, \text{J/g}\).

Final Answer

The energy released per atom of \({ }_{92}^{235} \mathrm{U}\) is \(3.487 \times 10^{-11} \, \text{J/atom}\) and the energy released per gram of \({ }_{92}^{235} \mathrm{U}\) is \(8.936 \times 10^{10} \, \text{J/g}\).

Key concepts

Energy Calculation

Calculating the energy released during nuclear fission is crucial in understanding how much energy can be obtained from a nuclear reaction. In the case of uranium-235 (\( _{92}^{235} \mathrm{U} \)), knowing that fission releases \(2.1 \times 10^{13}\) joules per mole helps to determine the efficiency and power output of a nuclear reactor.
To find how much energy is released per atom, we divide the total energy per mole by Avogadro's Number, because a mole contains exactly \(6.022 \times 10^{23}\) atoms. This division provides insight into the energy dispersed by each individual uranium atom when it undergoes fission.
The energy released per gram is another vital calculation. Here, we take the energy per mole and divide it by the molar mass of uranium-235. This calculation gives the amount of energy obtained from a given mass of uranium, helping us estimate fuel requirements and efficiency in practical applications.

Avogadro's Number

Avogadro's Number, \(6.022 \times 10^{23}\) atoms/mol, is a constant that allows us to convert between the macroscopic and microscopic worlds in chemistry and physics.
Named after Amedeo Avogadro, this constant represents the number of constituent particles, usually atoms or molecules, found in one mole of a substance. It's an essential bridge that connects the atomic scale with quantities we can measure and observe.
In our energy calculation, Avogadro's Number helps determine how much energy is released from a single atom of uranium-235. By dividing the total energy per mole by this constant, we make it possible to assess the energy impact of a single nuclear reaction, down to the atomic level. This is important for theoretical calculations and understanding the fundamental principles of nuclear energy.

Molar Mass of Uranium-235

The molar mass of \( _{92}^{235} \mathrm{U} \) is essential in energy calculations because it quantifies the mass of 1 mole of uranium-235. It is represented as \(235 \, \text{g/mol}\) and signifies that one mole of uranium-235, containing \(6.022 \times 10^{23}\) atoms, weighs 235 grams.
Understanding the molar mass helps us relate the fission energy per mole to the energy per gram of uranium. By dividing the energy per mole by the molar mass, we can predict the energy output from any given mass of fuel. This conversion is particularly vital in nuclear power generation.
Recognizing the molar mass in calculations ties together the mass of uranium-235 with practical outcomes like energy density and fuel usage efficiency, offering critical insights into the design and operation of nuclear reactors.