Problem 78 - University Physics with Modern Physics

Question

Calculate the energy released in the fission reaction \(_{92}^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow ^{140}_{54} \mathrm{Xe}+ ^{94}_{38} \mathrm{Sr}+2_{0}^{1} \mathrm{n}\) You can ignore the initial kinetic energy of the absorbed neutron. The atomic masses are \(_{92}^{235} \mathrm{U},\) 235.04392 \(\mathrm{u};\) \(^{140}_{54} \mathrm{Xe},\) 139.921636 \(\mathrm{u};\) and \(^{94}_{38} \mathrm{Sr},\) 93.915360 \(\mathrm{u}.\)

Step-by-Step Solution

Answer

The energy released is 185.0 MeV.

Step 1: Understand the Problem

In this problem, we're given a nuclear fission reaction and asked to calculate the energy released. The reaction is: \(_{92}^{235}\mathrm{U} + _{0}^{1}\mathrm{n} \, \rightarrow\, ^{140}_{54} \mathrm{Xe} + ^{94}_{38} \mathrm{Sr} + 2_{0}^{1} \mathrm{n}\). We have the masses of \(^A_{Z}X\) for uranium, xenon, strontium, and neutrons.

Step 2: Calculate Total Mass of Reactants

The total mass of the reactants includes the mass of the uranium isotope and one neutron. Mass of reactants = Mass of \(_{92}^{235}\mathrm{U}\) + Mass of \(_{0}^{1}\mathrm{n}\). Using given values: mass of \(_{92}^{235}\mathrm{U} = 235.04392\,\mathrm{u}\) and \(_{0}^{1}\mathrm{n} = 1.008665\,\mathrm{u}\). Hence, Total mass of reactants = \(235.04392 + 1.008665 = 236.052585\,\mathrm{u}\).

Step 3: Calculate Total Mass of Products

The total mass of the products includes the masses of xenon, strontium, and two neutrons. Mass of \(^A_{Z}X\) = \(^{140}_{54}\mathrm{Xe} = 139.921636\,\mathrm{u}\), \(^{94}_{38}\mathrm{Sr} = 93.915360\,\mathrm{u}\), and \(_{0}^{1}\mathrm{n} = 1.008665\,\mathrm{u}\). Total mass of products = \(139.921636 + 93.915360 + 2 \times 1.008665\) = \(139.921636 + 93.915360 + 2.017330 = 235.854326\,\mathrm{u}\).

Step 4: Calculate the Mass Defect

The mass defect (\(\Delta m\)) is the difference between the total mass of the reactants and the total mass of the products. \(\Delta m = 236.052585\,\mathrm{u} - 235.854326\,\mathrm{u} = 0.198259\,\mathrm{u}\).

Step 5: Convert Mass Defect to Energy Released

Use Einstein's energy-mass equivalence formula \(E = \Delta m \cdot c^2\) to calculate the energy released. First, convert \(\Delta m\) to kilograms: \(1\,\mathrm{u} = 1.660539040\times 10^{-27}\,\mathrm{kg}\). \(\Delta m = 0.198259\,\mathrm{u} \times 1.660539040\times 10^{-27}\,\mathrm{kg/u} = 3.2918\times 10^{-28}\,\mathrm{kg}\). Then use \(c = 3\times 10^8\,\mathrm{m/s}\): \(E = 3.2918\times 10^{-28}\,\mathrm{kg} \cdot (3\times 10^8\,\mathrm{m/s})^2 = 2.96262\times 10^{-11}\,\mathrm{J}\). Convert to MeV: \(1\,\mathrm{J} = 6.242\times 10^{12}\,\mathrm{MeV}\).Energy = \(2.96262\times 10^{-11}\,\mathrm{J} \times 6.242\times 10^{12}\,\mathrm{MeV/J} = 185.0\,\mathrm{MeV}\).

Key Concepts

Mass Defect

In nuclear physics, the term *mass defect* refers to the difference between the total mass of the nucleons that make up a nucleus and the actual mass of the nucleus itself. This discrepancy is pivotal in understanding how energy is released during a nuclear reaction.

During nuclear fission, a large nucleus splits into two smaller nuclei, along with some free neutrons. The mass of the products is usually less than the mass of the reactants. This difference is known as the mass defect ( "Delta m" ) and is central in calculating the energy released in nuclear reactions.

The mass defect arises because some of the mass is converted into energy, as described by Einstein's famous equation. This occurs because the mass within a bound nucleus is less due to the binding energy that holds the nucleons together being lost to the coulomb barrier during reactions. Understanding this concept helps in comprehending why nuclear reactions can release such tremendous amounts of energy.

Energy-Mass Equivalence

The principle of *energy-mass equivalence* is elegantly encapsulated in Einstein's famous equation: "E = mc^2" . This simple yet profound formula states that mass ( "m" ) can be converted into energy ( "E" ), where "c" is the speed of light in a vacuum.

In the context of nuclear fission, the mass defect is converted into energy. When a uranium nucleus splits, part of the mass is not accounted for in the mass of the products. This missing mass is transformed into the energy emitted from the fission process.

Energy-mass equivalence is a cornerstone in nuclear physics, making it indispensable for understanding how nuclear power is generated. This concept explains the enormous energy yield in nuclear reactions, which is exponentially greater than chemical reactions due to the higher binding energy involved at the nuclear level.

Atomic Mass Unit

The *atomic mass unit* (amu or just "u" ) is a standard unit of mass that quantifies mass on an atomic or molecular scale. One atomic mass unit is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its ground state.

When dealing with nuclear reactions, the atomic mass unit is extremely convenient for calculation purposes due to the small masses involved. The unit simplifies comparisons and analysis by providing a standardized measure.

In the fission reaction calculation, the atomic masses of the isotopes involved are given in atomic mass units. It facilitates the determination of the mass defect by adding the masses of nuclear reactants and products precisely in a measure that is easily relatable to energy through conversion factors, tying back to the concept of mass defect and energy-mass equivalence.

Nuclear Reaction Calculations

Nuclear reaction calculations involve determining several factors, such as reactant and product masses, energy release, and so on. The first step is to identify the reactants and products in the reaction and gather their respective atomic masses.

For the given fission reaction, start by summing up the masses of the uranium isotope and all neutrons. Then, compute the mass of the resulting elements and any ejected particles. This is essential for identifying the mass defect.

Once you've calculated this mass defect, convert it to energy using Einstein's energy-mass equivalence formula. The final result of this detailed computation gives the energy released in million electronvolts (MeV), showcasing the tremendous energy latent within the nucleus and underlining the powerful applications of nuclear fission.