Nuclear Fission Of U-235 For Ba-141 And N-141 Q-Det

Tut 2 Nuclear1 For The Fission235u N 141 Ba X 3 N Qdet

Determine the nuclide X and the energy Q released in the fission reaction 235U + n → 141 Ba + X + 3 n + Q. Assume the following atomic weights: 235U: 235 u, 141Ba: 140 u, neutron: 1 u, X: 91 u. Use the energy equivalent of 1 amu from data sheets. Additionally, analyze a nuclear reactor scenario with specified parameters: a reactor containing 100 tonnes of natural uranium, neutron flux of 10¹⁷ neutrons/m²/s, microscopic cross-sections for 235U (fission: 579 b; capture: 101 b), and 200 MeV released per fission. Use the electronic charge from data sheets.

First, to identify the nuclide X in the fission process, apply conservation of nucleons and mass energies. The initial nucleus, 235U, has 92 protons and 143 neutrons, totaling 235 nucleons. Upon fission with a neutron (1 neutron), the total nucleons are 236. Post-fission, the products are 141Ba (with 56 protons, 85 neutrons) and X, plus 3 neutrons. Since the total nucleons are conserved, the sum of nucleons in 141Ba, X, and 3 neutrons should equal 236.

Calculating the nucleons: 141 (Ba) nucleons + nucleons in X + (3 × 1) nucleons must equal 236. Thus, nucleons in X = 236 - 141 - 3 = 92. This indicates that X has 92 nucleons, which corresponds to the atomic number of uranium, meaning X must be 92U, specifically 92U.

Next, we determine the energy Q released during the fission. Use the atomic weights: 235U (235 u), 141Ba (140 u), neutron (1 u), X (92U, 92 u). However, since the atomic weight of 92U is given as 91 u in the problem, for consistency, we will use that value. The total mass before fission: (235 × 1 u) for 235U + 1 u for neutron = 236 u. The total mass after: (140 u for Ba) + (91 u for X) + (3 × 1 u for neutrons) = 140 + 91 + 3 = 234 u.

Mass defect = initial mass - final mass = 236 u - 234 u = 2 u. The energy released Q = (mass defect) × (energy equivalent of 1 u). According to data sheets, the energy equivalent of 1 u is approximately 931.5 MeV. Therefore, Q ≈ 2 × 931.5 MeV = 1863 MeV.

Now, moving to the reactor scenario, the primary goal is to compute the thermal power and the amount of 235U consumed over a day. The natural uranium contains 0.715% of 235U by weight, so for 100 tonnes (or 1×10⁵ kg) of natural uranium, the mass of 235U is: 1×10⁵ kg × 0.00715 ≈ 715 kg.

The number of atoms of 235U is given by N = (mass of 235U) / (atomic mass of 235U) × Avogadro’s number. The molar mass of 235U is 235 g/mol, so:

N = (715,000 g) / (235 g/mol) × 6.022×10²³ mol^-1 ≈ 1.83×10²⁴ atoms.

The fission rate R can be approximated by R = flux (Φ) × microscopic cross-section (σf) × number of target atoms. Convert cross-section from barns to m²: 1 b = 10^-28 m².

Calculate the macroscopic fission rate: R = Φ × σf × N = 10¹⁷ m^-2/s × 579×10^-28 m² × 1.83×10²⁴.

R ≈ 10¹⁷ × 579×10^-28 × 1.83×10²⁴ ≈ (10¹⁷ × 579×10^-28) × 1.83×10²⁴ ≈ 5.79×10^-9 × 10¹⁷ × 1.83×10²⁴.

Simplifying, R ≈ 1.06×10¹⁴ fissions per second.

The total energy produced per second (power) is then P = R × energy per fission = 1.06×10¹⁴ × 200 MeV.

Convert MeV to Joules: 1 MeV ≈ 1.602×10^-13 J.

P ≈ 1.06×10¹⁴ × 200 × 1.602×10^-13 J/s ≈ 1.06×10¹⁴ × 3.204×10^-11 ≈ 3.40×10³ Watts, or approximately 3.4 kW.

This indicates that under the given conditions, the reactor produces about 3.4 kilowatts of thermal power.

Finally, to estimate the reduction in 235U mass over one day, determine the total number of fissions in 24 hours:

Number of seconds in a day: 86400 s.

Total fissions per day = R × 86400 ≈ 1.06×10¹⁴ × 86400 ≈ 9.16×10¹⁸ fissions.

Mass of 235U consumed: number of fissions × mass of 1 atom of 235U (mass per atom = atomic mass / Avogadro’s number):

Mass per atom = 235 g / 6.022×10²³ ≈ 3.91×10^-22 g.

Mass used in one day = 9.16×10¹⁸ × 3.91×10^-22 g ≈ 0.358 g.

Proportion of original 235U used per day = 0.358 g / 715,000 g ≈ 5×10^-7.

Thus, in one day, approximately 0.000058% of the initial 235U is consumed in this reactor under the given neutron flux conditions.

Paper For Above instruction

The process of nuclear fission involves splitting heavy atomic nuclei into lighter fragments, releasing significant amounts of energy. A classic example of this process is the fission of uranium-235 (235U) when it absorbs a neutron. This reaction not only produces energy but also results in the formation of different fission products, including isotopes like barium-141 (141Ba) and a neutron flux of additional neutrons. Understanding the specifics of such reactions involves calculating the identity of the fission products and the energy released, which are fundamental to nuclear physics and reactor engineering.

In the given reaction, 235U absorbs a neutron, leading to a split into 141Ba, an unknown nuclide X, and three neutrons, along with energy Q. To identify X, we apply principles of atomic and nucleon number conservation. The initial nucleus has 92 protons and 143 neutrons; upon absorption of the incoming neutron, the total nucleons increase to 236. Post-fission, the products encompass 141Ba, X, and three neutrons. The sum of nucleons from 141Ba (140 nucleons) and the three neutrons (3 nucleons) amounts to 143, so the remaining nucleons in X must total 236 - 143 = 93. Considering proton number conservation, X must also contain 92 protons, making it uranium-92. Given the provided atomic weights, X is identified as uranium-92U.

The energy released in this fission event can be computed from the mass defect, the difference between the initial and final masses in atomic mass units (amu). Initial mass includes 235U plus one neutron: 235 u + 1 u = 236 u. The final masses are the sum of the products' masses: 140 u for Ba, 91 u for X, and 3 u for neutrons, totaling 234 u. The mass defect: 236 u - 234 u = 2 u. Using the energy equivalent of 1 u (931.5 MeV), the released energy Q is approximately 2 × 931.5 MeV ≈ 1863 MeV. This considerable energy release underpins nuclear power generation.

Moving to the practical scenario of a nuclear reactor containing 100 tonnes of natural uranium, the interest lies in calculating the reactor's thermal power output and the rate of uranium consumption. The proportion of fissile 235U in natural uranium is only 0.715%, which translates to a mass of approximately 715 kg for 100 tonnes. The number of uranium atoms can be derived from atomic weight and Avogadro's number, resulting in about 1.83×10²⁴ atoms. The neutron flux coupled with the microscopic fission cross-section (~579 barns or 579×10^-28 m²) gives a fission rate that leads to an estimated 1.06×10¹⁴ fissions per second.

Considering each fission releases 200 MeV (which equals approximately 3.204×10^-11 Joules), the reactor's power output becomes roughly 3.4 kW—significantly low for practical energy generation but illustrative of the calculation process. Over a 24-hour period, the total fissions performed are around 9.16×10¹⁸, consuming approximately 0.358 grams of 235U, which is a tiny fraction of the initial amount—about 5×10^-7. This demonstrates the efficiency and massive energy release potential per atom in nuclear fission reactions, underlying the principles of nuclear power plants and their fuel cycles.

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