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Nuclei - Nuclear Fission and Fusion

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Nuclear Fission is the process in which a heavy nucleus (such as 92235U_{92}^{235}\text{U}) splits into two or more smaller nuclei, releasing a large amount of energy and neutrons.

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A typical fission reaction: 92235U+01n→56141Ba+3692Kr+301n+Q_{92}^{235}\text{U} + _{0}^{1}\text{n} \rightarrow _{56}^{141}\text{Ba} + _{36}^{92}\text{Kr} + 3_{0}^{1}\text{n} + Q. The energy released QQ is approximately 200 MeV200 \text{ MeV} per fission.

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Nuclear Chain Reaction: If at least one neutron from a fission event causes another fission, a chain reaction occurs. The multiplication factor kk determines the state: k=1k=1 (critical/steady), k>1k>1 (supercritical/explosive), k<1k<1 (subcritical/dies out).

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Nuclear Reactor Components: Moderator (slows down fast neutrons using D2OD_2O or Graphite), Control Rods (absorbs neutrons using Cadmium or Boron), and Coolant (transfers heat to produce steam).

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Nuclear Fusion is the process where two light nuclei combine to form a heavier nucleus (e.g., fusion of Hydrogen isotopes into Helium). This occurs at very high temperatures (T≈107 KT \approx 10^7 \text{ K}) to overcome the strong electrostatic repulsion between nuclei.

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Binding Energy per Nucleon: Fusion occurs for light nuclei (A<30A < 30) and Fission occurs for heavy nuclei (A>170A > 170) because the resulting products have higher binding energy per nucleon, leading to increased stability and energy release.

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Mass-Energy Equivalence: In both processes, the total mass of the products is less than the total mass of the reactants. This mass defect Δm\Delta m is converted into energy according to Einstein's equation E=Δmc2E = \Delta m c^2.

📐Formulae

E=Δmc2E = \Delta m c^2

Δm=[∑mreactants]−[∑mproducts]\Delta m = [\sum m_{\text{reactants}}] - [\sum m_{\text{products}}]

Q=Δm×931.5 MeV (if Δm is in amu)Q = \Delta m \times 931.5 \text{ MeV} \text{ (if } \Delta m \text{ is in amu)}

k=number of neutrons present at the beginning of a generationnumber of neutrons present at the beginning of the previous generationk = \frac{\text{number of neutrons present at the beginning of a generation}}{\text{number of neutrons present at the beginning of the previous generation}}

12H+12H→13H+11H+4.03 MeV_{1}^{2}\text{H} + _{1}^{2}\text{H} \rightarrow _{1}^{3}\text{H} + _{1}^{1}\text{H} + 4.03 \text{ MeV}

💡Examples

Problem 1:

Calculate the energy released in the fission of 1 kg1 \text{ kg} of 92235U_{92}^{235}\text{U}, assuming 200 MeV200 \text{ MeV} is released per fission event.

Solution:

  1. Number of atoms in 1 kg1 \text{ kg} of 92235U=MassMolar Mass×NA=1000235×6.022×1023≈2.56×1024 atoms_{92}^{235}\text{U} = \frac{\text{Mass}}{\text{Molar Mass}} \times N_A = \frac{1000}{235} \times 6.022 \times 10^{23} \approx 2.56 \times 10^{24} \text{ atoms}.
  2. Total energy E=N×200 MeV=2.56×1024×200×106×1.6×10−19 JoulesE = N \times 200 \text{ MeV} = 2.56 \times 10^{24} \times 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ Joules}.
  3. E≈8.2×1013 JE \approx 8.2 \times 10^{13} \text{ J}.

Explanation:

We first find the total number of nuclei in the given mass using Avogadro's number and then multiply by the energy released per nucleus, converting units from MeV\text{MeV} to Joules.

Problem 2:

Determine the mass defect and energy released in the fusion reaction: 12H+13H→24He+01n_{1}^{2}\text{H} + _{1}^{3}\text{H} \rightarrow _{2}^{4}\text{He} + _{0}^{1}\text{n}. (Given: m(12H)=2.0141 um(_{1}^{2}\text{H}) = 2.0141 \text{ u}, m(13H)=3.0160 um(_{1}^{3}\text{H}) = 3.0160 \text{ u}, m(24He)=4.0026 um(_{2}^{4}\text{He}) = 4.0026 \text{ u}, m(01n)=1.0087 um(_{0}^{1}\text{n}) = 1.0087 \text{ u})

Solution:

  1. Mass of reactants =2.0141+3.0160=5.0301 u= 2.0141 + 3.0160 = 5.0301 \text{ u}.
  2. Mass of products =4.0026+1.0087=5.0113 u= 4.0026 + 1.0087 = 5.0113 \text{ u}.
  3. Mass defect Δm=5.0301−5.0113=0.0188 u\Delta m = 5.0301 - 5.0113 = 0.0188 \text{ u}.
  4. Energy Q=0.0188×931.5 MeV≈17.51 MeVQ = 0.0188 \times 931.5 \text{ MeV} \approx 17.51 \text{ MeV}.

Explanation:

The mass defect is the difference between the total initial mass and total final mass. This missing mass is converted into the QQ-value of the fusion reaction.

Nuclear Fission and Fusion - Revision Notes & Key Formulas | CBSE Class 12 Physics