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Nuclear and Quantum Physics - Fission and Fusion

Grade 11IBPhysics

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

πŸ”‘Concepts

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Mass Defect (Ξ”m\Delta m): This is the difference between the total mass of the individual constituent nucleons (protons and neutrons) and the actual mass of the nucleus. It represents the mass converted into energy when the nucleus is formed: Ξ”m=(Zmp+Nmn)βˆ’Mnucleus\Delta m = (Z m_p + N m_n) - M_{nucleus}.

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Binding Energy (BEBE): The energy required to completely separate the nucleons of a nucleus into individual particles. It is equivalent to the energy released during the formation of the nucleus, calculated using E=Ξ”mc2E = \Delta m c^2.

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Binding Energy per Nucleon (BEA\frac{BE}{A}): A measure of the stability of a nucleus. The higher the binding energy per nucleon, the more stable the nucleus. The maximum value occurs at Iron-56 (2656Fe^{56}_{26}Fe).

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Nuclear Fission: The process in which a heavy nucleus (e.g., 92235U^{235}_{92}U) captures a neutron and splits into two smaller 'daughter' nuclei of roughly equal mass, plus several neutrons and energy. This occurs for elements with A>56A > 56 as they move toward the peak of stability.

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Nuclear Fusion: The process in which two light nuclei (e.g., 12H^{2}_{1}H and 13H^{3}_{1}H) combine to form a heavier, more stable nucleus. This requires extremely high temperatures and pressures to overcome the electrostatic Coulomb repulsion between the positively charged nuclei.

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Unified Atomic Mass Unit (uu): Defined as 112\frac{1}{12} of the mass of a neutral carbon-12 atom. 1uβ‰ˆ1.661Γ—10βˆ’27kg1 u \approx 1.661 \times 10^{-27} kg, which corresponds to an energy of approximately 931.5MeV931.5 MeV.

πŸ“Formulae

E=Ξ”mc2E = \Delta m c^2

Ξ”m=[Zmp+(Aβˆ’Z)mn]βˆ’Mnucleus\Delta m = [Z m_p + (A - Z) m_n] - M_{nucleus}

Q=(mreactantsβˆ’mproducts)c2Q = (m_{reactants} - m_{products}) c^2

1u=931.5Β MeVΒ cβˆ’21 u = 931.5 \text{ MeV } c^{-2}

πŸ’‘Examples

Problem 1:

Calculate the energy released in the following fission reaction: 92235U+01n→56141Ba+3692Kr+301n^{235}_{92}U + ^{1}_{0}n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3 ^{1}_{0}n. Given masses: m(U)=235.0439um(U) = 235.0439 u, m(n)=1.0087um(n) = 1.0087 u, m(Ba)=140.9144um(Ba) = 140.9144 u, and m(Kr)=91.9262um(Kr) = 91.9262 u. Use 1u=931.5MeV1 u = 931.5 MeV.

Solution:

  1. Calculate mass of reactants: Mreact=235.0439u+1.0087u=236.0526uM_{react} = 235.0439 u + 1.0087 u = 236.0526 u.
  2. Calculate mass of products: Mprod=140.9144u+91.9262u+3(1.0087u)=235.8667uM_{prod} = 140.9144 u + 91.9262 u + 3(1.0087 u) = 235.8667 u.
  3. Calculate mass defect: Ξ”m=236.0526uβˆ’235.8667u=0.1859u\Delta m = 236.0526 u - 235.8667 u = 0.1859 u.
  4. Energy released: E=0.1859uΓ—931.5MeV/uβ‰ˆ173.17MeVE = 0.1859 u \times 931.5 MeV/u \approx 173.17 MeV.

Explanation:

The energy released is the result of the mass difference between the initial reactants and the final products, converted to energy according to mass-energy equivalence.

Problem 2:

In a fusion reaction, Deuterium (12H^{2}_{1}H) and Tritium (13H^{3}_{1}H) fuse to form Helium (24He^{4}_{2}He) and a neutron (01n^{1}_{0}n). If the energy released is 17.6MeV17.6 MeV, calculate the mass defect in atomic mass units (uu).

Solution:

  1. Use the conversion factor 1u=931.5MeV1 u = 931.5 MeV.
  2. The mass defect Ξ”m\Delta m is given by Ξ”m=E931.5\Delta m = \frac{E}{931.5}.
  3. Ξ”m=17.6MeV931.5MeV/uβ‰ˆ0.01889u\Delta m = \frac{17.6 MeV}{931.5 MeV/u} \approx 0.01889 u.

Explanation:

This demonstrates the calculation of mass defect starting from the known energy output of a nuclear reaction using the standard conversion factor for IB Physics.

Fission and Fusion - Revision Notes & Key Formulas | IB Grade 11 Physics