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Nuclear and Quantum Physics - Fission and Fusion

Grade 12IBPhysics

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

🔑Concepts

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Binding Energy per Nucleon: The stability of a nucleus is determined by its binding energy per nucleon (BE/ABE/A). Nuclei with higher BE/ABE/A are more stable. The curve peaks at 2656Fe^{56}_{26}\text{Fe}, making it the most stable isotope.

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Nuclear Fission: The process where a heavy nucleus (typically A>200A > 200, such as 92235U^{235}_{92}\text{U}) splits into two lighter daughter nuclei of roughly equal mass, usually triggered by the absorption of a slow-moving 'thermal' neutron. This process releases energy because the product nuclei have a higher binding energy per nucleon than the parent.

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Nuclear Fusion: The process where two light nuclei (typically A<10A < 10, such as 12H^{2}_{1}\text{H} and 13H^{3}_{1}\text{H}) combine to form a heavier, more stable nucleus. This requires extremely high temperatures (approx. 108 K10^8 \text{ K}) and pressure to overcome the electrostatic (Coulomb) repulsion between the positively charged nuclei.

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Mass Defect (Δm\Delta m): The difference between the total mass of the individual nucleons and the actual mass of the nucleus. This 'missing mass' is equivalent to the energy released during the formation of the nucleus: ΔE=Δmc2\Delta E = \Delta m c^2.

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Energy Release Calculation: For both fission and fusion, the energy released (ΔE\Delta E) is calculated using the change in rest mass: ΔE=(mass of reactants−mass of products)×c2\Delta E = (\text{mass of reactants} - \text{mass of products}) \times c^2.

📐Formulae

ΔE=Δmc2\Delta E = \Delta m c^2

BE=[Zmp+(A−Z)mn−Mnucleus]c2BE = [Z m_p + (A-Z) m_n - M_{nucleus}] c^2

1 u=931.5 MeV c−2=1.661×10−27 kg1 \text{ u} = 931.5 \text{ MeV } c^{-2} = 1.661 \times 10^{-27} \text{ kg}

Binding Energy per Nucleon=BEA\text{Binding Energy per Nucleon} = \frac{BE}{A}

💡Examples

Problem 1:

Calculate the energy released in the fusion reaction: 12H+13H→24He+01n^{2}_{1}\text{H} + ^{3}_{1}\text{H} \rightarrow ^{4}_{2}\text{He} + ^{1}_{0}\text{n}. Given masses: m(12H)=2.014102 um(^{2}_{1}\text{H}) = 2.014102 \text{ u}, m(13H)=3.016049 um(^{3}_{1}\text{H}) = 3.016049 \text{ u}, m(24He)=4.002603 um(^{4}_{2}\text{He}) = 4.002603 \text{ u}, m(01n)=1.008665 um(^{1}_{0}\text{n}) = 1.008665 \text{ u}.

Solution:

  1. Calculate total initial mass: minitial=2.014102+3.016049=5.030151 um_{initial} = 2.014102 + 3.016049 = 5.030151 \text{ u}.
  2. Calculate total final mass: mfinal=4.002603+1.008665=5.011268 um_{final} = 4.002603 + 1.008665 = 5.011268 \text{ u}.
  3. Find mass defect: Δm=5.030151−5.011268=0.018883 u\Delta m = 5.030151 - 5.011268 = 0.018883 \text{ u}.
  4. Convert to energy: E=0.018883×931.5 MeV≈17.59 MeVE = 0.018883 \times 931.5 \text{ MeV} \approx 17.59 \text{ MeV}.

Explanation:

The energy released in the reaction corresponds to the mass defect converted into kinetic energy of the products. Since the products have a lower total mass than the reactants, energy is liberated.

Problem 2:

In a fission reaction, a 92235U^{235}_{92}\text{U} nucleus absorbs a neutron and splits into 56141Ba^{141}_{56}\text{Ba}, 3692Kr^{92}_{36}\text{Kr}, and 3 neutrons. Write the balanced nuclear equation.

Solution:

01n+92235U→56141Ba+3692Kr+301n^{1}_{0}\text{n} + ^{235}_{92}\text{U} \rightarrow ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3^{1}_{0}\text{n}

Explanation:

In any nuclear equation, both the atomic number (ZZ) and the mass number (AA) must be conserved. Reactant total A=1+235=236A = 1 + 235 = 236; Product total A=141+92+3(1)=236A = 141 + 92 + 3(1) = 236. Reactant total Z=0+92=92Z = 0 + 92 = 92; Product total Z=56+36+3(0)=92Z = 56 + 36 + 3(0) = 92.

Fission and Fusion - Revision Notes & Key Formulas | IB Grade 12 Physics