krit.club logo

Atoms and Nuclei - Nuclear Fission and Fusion

Grade 12ICSEPhysics

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

πŸ”‘Concepts

β€’

Nuclear Fission: The process in which a heavy nucleus (like 92235U_{92}^{235}U) splits into two or more lighter nuclei of comparable masses, typically after absorbing a slow neutron. This process is accompanied by the release of a massive amount of energy (QQ-value) and additional neutrons.

β€’

Nuclear Fusion: The process where two or more lighter nuclei (like 12H_{1}^{2}H and 13H_{1}^{3}H) combine to form a heavier, more stable nucleus. This requires extremely high temperatures (around 10710^7 K) to overcome the strong electrostatic repulsion between nuclei, known as the Coulomb barrier.

β€’

Mass Defect (Deltam\\Delta m): The difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus. This 'lost' mass is converted into binding energy as per Einstein's equation E=Ξ”mc2E = \Delta m c^2.

β€’

Binding Energy per Nucleon (BE/ABE/A): This is a measure of nuclear stability. Nuclei with a higher binding energy per nucleon (peaking at 2656Fe_{26}^{56}Fe) are more stable. Fission occurs for A>120A > 120 and fusion for A<20A < 20 to move towards the peak of stability.

β€’

Chain Reaction: In fission, the neutrons released can induce further fission in neighboring nuclei. If at least one neutron per fission triggers another, it is a self-sustaining reaction. Controlled chain reactions are used in nuclear reactors, while uncontrolled ones are used in atomic bombs.

β€’

Critical Mass: The minimum mass of fissile material required to maintain a self-sustaining nuclear chain reaction.

πŸ“Formulae

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

E=Ξ”mΓ—c2E = \Delta m \times c^2

BE=Ξ”m(u)Γ—931.5 MeVBE = \Delta m (u) \times 931.5 \, MeV

Bˉ=BEA\bar{B} = \frac{BE}{A}

Q=[Ξ£Mreactantsβˆ’Ξ£Mproducts]Γ—c2Q = [\Sigma M_{reactants} - \Sigma M_{products}] \times c^2

πŸ’‘Examples

Problem 1:

Calculate the energy released (in MeVMeV) in the following fission reaction: 92235U+01nβ†’56141Ba+3692Kr+301n_{92}^{235}U + _{0}^{1}n \rightarrow _{56}^{141}Ba + _{36}^{92}Kr + 3_{0}^{1}n. Given masses: m(92235U)=235.0439 um(_{92}^{235}U) = 235.0439 \, u, m(56141Ba)=140.9177 um(_{56}^{141}Ba) = 140.9177 \, u, m(3692Kr)=91.8954 um(_{36}^{92}Kr) = 91.8954 \, u, and m(01n)=1.00866 um(_{0}^{1}n) = 1.00866 \, u.

Solution:

  1. Calculate mass of reactants: M1=235.0439+1.00866=236.05256 uM_1 = 235.0439 + 1.00866 = 236.05256 \, u.
  2. Calculate mass of products: M2=140.9177+91.8954+3(1.00866)=235.83908 uM_2 = 140.9177 + 91.8954 + 3(1.00866) = 235.83908 \, u.
  3. Find mass defect: Ξ”m=M1βˆ’M2=236.05256βˆ’235.83908=0.21348 u\Delta m = M_1 - M_2 = 236.05256 - 235.83908 = 0.21348 \, u.
  4. Energy released: Q=0.21348Γ—931.5 MeVβ‰ˆ198.86 MeVQ = 0.21348 \times 931.5 \, MeV \approx 198.86 \, MeV.

Explanation:

The energy released in a nuclear reaction is the product of the mass defect (difference between total reactant mass and total product mass) and the energy equivalent of one atomic mass unit (931.5 MeV931.5 \, MeV).

Problem 2:

In a hydrogen fusion reaction, two deuterons (12H_{1}^{2}H) fuse to form a helium nucleus (24He_{2}^{4}He). If the mass of 12H_{1}^{2}H is 2.0141 u2.0141 \, u and 24He_{2}^{4}He is 4.0026 u4.0026 \, u, find the energy released.

Solution:

  1. Reaction: 212H→24He2 _{1}^{2}H \rightarrow _{2}^{4}He.
  2. Mass of reactants: 2Γ—2.0141=4.0282 u2 \times 2.0141 = 4.0282 \, u.
  3. Mass of product: 4.0026 u4.0026 \, u.
  4. Ξ”m=4.0282βˆ’4.0026=0.0256 u\Delta m = 4.0282 - 4.0026 = 0.0256 \, u.
  5. Energy E=0.0256Γ—931.5 MeV=23.8464 MeVE = 0.0256 \times 931.5 \, MeV = 23.8464 \, MeV.

Explanation:

Fusion of light nuclei results in a more stable configuration with a higher binding energy per nucleon, leading to a significant release of energy equivalent to the mass lost during the process.

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