#nuclear fusion

Nuclear Reactions - Fission, Fusion and Energy Release

What is the primary source of energy in stars like the Sun?

**Nuclear fusion** source of energy in Sun, proton-proton cycle 4p→He+2e⁺+2ν+26.7 MeV, high temperature ~10⁷ K needed to give kinetic energy to overcome repulsion, thermal motion at high T allows tunneling, energy release because He BE/A higher than H. Fission releases energy because heavy nucleus BE/A ~7.6 MeV splits to intermediate ~8.5 MeV. The energy in stars like the Sun is generated through nuclear fusion, where light nuclei (e.g., hydrogen) combine to form heavier nuclei (e.g., helium), releasing energy due to increased binding energy per nucleon. Using E_n = -13.6/n² eV, r_n = n² a₀, L

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Reactions - Fission, Fusion and Energy Release

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Nuclear Reactions - Fission, Fusion and Energy Release

What is the role of high temperature in nuclear fusion?

**Nuclear fusion** source of energy in Sun, proton-proton cycle 4p→He+2e⁺+2ν+26.7 MeV, high temperature ~10⁷ K needed to give kinetic energy to overcome repulsion, thermal motion at high T allows tunneling, energy release because He BE/A higher than H. Fission releases energy because heavy nucleus BE/A ~7.6 MeV splits to intermediate ~8.5 MeV. High temperatures provide nuclei with sufficient kinetic energy to overcome the Coulomb barrier (electrostatic repulsion), allowing them to come close enough for the nuclear force to cause fusion. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀

Ref: NCERT > Physics Book > Atoms and Nuclei > Nuclear Reactions - Fission, Fusion and Energy Release

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Hydrogen Atom Properties - Radius, Speed and Energy

Why does nuclear fusion require extremely high temperatures?

**Hydrogen atom radius** r_n = n² a₀, a₀=5.3×10⁻¹¹ m first Bohr radius, r₂=4a₀=2.12×10⁻¹⁰ m, ratio r₄/r₂ =16/4=4, r₃=9a₀, circumference 2πr_n =2π n² a₀, for n=3 circumference=2π×9×5.3×10⁻¹¹=3×10⁻⁹ m. Orbital period T =2πr/v, v_n = v₁/n, v₁=2.2×10⁶ m/s, T₂=2πr₂/v₂, v₂=1.1×10⁶ m/s, T₂≈1.21×10⁻¹⁵ s. High temperatures provide nuclei with enough kinetic energy to overcome the electrostatic repulsion (Coulomb barrier) between positively charged nuclei, enabling fusion via the nuclear force. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yie

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Mass Defect, Binding Energy and Binding Energy per Nucleon

Why is nuclear fusion more likely to release energy when light nuclei are involved?

**Binding energy calculation** from mass defect, for nucleus mass number 28 BE 224 MeV BE/A=8 MeV, A=18 BE 144 MeV BE/A=8 MeV, BE/A indicates stability, fusion of light nuclei and fission of heavy release energy because product has higher BE/A, difference released. In light nuclei (A < 30), the binding energy per nucleon is lower. When they fuse into a heavier nucleus, the binding energy per nucleon increases, releasing energy as the final system is more tightly bound. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV,

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

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Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the outcome when two light nuclei fuse into a heavier nucleus?

**Energy equivalent** E= m c², 0.005 kg matter E=0.005×9×10¹⁶=4.5×10¹⁴ J, 0.01 kg 9×10¹⁴ J, mass defect 0.1 u => BE=0.1×931.5=93.15 MeV, mass defect from BE 149.04 MeV => Δm=149.04/931.5=0.16 u, BE per nucleon 8.5 MeV A=20 total BE=170 MeV. Fusion of light nuclei increases the binding energy per nucleon, making the resulting nucleus more tightly bound and releasing energy as a result of this increased stability. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields Energy release, consistent with Bohr model and nuclear binding energy

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

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Hydrogen Atom Properties - Radius, Speed and Energy

What is the key requirement for nuclei to undergo fusion?

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. Fusion requires nuclei to overcome the Coulomb barrier (electrostatic repulsion between positively charged nuclei), which is achieved by providing high kinetic energy through elevated temperatures. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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PHYSICS

Why does nuclear fusion require extremely high temperatures?

High temperatures provide nuclei with enough kinetic energy to overcome the electrostatic repulsion (Coulomb barrier) between positively charged nuclei, enabling fusion via the nuclear force.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27), Chapter: Kinetic Theory (Latest NCERT 2026-27), Topic: RMS speed v_rms ∝ √T, temperature dependence, ratio v₂/v₁ = √(T₂/T₁) and calculation

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