Chemistry · Ch 13 — Nuclear Chemistry and Radioactivity
Nuclear Fusion
Nuclear Fusion
Nuclear fusion is the combination of two lighter nuclei into a single, heavier nucleus, accompanied by the release of an enormous amount of energy -- it is fusion, not fission, that powers the Sun and other stars, and the energy Earth receives from sunlight is ultimately fusion energy.
Four representative fusion reactions occurring in the Sun and stars: (i) ------; (ii) ------; (iii) --------; (iv) ------. Together, these form a simplified version of the proton-proton chain, the dominant pathway by which stars like the Sun convert hydrogen into helium.
Fusion has one clear advantage over fission: it produces relatively more energy per unit mass of fuel consumed. But it also has one major practical obstacle: sustaining fusion requires an extremely high temperature, typically of order K, in order for the positively charged light nuclei to get close enough together to fuse despite their mutual electrostatic repulsion -- this is why controlled, energy-positive fusion power on Earth has proved far harder to achieve in practice than fission power. …
Worked out. (i) 1-1-H + 1-1-H → 2-1-H + 0-1-e (two protons fuse to deuterium, emitting a positron). (ii) 1-1-H + 2-1-H → 3-2-He (a proton fuses with deuterium to give helium-3). (iii) 3-2-He + 3-2-He → 4-2-He + 2(1-1-H) (two helium-3 nuclei fuse to give helium-4 plus two protons). (iv) 3-2-He + 1-1-H → 4-2-He + 0-1-e (helium-3 fuses with a proton to give helium-4, emitting a positron). Together these form (a simplified version of) the proton-proton chain by which the Sun converts hydrogen into helium, releasing the energy that reaches Earth as sunlig …
Worked out. Worked example for the fusion reaction 2-1-H + 3-2-He → 4-2-He + 1-1-H. Given atomic masses: 3-2-He = 3.0160 u, 4-2-He = 4.0026 u, 1-1-H = 1.0078 u. FIDELITY NOTE: the source's own problem statement prints the mass of 2-1-H (deuterium) as '2.041 u', but its own worked solution two lines later uses '2.0141 u' -- the standard, correct atomic mass of deuterium is 2.0141 u, so '2.041 u' in the problem statement is treated as a digit-dropped extraction artifact of the same '2.0141' value, not a different, deliberately-set number; the solution's own 2.0141 u is used here. Mass defect: Δm = (mass of 2-1-H + mass of 3-2-He) - (mass of 4-2-He + mass of 1-1-H) = (2.0141 + 3.0160) - (4.0026 + 1.0078) = 5.0301 - 5.0104 = 0.0197 u. Energy releas …