☰ Contents · Physics

Nuclear reactions and the displacement rule

Lessons 42 · 1 lessons · N. Sh. Turdiyev, K. A. Tursunmetov, A. G. Ganiyev, K. T. Suyarov, J. E. Usarov, A. K. Avliyoqulov. Physics Grade 11, 1st edition. Niso Poligraf Publishing House, Tashkent, 2018
42

Nuclear reactions. The displacement law

Textbook: pp. 170–172
GoalKnow the conservation laws in nuclear reactions and the displacement rules for α, β and γ radiation; complete reaction equations and calculate the energy release by Q = Δm·c².
New words
nuclear reaction · yadro reaksiyasidisplacement rule · siljish qoidasiparent and daughter nucleus · ona va bola yadroenergy release · energiya chiqishi
Explanation

A nuclear reaction is the interaction of a nucleus with another nucleus or particle that transforms it into other nuclei. In every reaction electric charge, the number of nucleons (mass number), total energy, momentum and angular momentum are conserved; so in an equation the sums of the lower indices (Z) and of the upper indices (A) must be equal on both sides. Displacement rules: in α-decay (the nucleus emits ⁴₂He) A decreases by 4 and Z by 2, so the element moves two places to the left in the periodic table; in β⁻-decay a neutron in the nucleus turns into a proton, emitting an electron and an antineutrino: A is unchanged, Z increases by 1 (one place to the right); in β⁺-decay a positron is emitted and Z decreases by 1; in γ-radiation an excited nucleus drops to its ground state and emits a photon, so neither A nor Z changes. The energy release is Q = (Σm_initial − Σm_final)·c²: if Q > 0 the reaction is exothermic (energy is released), if Q < 0 endothermic (energy is absorbed). With masses in u, Q (MeV) = Δm·931.5. Artificial radioactivity was discovered in 1934 by I. and F. Joliot-Curie by bombarding aluminium with α-particles.

Worked examples
²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He: A: 226 = 222 + 4, Z: 88 = 86 + 2, correct. β⁻: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e + antineutrino: A: 14 = 14 + 0, Z: 6 = 7 − 1.
²₁H + ²₁H → ³₂He + ¹₀n. Atomic masses: 2·2.01410 = 4.02820 u at the start; at the end 3.01603 + 1.00866 = 4.02469. Δm = 0.00351 u > 0, Q = 0.00351·931.5 ≈ 3.27 MeV, so the reaction is exothermic.
Class activity

In your notebook decide whether each step in the chain ²³⁸U → ²³⁴Th → ²³⁴Pa → ²³⁴U is α or β⁻ by checking the sums of A and Z (Th Z = 90, Pa Z = 91, U Z = 92). A paper exercise.

Practice
1
Which quantities are conserved in nuclear reactions?
2
How many α-decays must occur as ²³⁵U turns into stable ²⁰⁷Pb? (from mass number alone)
3
What is Z after the α-decay of ²²⁶Ra (Z = 88)?
4
Why does the element not change in γ-radiation?