Nuclear equations Knowledge-Based Study

Study notes

1. A nuclear equation represents changes that occur within atomic nuclei during radioactive decay, nuclear fission, fusion, or artificial transmutation. Unlike chemical reactions, where atoms rearrange while preserving elemental identities, nuclear reactions often convert one element into another through changes in the nucleus.

   Notation of Nuclides and Subatomic Particles

   Nuclides and particles are represented using standard isotopic notation: $^{A}_{Z}\text{X}$, where:

   • $A$ (Mass Number): The total number of nucleons (protons + neutrons) in the nucleus.
   • $Z$ (Atomic Number): The number of protons, which determines the chemical identity of the element.
   • $\text{X}$: The chemical symbol of the element or particle.

   Fundamental Conservation Laws

   Every valid nuclear equation must satisfy two strict conservation principles:

   1. Conservation of Mass Number ($\sum A$): The sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products.
   2. Conservation of Atomic Number or Nuclear Charge ($\sum Z$): The sum of the atomic numbers (charges) of the reactants must equal the sum of the atomic numbers (charges) of the products.

   Balancing nuclear equations requires familiarity with the standard symbols, mass numbers, and charges of emitted or absorbed particles:

   • Alpha Particle ($\alpha$): A helium-4 nucleus consisting of two protons and two neutrons, denoted as $^{4}_{2}\text{He}$ or $^{4}_{2}\alpha$.
   • Beta-Minus Particle ($\beta^-$): A high-speed electron ejected from the nucleus when a neutron converts into a proton, denoted as $^{0}_{-1}\text{e}$ or $^{0}_{-1}\beta$.
   • Positron ($\beta^+$): The antimatter counterpart of an electron, carrying a $+1$ charge, denoted as $^{0}_{+1}\text{e}$ or $^{0}_{+1}\beta$.
   • Gamma Ray ($\gamma$): High-energy electromagnetic radiation with no mass and no charge, denoted as $^{0}_{0}\gamma$.
   • Neutron: A neutral nucleon denoted as $^{1}_{0}\text{n}$.
   • Proton: A hydrogen-1 nucleus denoted as $^{1}_{1}\text{p}$ or $^{1}_{1}\text{H}$.

   When unstable parent nuclei spontaneously decay into more stable daughter nuclei, they follow predictable pathways governed by nuclear conservation laws:

   1. Alpha ($\alpha$) Decay

   Heavy, unstable nuclei often reduce mass by ejecting an alpha particle. The mass number decreases by 4, and the atomic number decreases by 2:

   $$^{238}_{92}\text{U} \rightarrow\, ^{234}_{90}\text{Th} +\, ^{4}_{2}\text{He}$$

   2. Beta-Minus ($\beta^-$) Decay

   In neutron-rich nuclei, a neutron transforms into a proton and an electron. The mass number remains unchanged, while the atomic number increases by 1:

   $$^{14}_{6}\text{C} \rightarrow\, ^{14}_{7}\text{N} +\, ^{0}_{-1}\text{e}$$

   3. Positron Emission ($\beta^+$)

   In proton-rich nuclei, a proton converts into a neutron, ejecting a positron. The mass number remains constant, while the atomic number decreases by 1:

   $$^{11}_{6}\text{C} \rightarrow\, ^{11}_{5}\text{B} +\, ^{0}_{+1}\text{e}$$

   4. Electron Capture

   An inner-orbital electron is captured by the nucleus, combining with a proton to form a neutron:

   $$^{81}_{37}\text{Rb} +\, ^{0}_{-1}\text{e} \rightarrow\, ^{81}_{36}\text{Kr}$$

   To determine an unknown nuclide or particle in a nuclear reaction, set up algebraic equations for the mass numbers and atomic numbers separately.

   Step-by-Step Example

   Determine the missing nuclide $^{A}_{Z}\text{X}$ in the following neutron bombardment reaction:

   $$^{27}_{13}\text{Al} +\, ^{1}_{0}\text{n} \rightarrow\, ^{A}_{Z}\text{X} +\, ^{4}_{2}\text{He}$$

   1. Balance Mass Numbers ($A$):
      Reactant sum: $27 + 1 = 28$
      Product sum: $A + 4 = 28 \implies A = 24$
   2. Balance Atomic Numbers ($Z$):
      Reactant sum: $13 + 0 = 13$
      Product sum: $Z + 2 = 13 \implies Z = 11$
   3. Identify the Element:
      Locate atomic number $Z = 11$ on the periodic table, which corresponds to sodium ($\text{Na}$).

   The completed and balanced nuclear equation is:

   $$^{27}_{13}\text{Al} +\, ^{1}_{0}\text{n} \rightarrow\, ^{24}_{11}\text{Na} +\, ^{4}_{2}\text{He}$$