Nuclear Chemistry: Binding Energy, Decay Modes and Half-Lives

Neutron-to-proton ratio and the band of stability

A nucleus contains protons and neutrons, together called nucleons. The protons repel one another electrically, and the strong nuclear force, which acts only over very short distances, holds all the nucleons together. Whether a nucleus is stable depends mainly on the balance between protons and neutrons.

If you plot neutrons against protons for the stable nuclides, they fall in a narrow band of stability. Light stable nuclei have roughly equal numbers of each: nitrogen-14 has seven of each. Heavier nuclei need extra neutrons to offset the growing proton repulsion. OpenStax gives an n:p ratio of 1.15 for iron-56 and 1.52 for lead-207. Every isotope of every element beyond bismuth (Z = 83) is radioactive.

A nuclide off the band decays in a way that moves it toward the band. That single idea predicts most decay modes, as the decay-mode list below shows.

Mass defect and nuclear binding energy

A nucleus has less mass than the separate protons and neutrons it is made from. The missing mass, the mass defect, corresponds to the nuclear binding energy through E = mc². In atomic mass units, 1 u is equivalent to 931.494 MeV (CODATA value from NIST).

Worked example 1. Find the binding energy of helium-4. Using atomic masses keeps the electrons balanced: two hydrogen-1 atoms supply the two protons and two electrons. NIST lists 1.007 825 032 u for ¹H and 4.002 603 254 u for ⁴He, and CODATA gives 1.008 664 916 u for the neutron. Separate particles: 2(1.007 825 032) + 2(1.008 664 916) = 4.032 979 896 u. Mass defect: 4.032 979 896 − 4.002 603 254 = 0.030 376 642 u. Binding energy: 0.030 376 642 u × 931.494 MeV/u = 28.30 MeV, or 7.074 MeV per nucleon. OpenStax quotes 28.4 MeV because it uses masses rounded to four decimal places, which shows how sensitive a small difference between large numbers is to rounding.

Scaled to a mole, 28.30 MeV per nucleus is 2.73 × 10¹² J/mol. Burning a mole of methane releases about 890 kJ (from Appendix G formation enthalpies), so forming helium-4 from its nucleons releases roughly three million times as much energy per mole. Binding energy per nucleon is largest near mass number 56; OpenStax calculates 8.820 MeV per nucleon for iron-56. That is why fusing light nuclei and splitting very heavy ones both release energy.

Five decay modes and what each one changes

Each decay mode changes the mass number A and the atomic number Z in a fixed way. Knowing which side of the band a nuclide lies on tells you which modes are likely.

• Alpha decay emits a helium-4 nucleus, ⁴₂He. A falls by 4 and Z by 2. It is common for heavy nuclei beyond bismuth.
• Beta decay emits an electron, ⁰₋₁e, as a neutron becomes a proton. A is unchanged and Z rises by 1. It occurs in neutron-rich nuclides above the band.
• Positron emission emits a positron, ⁰₊₁e, as a proton becomes a neutron. A is unchanged and Z falls by 1. It occurs in proton-rich nuclides below the band.
• Electron capture pulls an inner electron into the nucleus, where it combines with a proton to make a neutron. A is unchanged and Z falls by 1, the same change as positron emission.
• Gamma emission releases a high-energy photon, ⁰₀γ, from a nucleus in an excited state. A and Z are unchanged. It often accompanies the other modes.

Balancing a nuclear equation by conserving A and Z

A nuclear equation is balanced when the mass numbers (superscripts) add to the same total on both sides and the atomic numbers (subscripts) do too. The atomic number of the product then identifies the element.

Uranium-238 undergoes alpha decay: ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He. Check: 238 = 234 + 4 and 92 = 90 + 2. Carbon-14 undergoes beta decay: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e, where 6 = 7 + (−1). Fluorine-18, used in PET scans, undergoes positron emission: ¹⁸₉F → ¹⁸₈O + ⁰₊₁e. Potassium-40 can undergo electron capture, with the electron written as a reactant: ⁴⁰₁₉K + ⁰₋₁e → ⁴⁰₁₈Ar.

To find an unknown particle, subtract. Thorium-234 decays to protactinium-234: ²³⁴₉₀Th → ²³⁴₉₁Pa + X. For X, A = 234 − 234 = 0 and Z = 90 − 91 = −1, so X is ⁰₋₁e and this is beta decay. The predictions fit the band: carbon-14 has an n:p ratio of 1.33, high for so light a nucleus, and beta decay brings it to 1.00 in nitrogen-14.

Half-life calculations for first-order decay

Radioactive decay is a first-order process: in a given time, the same fraction of the remaining nuclei decays, whatever the starting amount. The half-life, t½, is the time for half of the nuclei to decay. It is essentially independent of temperature, pressure and chemical form, which is what makes it useful for dating. The amount remaining after time t is N = N₀ × (½)^(t/t½). Equivalently, ln(N/N₀) = −kt with decay constant k = 0.693/t½.

Worked example 2. Iodine-131, used to treat thyroid disease, has t½ = 8.02 days (OpenStax Table 21.2). How much of a 50.0 mg sample remains after 30.0 days? t/t½ = 30.0 ÷ 8.02 = 3.741 half-lives. N = 50.0 mg × (½)^3.741 = 3.74 mg. A quick estimate confirms it: after 3 half-lives 6.25 mg would remain and after 4 half-lives 3.13 mg, so the answer must lie between them.

Worked example 3. Fluorine-18 has t½ = 110. minutes. How long until its activity falls to 10.0 percent of the starting value? Activity is proportional to the number of nuclei, so solve (½)^(t/110.) = 0.100. t = 110. min × ln(10.0) ÷ ln 2 = 365 min, about 6.1 hours. That short window is why fluorine-18 has to be made close to where it is used.

Worked example: radiocarbon dating a wooden artifact

Living organisms keep a steady ratio of carbon-14 to carbon-12 by exchanging carbon with their environment. After death, carbon-14 decays with t½ = 5730 years (OpenStax section 21.3) and is not replaced, so the ratio falls in a predictable way.

A wooden tool contains 28.0 percent of the carbon-14 found in living wood. Solve 0.280 = (½)^(t/5730 y). Taking logarithms gives t = 5730 y × ln(1/0.280) ÷ ln 2 = 1.05 × 10⁴ years, or about 1.84 half-lives. Check the estimate: after one half-life 50 percent remains and after two 25 percent, and 28.0 percent lies just short of two.

OpenStax states that the method is accurate to about 30 000 years and gives reasonable dates to a maximum of about 50 000 years. Beyond that, too little carbon-14 remains to measure reliably. Real radiocarbon ages are also corrected for past changes in atmospheric carbon-14.

Errors in nuclear equations and decay arithmetic

Two checks catch most problems: balance A and Z separately in every equation, and estimate how many half-lives have passed before using a calculator.

• Swapping beta and positron changes. Beta raises Z by 1; positron emission and electron capture lower it by 1.
• Balancing mass numbers but not atomic numbers, or the reverse.
• Subtracting the masses of bare protons from an atomic mass without accounting for electrons. Using hydrogen-1 atomic masses avoids the problem.
• Dividing the starting amount by the number of half-lives instead of halving it repeatedly.
• Mixing time units, such as a half-life in days with an elapsed time in hours.
• Rounding masses before taking the difference in a mass-defect calculation.