☰ Contents · Physics

Nuclear composition and binding energy

Lessons 39 · 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
39

Composition of the nucleus. Binding energy. Mass defect

Textbook: pp. 160–163
GoalKnow the composition of the nucleus (proton, neutron, Z, N, A), isotopes and isobars, nuclear forces, mass defect and binding energy E = Δm·c²; carry out calculations.
New words
nucleon · nuklonmass number · massa sonimass defect · massa defektibinding energy · bog‘lanish energiyasi
Explanation

The nucleus consists of protons (positive, charge +e) and neutrons (uncharged), the nucleons; it is about 10⁴–10⁵ times smaller than the atom (≈ 10⁻¹⁵–10⁻¹⁴ m) but holds almost all the atom's mass (over 99.9 %). Z is the number of protons (the atomic number), N the number of neutrons, A = Z + N the mass number; a nucleus is written ᴬ_Z X. Nuclei with the same Z but different A are isotopes; nuclei with the same A but different Z are isobars. The nuclear radius is R ≈ R₀·A^(1/3) with R₀ ≈ 1.2·10⁻¹⁵ m, so the density of nuclear matter is nearly the same in all nuclei, about 2·10¹⁷ kg/m³. Protons repel one another yet the nucleus holds, because between nucleons act nuclear forces, strong short-range (≈ 10⁻¹⁵ m) attractive forces that do not depend on charge. The mass of a nucleus is smaller than the sum of the masses of its nucleons: Δm = Z·m_p + (A − Z)·m_n − m_nucleus, the mass defect. The corresponding energy E_bind = Δm·c² is the energy needed to split the nucleus into nucleons; 1 u corresponds to ≈ 931.5 MeV/c². The binding energy per nucleon E_bind/A is greatest for medium-mass nuclei (≈ 8–9 MeV) and ≈ 7.6 MeV for uranium, so both fusion of light nuclei and fission of heavy ones release energy.

Worked examples
Helium-4 nucleus: Z = 2, N = 2. m_p = 1.00728 u, m_n = 1.00866 u, nuclear mass 4.00151 u: Δm = 2·1.00728 + 2·1.00866 − 4.00151 = 0.03037 u; E_bind = 0.03037·931.5 ≈ 28.3 MeV, about 7.1 MeV per nucleon.
Aluminium ²⁷₁₃Al: 13 protons, N = 27 − 13 = 14 neutrons; radius R = 1.2·10⁻¹⁵·∛27 = 1.2·3·10⁻¹⁵ = 3.6·10⁻¹⁵ m. Density ρ = m/V ≈ 27·1.67·10⁻²⁷/(4.19·(3.6·10⁻¹⁵)³) ≈ 2.3·10¹⁷ kg/m³.
Class activity

In your notebook tabulate Z, N and A for four nuclei (for example ¹²C, ¹⁶O, ⁵⁶Fe, ²³⁸U) and look for isotope and isobar pairs (¹⁴C and ¹⁴N are isobars). Paper work only, no experiment.

Practice
1
What is the mass defect?
2
The ²³⁵U nucleus has Z = 92. What is the number of neutrons?
3
If Δm = 0.01 u, what is E_bind approximately? (1 u = 931.5 MeV)
4
A nucleus with A = 16 has binding energy 128 MeV. How many MeV per nucleon?
5
Why is the mass of a nucleus smaller than the sum of the masses of its nucleons?