A brief note on The Nucleus . Read it before a test. Try each question at the end before you open its answer.
The nucleus and its size Nucleons: protons and neutrons. The atomic number Z Z Z counts protons, the neutron number is N N N , and the mass number is A = Z + N A = Z + N A = Z + N . Isotopes: same Z Z Z . Isobars: same A A A . Isotones: same N N N . Atomic mass unit: 1 u = 1 12 1\ \text{u} = \dfrac{1}{12} 1 u = 12 1 of a carbon-12 atom = 1.660539 × 10 − 27 = 1.660539 \times 10^{-27} = 1.660539 × 1 0 − 27 kg. Radius: R = R 0 A 1 / 3 R = R_0A^{1/3} R = R 0 A 1/3 , with R 0 = 1.2 R_0 = 1.2 R 0 = 1.2 fm and 1 fm = 10 − 15 1\ \text{fm} = 10^{-15} 1 fm = 1 0 − 15 m. Density: about 2.3 × 10 17 kg/m 3 2.3 \times 10^{17}\ \text{kg/m}^3 2.3 × 1 0 17 kg/m 3 for every nucleus, because the mass and the volume are both proportional to A A A . Mass defect and binding energy Mass defect: Δ m = Z m H + N m n − M \Delta m = Zm_H + Nm_n - M Δ m = Z m H + N m n − M , with atomic masses in u. Mass and energy: E = m c 2 E = mc^2 E = m c 2 , and 1 u 1\ \text{u} 1 u is equivalent to 931.494 931.494 931.494 MeV. NCERT rounds this to 931.5 931.5 931.5 MeV. Binding energy: B = Δ m × 931.494 MeV/u B = \Delta m \times 931.494\ \text{MeV/u} B = Δ m × 931.494 MeV/u . It is the energy needed to pull the nucleus apart. Per nucleon: B / A B/A B / A . Helium-4 has 7.07 7.07 7.07 MeV, and iron-56 has 8.79 8.79 8.79 MeV. The binding energy curve B / A B/A B / A rises steeply for light nuclei, peaks near A = 56 A = 56 A = 56 to 62 62 62 at about 8.8 8.8 8.8 MeV, and falls to 7.6 7.6 7.6 MeV at uranium. Fusion of light nuclei and fission of heavy nuclei both climb the curve, so both release energy: E = A ( b after − b before ) E = A\,(b_\text{after} - b_\text{before}) E = A ( b after − b before ) . Nuclear force: strong, attractive, short range, charge independent, and saturating. Heavy nuclei have more neutrons than protons, because the proton repulsion has a long range. A worked example Nitrogen-14 has an atomic mass of 14.003074 14.003074 14.003074 u, with 7 7 7 protons and 7 7 7 neutrons. Find its binding energy per nucleon.
The parts. 7 × 1.007825 u + 7 × 1.008665 u = 14.115430 7 \times 1.007825\ \text{u} + 7 \times 1.008665\ \text{u} = 14.115430 7 × 1.007825 u + 7 × 1.008665 u = 14.115430 u. The mass defect. 14.115430 u − 14.003074 u = 0.112356 14.115430\ \text{u} - 14.003074\ \text{u} = 0.112356 14.115430 u − 14.003074 u = 0.112356 u. The binding energy. 0.112356 u × 931.494 MeV/u = 104.66 0.112356\ \text{u} \times 931.494\ \text{MeV/u} = 104.66 0.112356 u × 931.494 MeV/u = 104.66 MeV. Per nucleon. 104.66 MeV 14 = 7.476 \dfrac{104.66\ \text{MeV}}{14} = 7.476 14 104.66 MeV = 7.476 MeV. Practise with new numbers
Check yourself A nucleus has Z = 82 Z = 82 Z = 82 and A = 208 A = 208 A = 208 . How many neutrons does it have? Stuck? Show a hint Show the answerA mass defect of 0.050 0.050 0.050 u is found for a nucleus of 40 40 40 nucleons. What is the binding energy per nucleon? Stuck? Show a hint Show the answerWhy does fission of a heavy nucleus release energy? Stuck? Show a hint Show the answerSources NCERT Physics Part II (Class 12, rationalised edition), Chapter 13 Nuclei. OpenStax University Physics Volume 3, Chapter 10 Nuclear Physics. AME2020 atomic mass evaluation, for the atomic masses.