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Modern Physics Unit 15: Nuclear Structure and Decay Lecture 15.1: Nuclear Characteristics

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Modern Physics Unit 15: Nuclear Structure and Decay Lecture 15.1: Nuclear Characteristics Modern Physics Unit 15: Nuclear Structure and Decay Lecture 15.1: Nuclear Characteristics Ron Reifenberger Professor of Physics Purdue University 1 Nucleons - the basic building blocks of the nucleus Also written as: 7 3 Li ~10-15 m Examples: A X=Chemical Element Z X Z = number of protons = atomic number 12 A = atomic mass number = Z+N 6 C N= A-Z = number of neutrons 35 17 Cl 2 What is the size of a nucleus? Three possibilities • Range of nuclear force? • Mass radius? • Charge radius? It turns out that for nuclear matter Nuclear force radius ≈ mass radius ≈ charge radius defines nuclear force range defines nuclear surface 3 Nuclear Charge Density The size of the lighter nuclei can be approximated by modeling the nuclear charge density ρ (C/m3): t ≈ 4.4a; a=0.54 fm 90% 10% Usually infer the best values for ρo, R and a for a given nucleus from scattering experiments 4 Nuclear mass density Scattering experiments indicate the nucleus is roughly spherical with a radius given by 1 3 −15 R = RRooA ; =1.07 × 10meters= 1.07 fm = 1.07 fermis A=atomic mass number What is the nuclear mass density of the most common isotope of iron? 56 26 Fe⇒= A56; Z = 26, N = 30 m Am⋅⋅3 Am 33A ⋅ m m ρ = nuc p= p= pp = o 1 33 VR4 3 3 3 44ππA R nuc π R 4(π RAo ) o o 3 3×× 1.66 10−27 kg = 3.2× 1017kg / m 3 4×× 3.14 (1.07 × 10−15m ) 3 The mass density is constant, independent of A! 5 Nuclear mass density for 27Al, 97Mo, 238U (from scattering experiments) (kg/m3) ρo heavy mass nucleus light mass nucleus middle mass nucleus 6 Typical Densities Material Density Helium 0.18 kg/m3 Air (dry) 1.2 kg/m3 Styrofoam ~100 kg/m3 Water 1000 kg/m3 Iron 7870 kg/m3 Lead 11,340 kg/m3 17 3 Nuclear Matter ~10 kg/m 7 Isotopes - same chemical element but different mass (J.J. Thomson 1913 and F.W. Aston 1919) Nucleus protons neutrons electrons 1 1 H 1 0 1 Hydrogen 2 1 H 1 1 1 Deuterium (1 in 6000) 3 17 1 H 1 2 1 Tritium (1 in 10 ) 235 92 U 92 143 92 U-235 (0.72%) 238U 92 92 146 92 U-238 (99.27%) More than 3000 isotopes have been identified. About 400 are considered stable. 8 Nuclear Mass – a new unit of mass (1961) Nuclear masses are very small (~10-27 kg); invent a new unit of mass - the atomic mass unit (amu) or unified mass unit (u) or the Dalton (Da). Consider Carbon-12: Define new mass unit as 12 1 u = 1.660 540×10-27 kg 6 C These numbers are VERY precise. You cannot approximate c by 3x108 m/s ! E=mc2 → E=(1.660 540×10-27 kg) (2.998x108 m/s)2 =1.492x10-10J = 931.5 MeV → 1u =931.5 MeV/c2 (not 933.7 MeV/c2) With this definition, 12 the mass of6 C is exactly12.000u 9 mp=1.007 276 u; mn=1.008 665u; me=0.000 549u The deuteron – 1 proton bound to 1 neutron - the simplest compound nucleus 2 1 H ISSUES: • How are the neutron and proton bound together to form a deuteron? • The deuteron nuclear mass = 2.013 553 u (from experiment) • The deuterium atomic mass = 2.014 102 u (from experiment) • The difference = 0.000 549 u, the mass of an electron • In what follows, the electron binding energy (13.6 eV for deuterium) is systematically neglected because the nuclear binding energy is about one million times greater 10 The mass of the parts is more than the whole?? mp=1.007 276 u mn=1.008 665 u 2.015 941 u The mass of a nucleus is NOT equal to the sum of the masses of its deuterium nucleus parts! (from experiment) 2.013 553 u 1 proton + 1 neutron 2.015 941 u This implies that mmmd= pn + −∆ m 2 =+−mpn m Bc/ The “lost” mass is converted to a “nucleon binding energy” equal to Defines binding energy B ΔE=B=∆mc2 11 of the nucleons 11 Calculate the binding energy of the deuteron mmm=+−B d pn c2 1. Add an electron mass to each side. WHY? mmmmm+= ++−B de pne c2 2. (md + me ) is now the atomic deuterium mass 2 M( H) and (mp + me ) is now the atomic hydrogen mass M(1H). So MH21=+− m MH B ( ) n ( ) c2 3. When we convert nuclear masses to atomic masses, we neglect the binding energy of the electrons. WHY? 12 4. Electron binding energies appear on both sides of the equation and they tend to cancel in almost all nuclear-mass calculations. For this reason we use atomic masses (easy to measure) rather than nuclear masses. B =+−=m MH12 MH0.002 388 u c2 n ( ) ( ) 5. Convert u to energy using 1 u = 931.5 MeV /c2 B 931.5Mev / c2 =×=0.002 388u 2.22Mev / c2 cu2 1 13 What does it mean? The nuclear potential energy of two nucleons vs. separation Energy (MeV) ~0.4fm radial separation, r 2.2 MeV In 3-d, ALL the nucleons are packed ~20 MeV into a sphere of radius R R 1 3 R RAo 1 =1.07 × 10−15 m( 2) 3 R~1.35 fm =×=1.07 10−15 m( 1.26) 1.35 fm p n (no Coulomb repulsion for this case) 14 Up Next – The strong nuclear force 15 .
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