The Nucleus. PHY 3101 D. Acosta

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1 The Nucleus PHY 30 D. Acosta

2 Rutherford Scattering Experiments by Geiger & Marsden in 909 /5/005 PHY D. Acosta

3 Rutherford Model of the Atom Conclusion: the atom contains a positive nucleus < 0 fm in size ( fm = 0-5 m) /5/005 PHY D. Acosta 3

4 The Neutron The neutron was discovered in 93 by James Chadwick α-particles accelerated in a small accelerator and collided with Be nuclei Neutral, very penetrating radiation Found by elastic scattering off protons in paraffin wax By the way, the positron (anti-electron) also was discovered in 93 by Carl Anderson in cosmic rays Anti-matter predicted by P.A.M. Dirac in his relativistic version of the Schrodinger Equation /5/005 PHY D. Acosta 4

5 The Periodic Table All elements composed of just electrons, neutrons, and protons Elements of the same group have nearly the same chemical property Chemical periodicity depends on the atomic number Z Any other fundamental particles? Next chapter /5/005 PHY D. Acosta 5

6 Nomenclature Z A X X is the element A is the atomic mass (Z+N) Z is the atomic number (number of protons) N is the number of neutrons Atoms are neutral. Number of electrons equals number of protons = Z Chemical properties depend on Z Ordering of Periodic Table given by valence configuration of electrons Isotopes: Same Z, different A Isobars: Same A, different Z Isotones: Same N, different A 4 He 3 He 3 H 3 He 3 4 C N 6 7 /5/005 PHY D. Acosta 6

7 Atomic Mass Units (u) mass of C u The atomic mass is the mass of an atomic isotope, including electrons 7 u = kg = MeV / c m m m p n e = u = MeV / c = u = MeV / c 4 = u = 05. MeV / c Note that mass of C is 6 m p + 6m n + 6m e =. u >.0 u The nucleus is bound Binding energy is 0. u = 90 MeV It takes energy to liberate all particles Should not think of mass as measuring the number of particles, only the rest energy of the system: Mass is a measure of inertia (a = F/m) not contents /5/005 PHY D. Acosta 7

8 Binding Energy a f B= m separate m( combined) c Take the mass of all particles individually, including electrons, and subtract the mass of the combined system A system is bound if the binding energy is positive. Example: Deuterium Note that e - mass cancels B = M H + M n M H c = b g b g b g u u. 040u c = u MeV / u = 4. MeV If the binding energy is negative, the system will decay. The energy released is a f Q= m( combined) m separate c = B /5/005 PHY D. Acosta 8

9 Atomic Binding Energies The Coulomb potential for an electron in a hydrogen-like atom can be written in terms of the dimensionless fine structure constant cz e Vr ()= ( ) α α = r 4πε c 37 The energy levels are given by E c Z F n = α µ µ = + n m m Hydrogen: µ = me E = 3. 6 ev Positronium (e + e - ): me µ = E = 68. ev These are the binding energies! e.g. mass of H is less than mass of e+p 0 HG e N I KJ The Bohr radii are rn = n r = µ c α 0 m /5/005 PHY D. Acosta 9

10 Nuclear Binding Energies Consider the binding energy of the deuteron proton neutron bound state The binding potential is roughly similar to that of the Coulomb potential, but with a dimensionless constant characteristic of the Strong Nuclear Force rather than EM af af s c qs V r = α α s = 0. > 0α r 4πε 0 c The energy levels are given by En = α s µ c n F I mp = HG µ + KJ = 470 MeV / c mp mn E = ( )( ) MeV.. 3 MeV Agrees with measured value of. MeV million times larger than atomic energies! Nuclear radius is 0,000 times smaller: rn = n r = m µ c α s /5/005 PHY D. Acosta 0

11 Nuclear Potential Well Rutherford concludes from Geiger and Marsden that the range of the Strong Nuclear Force is < 0-4 m No deviation in the scattering rate of the highest-energy α-particles off nuclei from that predicted by electromagnetic Coulomb scattering Thus, the Strong Nuclear Force is shortranged, and does not extend to infinity To probe the size of nuclei, need higher energies than α-particles from radioactive decay The nuclear potential well resembles a semiinfinite potential well α-particles inside the nucleus must tunnel to escape! Higher rate for higher energy α- particles /5/005 PHY D. Acosta

Size of Nuclei Robert Hofstadter performs experiment at Stanford using a new linear accelerator for electrons in 950s E = 00 -- 500 MeV λ = h / p =.5 fm The proton is not a point!

12 Size of Nuclei Robert Hofstadter performs experiment at Stanford using a new linear accelerator for electrons in 950s E = MeV λ = h / p =.5 fm The proton is not a point! (Deviation of elastic scattering rate from Rutherford Scattering prediction) Proton and nuclei have extended charge distributions Nobel prize in 96 nucleus af a f ρ r ρ0 = + exp r R / a R r0 A 3 / 4 V = πr 3 A= 3 r 0 a =. 0 = 05. fm 5 m =. fm #nucleons /5/005 PHY D. Acosta

Nuclear Properties. Thornton and Rex, Ch. 12

Nuclear Properties. Thornton and Rex, Ch. 12 Nuclear Properties Thornton and Rex, Ch. 12 A pre-history 1896 Radioactivity discovered - Becquerel a rays + (Helium) b rays - (electrons) g rays 0 (EM waves) 1902 Transmutation observed - Rutherford and