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Nuclear Systematics and Rutherford scattering. Terminology. Atomic number (Z) is the number of protons in the nucleus of an atom, and also the number of electrons in a neutral atom Nucleon : proton (Z) or neutron (N) Nuclide : nucleus uniquely specified by the values of N & Z

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## Nuclear Systematics and Rutherford scattering

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**Terminology**• Atomic number (Z) is the number of protons in the nucleus of an atom, and also the number of electrons in a neutral atom • Nucleon: proton (Z) or neutron (N) • Nuclide: nucleus uniquely specified by the values of N & Z • Mass number (A) is the total number of nucleons in a nucleus (A=Z+N) • Isotopes: nuclides with the same protons (Z) e.g. 235U and 238U • Isotones: nuclides with the same neutrons (N) e.g. 2H (d) and 3He • Isobars: nuclides with the same A • Atomic mass unit (u): one-twelfth of the mass of a neutral atom of 12C (six protons, six neutrons, and six electrons). 1 u = 1.66 x 10–27 kg = 931.5 MeV/c2 • Atomic mass is the mass of a neutral atom and includes the masses of protons, neutrons, and electrons as well as all the binding energy. • Nuclear mass is the mass of the nucleus and includes the masses of the protons and neutrons as well as the nuclear binding energy, but does not include the mass of the atomic electrons or electronic binding energy. • Radioisotopes: members of a family of unstable nuclides with a common value of Z**Units**• SI units are fine for macroscopic objects like footballs but are very inconvenient for nuclei and particles use natural units. • Energy: 1 eV = energy gained by electron in being accelerated by 1V. • Mass: MeV/c2 (or GeV/c2) 1 MeV/c2 = 1.78X10-30 kg. 1 GeV/c2 = 1.78X10-27 kg. Or use Atomic Mass Unit defined by mass of 12C= 12 u • Momentum: MeV/c (or GeV/c) 1 eV/c = e/c kg m s-1 • Cross sections: 1 barn =10-28 m2 • Length: fermi (fm) 1 fm = 10-15 m.**Properties of nucleons**http://www.phys.unsw.edu.au/PHYS3050**Chart of the Nuclides**Z=N Proton number * Unstable nuclei on either side of stable ones * Small Z tendency for Z=N * Large Z characterized by N>Z Neutron number**Behaviour of nucleus is determined by the combinations of**protons and neutrons Behaviour of nucleus is determined by strong and electromagnetic interactions At first sight we would expect the more neutrons the more strongly bound the nucleus – but in fact there is a tendency for Z = N Further from Z=N, the more unstable the nuclide becomes**The number of unstable nuclei is around 2000 but is always**increasing. * Tendency for even Z – even N to be the most stable nuclei * Even – Odd, and Odd – Even configurations are equally likely * Almost no Odd Z – Odd N are stable, and these are interesting small nuclei such as**Rutherford classified radioactivity**Awarded Nobel prize in chemistry 1908 “for investigations into the disintegration of the elements and the chemistry of radioactive substances” Together with Geiger and Marsden scattered alpha particles from atomic nuclei and produced the theory of “Rutherford Scattering”. Postulated the existence of the neutron Ernest Rutherford (1871-1937)**The Geiger – Marsden Experiment**glass 214Po Lead lining Au microscope ZnS**The Rutherford Cross – Section Formula**∆p p p vo m s Secret of deriving the formula quickly is to express the momentum transfer in two ways. The first way you can see in the top parallelogram diagram. (1)**The Rutherford Cross – Section Formula**φ The second way is by integrating the force on the alpha across the trajectory: since we know by Newton’s law and Coulomb’s law:**Eqn 3.5**At first sight this integral looks impossible because both and r are both functions of t. However the conservation of angular momentum helps: Eqn 3.1 From which one sees that: So that (3.5) becomes Now we can equate this with the first method of finding Δp**This allows us to get the impact parameter b as a function**of θ Where S0 is the distance of closest approach for head on collision Coulomb potential so * All particles scattered by more than some value of must have impact parameters less than b. So that cross-section for scattering into any angle greater than must be: Eqn(3.9)**b**Particles fly off into solid angle given by: The differential scattering cross-section is defined as:**Distance of closest approach**d 0 o o**The distance of closest approach “d” will be determined**by: Using: we get - Solution of this quadratic gives:**Failure of the Rutherford Formula**Increasing energy and constant angle Increasing angle and constant energy Failure of the formula occurs because the distance of closest approach is less than the diameter of the nucleus. This can happen if (a) the angle of scatter is large or (b) the energy of the particle is large enough. With Alpha particles from radioactive sources this is difficult. But with those from accelerators it becomes possible to touch the nucleus and find out its size because the distance of closest approach is given by:

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