Tony WeidbergNuclear Physics Lectures1 Today’s Menu Why study nuclear physics Why nuclear physics is difficult Course synopsis. Notation & Units

Tony WeidbergNuclear Physics Lectures2 What is the use of lectures Definition of a lecture: a process whereby notes are transferred from the pages of a lecturer to the pages of the student without passing through the head of either. Conclusion: to make lectures useful YOU have to participate, ask questions ! If you don’t understand something the chances are >50% of the audience doesn’t as well, so don’t be shy !

Tony WeidbergNuclear Physics Lectures3 Why Study Nuclear Physics? Understand origin of different nuclei –Big bang: H, He and Li –Stars: elements up to Fe –Supernova: heavy elements We are all made of stardust Need to know nuclear cross sections  experimental nuclear astrophysics is a hot topic.

Tony WeidbergNuclear Physics Lectures4 Practical Applications Nuclear fission for energy generation. –No greenhouse gasses –Safety and storage of radioactive material. Nuclear fusion –No safety issue (not a bomb) –Less radioactive material but still some. Nuclear transmutation of radioactive waste with neutrons. –Turn long lived isotopes  stable or short lived. Every physicist should have an informed opinion on these important issues!

Tony WeidbergNuclear Physics Lectures5 Medical Applications Radiotherapy for cancer –Kill cancer cells. –Used for 100 years but can be improved by better delivery and dosimetery –Heavy ion beams can give more localised energy deposition. Medical Imaging –MRI (Nuclear magnetic resonance) –X-rays (better detectors  lower doses) –PET –Many others…see Medical & Environmental short option.

Tony WeidbergNuclear Physics Lectures6 Other Applications Radioactive Dating –C 14 /C 12 gives ages for dead plants/animals/people. – Rb/Sr gives age of earth as 4.5 Gyr. Element analysis –Forenesic (eg date As in hair). –Biology (eg elements in blood cells) –Archaeology (eg provenance via isotope ratios).

Tony WeidbergNuclear Physics Lectures7

Tony WeidbergNuclear Physics Lectures8 Why is Nuclear Physics Hard? QCD theory of strong interactions  just solve the equations … At short distance/large Q coupling constant small  perturbation theory ok but long distance/small Q, q  large Not on syllabus !

Tony WeidbergNuclear Physics Lectures9 Nuclear Physics Models Progress with understanding nuclear physics from QCD=0  use simple, approximate, phenomenological models. Liquid Drop Model: phenomenology + QM + EM. Shell Model: look at quantum states of individual nucleons  understand spin/parity magnetic moments and deviations from SEMF for binding energy.

Tony WeidbergNuclear Physics Lectures10 Course Synopsis - 1 Liquid Drop Model and SEMF. Applications of SEMF –Valley of stability. – a  decays. –Fission & fusion. Limits of validity of liquid drop model (shell model effects)

Tony WeidbergNuclear Physics Lectures11 Course Synopsis - 2 Cross Sections –Experimental definition –FGR theory –Rutherford scattering –Breit-Wigner resonances Theory of  decays. Particle interactions in matter –Simple detectors for nuclear/particle physics.

Tony WeidbergNuclear Physics Lectures12 Corrections To err is human … and this is a new course  lots of mistakes. Please tell me about any mistakes you find in the notes (I will donate a bottle of wine to the person who finds the most mistakes!).

Tony WeidbergNuclear Physics Lectures13 The Minister of Science This is a true story honest. Once upon a time the government science minister visited the Rutherford Lab (UK national lab) and after a days visit of the lab was discussing his visit with the lab director and he said … I hope that you all have a slightly better grasp of the subject by the end!

Tony WeidbergNuclear Physics Lectures14 Notation Nuclei are labelled where El is the chemical symbol of the element, mass number A = number of neutrons N + number of protons Z. eg Excited states labelled by * or m if they are metastable (long lived).

Tony WeidbergNuclear Physics Lectures15 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. –1 eV= e J. Mass: MeV/c 2 (or GeV/c 2 ) –1 eV/c 2 = e/c 2 kg. –Or use AMU defined by mass of 12 C= 12 u Momentum: MeV/c (or GeV/c) –1 eV/c = e/c kg m s -1 Cross sections: (as big as a barn door) –1 barn = m 2 Length: fermi 1 fm = m.

Tony WeidbergNuclear Physics Lectures16 Nuclear Masses and Sizes Masses and binding energies –Absolute values measured with mass spectrometers. –Relative values from reactions and decays. Nuclear Sizes –Measured with scattering experiments (leave discussion until after we have looked at Rutherford scattering). –Isotope shifts

Tony WeidbergNuclear Physics Lectures17 Nuclear Mass Measurements Measure relative masses by energy released in decays or reactions. –X  Y +Z +  E –Mass difference between X and Y+Z is  E/c 2. Absolute mass by mass spectrometers (next transparency). Mass and Binding energy: B = [Z M H + N M n – M(A,Z)]/c 2

Tony WeidbergNuclear Physics Lectures18 Mass Spectrometer Ion Source Velocity selector  electric and magnetic forces equal and opposite –qE=qvB  v=E/B Momentum selector, circular orbit satisfies: –Mv=qBr –Measurement r gives M. Ion Source Velocity selector Detector

Tony WeidbergNuclear Physics Lectures19 Binding Energy vs A B increases with A up to 56 Fe and then slowly decreases. Why? Lower values and not smooth at small A.

Tony WeidbergNuclear Physics Lectures20 Nuclear Sizes & Isotope Shift Coulomb field modified by finite size of nucleus. Assume a uniform charge distribution in the nucleus. Gauss’s law  integrate and apply boundary conditions Difference between actual potential and Coulomb Use 1 st order perturbation theory

Tony WeidbergNuclear Physics Lectures21 Isotope Shifts

Tony WeidbergNuclear Physics Lectures22 Isotope Shifts Isotope shift for optical spectra Isotope shift for X-ray spectra (bigger effect because electrons closer to nucleus) Isotope shift for X-ray spectra for muonic atoms. Effect greatly enhanced because m  ~ 207 m e and a 0 ~1/m. All data consistent with R=R 0 A 1/3 with R 0 =1.25fm.

Tony WeidbergNuclear Physics Lectures23 Frequency shift of an optical transition in Hg at =253.7nm for different A relative to A=198. Data obtained by laser spectroscopy. The effect is about 1 in (Note the even/odd structure.) Bonn et al Z Phys A 276, 203 (1976) A 2/3 Isotope Shift in Optical Spectra  E/h (GHz)

Tony WeidbergNuclear Physics Lectures24 Data on the isotope shift of K X ray lines in Hg. The effect is about 1 in Again the data show the R 2 = A 2/3 dependence and the even/odd effect. Lee et al, Phys Rev C 17, 1859 (1978)

Tony WeidbergNuclear Physics Lectures25 Data on Isotope Shift of K Xrays from muonic atoms [in which a muon with m=207m e takes the place of the atomic electron]. Because a 0 ~ 1/m the effect is ~0.4%, much larger than for an electron. The large peak is 2p 3/2 to 1s 1/2. The small peak is 2p 1/2 to 1s 1/2. The size comes from the 2j+1 statistical weight. Shera et al Phys Rev C 14, 731 (1976) 58 Fe 56 Fe 54 Fe Energy (keV)

Tony WeidbergNuclear Physics Lectures26 SEMF Aim: phenomenological understanding of nuclear binding energies as function of A & Z. Nuclear density constant (see lecture 1). Model effect of short range attraction due to strong interaction by liquid drop model. Coulomb corrections. Fermi gas model  asymmetry term. QM  pairing term. Compare with experiment: success & failure!

Tony WeidbergNuclear Physics Lectures27 Liquid Drop Model Nucleus Phenomenological model to understand binding energies. Consider a liquid drop –Ignore gravity and assume no rotation –Intermolecular force repulsive at short distances, attractive at intermediate distances and negligible at large distances  constant density. E=-  n + 4  R 2 T  B=  n-  n 2/3 Analogy with nucleus –Nucleus has constant density –From nucleon nucleon scattering experiments: Nuclear force has short range repulsion and attractive at intermediate distances. –Assume charge independence of nuclear force, neutrons and protons have same strong interactions  check with experiment!

Tony WeidbergNuclear Physics Lectures28 Mirror Nuclei Compare binding energies of mirror nuclei (nuclei n  p). Eg 7 3 Li and 7 4 Be. Mass difference due to n/p mass and Coulomb energy.

Tony WeidbergNuclear Physics Lectures29 nn and pp interaction same (apart from Coulomb) “Charge symmetry”

Tony WeidbergNuclear Physics Lectures30 Charge Symmetry and Charge Independence Mirror nuclei showed that strong interaction is the same for nn and pp. What about np ? Compare energy levels in “triplets” with same A, different number of n and p. e.g. Same energy levels for the same spin states  SI same for np as nn and pp.

Tony WeidbergNuclear Physics Lectures31 Charge Independence Is np force is same as nn and pp? Compare energy levels in nuclei with same A. Same spin/parity states have same energy. np=nn=pp Na Mg Mg Na Ne

Tony WeidbergNuclear Physics Lectures32 Charge Independence of Strong Interaction If we correct for n/p mass difference and Coulomb interaction, then energy levels same under n  p. Conclusion: strong interaction same for pp, pn and nn if nucleons are in the same quantum state. Beware of Pauli exclusion principle! eg why do we have bound state of pn but not pp or nn?

Tony WeidbergNuclear Physics Lectures33 Asymmetry Term Neutrons and protons are spin ½ fermions  obey Pauli exclusion principle. If other factors were equal  ground state would have equal numbers of n & p. Illustration Neutron and proton states with same spacing . Crosses represent initially occupied states in ground state. If three protons were turned into neutrons the extra energy required would be 3×3 . In general if there are Z-N excess protons over neutrons the extra energy is ((Z-N)/2) 2 . relative to Z=N.

Tony WeidbergNuclear Physics Lectures34 Asymmetry Term From stat. mech. density of states in 6d phase space = 1/h 3 Integrate to get total number of protons Z, & Fermi Energy (all states filled up to this energy level). Change variables p  E

Tony WeidbergNuclear Physics Lectures35 Asymmetry Term Binomial expansion keep lowest term in y/A Correct functional form but too small by factor of 2. Why?

Tony WeidbergNuclear Physics Lectures36 Pairing Term Nuclei with even number of n or even number of p more tightly bound  fig. Only 4 stable o-o nuclei cf 153 e-e. p and n have different energy levels  small overlap of wave functions. Two p(n) in same level with opposite values of j z have AS spin state  sym spatial w.f.  maximum overlap  maximum binding energy because of short range attraction. Neutron number Neutron separation energy in Ba

Tony WeidbergNuclear Physics Lectures37 Pairing Term Phenomenological fit to A dependence Effect smaller for larger A  e-e+ive e-o0 o-o-ive

Tony WeidbergNuclear Physics Lectures38 Semi Empirical Mass Formula Put everything together: Fit to measured binding energy. –Fit not too bad (good to <1%). –Deviations are interesting  shell effects. –Coulomb term agrees with calculation. –Asymmetry term larger ? –Explain valley of stability. –Explains energetics of radioactive decays, fission and fusion.

Tony WeidbergNuclear Physics Lectures39 The Binding Energy per nucleon of beta-stable (odd A) nuclei. Fit values in MeV a15.56 b17.23 c d0.697  +12 (o-o)  0 (o-e)  -12 (e-e) A B/A (MeV)

Tony WeidbergNuclear Physics Lectures40 Valley of Stability SEMF allows us to understand valley of stability. Low Z, asymmetry term  Z=N Higher Z, Coulomb term  N>Z.