Lecture 0 Phys. 641: Nuclear Physics 1 Physics Department Yarmouk University 21163 Irbid Jordan مقدمة Overview Dr. Nidal Ershaidat http://ctaps.yu.edu.jo/physics/courses/phys641/lec1-1 ٢٠٠٨ /٢٠٠٧ ٦٤١ فيزياء الفصل الا ول 2 Nuclear Physics 1 ١ نووية فيزياء : المساق مدرس د. نضال الرشيدات محمد Lecturer s site on CTAPS http://ctaps.yu.edu.jo/physics/nershaidat ١٧:٠٠-١٥:٣٠ ا لمحاضرة: موعد
References 1. Nuclear Structure (3 Volumes) A. Bohr and B. R. Mottelson W.A. Benjamin Inc., Advanced Book Program, 1969-1975 2. Theoretical Nuclear Physics, J. M. Blatt and V.F. Weisskopf, Springer-Verlag, 1979 3. Nuclear and Particle Physics, W.S.C. Williams, Clarendon Press. Oxford, 1992 4. Physics of Nuclei and Particles, P. Mermier and E. Sheldon, Academic Press 1970 5. Introductory Nuclear Physics Kenneth S. Krane John Wiley & Sons, 1988 3 Quantum Mechanics References 4 1. Quantum Mechanics (2 volumes) C. Cohen-Tannoudji, B. Diu and F. Laloë Wiley-VCH, Berlin 2.Quantum Physics Stephen Gasiorowicz 3 rd Edition (2003) John Wiley & Sons Inc.
Syllabus المساق محتوى 5 Nucleon - Nucleon Scattering Nuclear Structure and Nuclear Volume Multipole Moments Shell model, Collective States Instrumentation and Methods in Nuclear Physics. Syllabus المساق محتوى Overview Review of Nuclear Structure and Nuclear Properties Nuclear Models Shell model, Collective States Elements of Scattering Theory Nucleon-Nucleon Scattering Instrumentation and Methods in Nuclear Physics. 6
Exams - Grades 7 11:30 1,2 and 3 الا ول 25% الساعة 2007/11/03 السبت 11:30 4,5,6 الثاني 25% الساعة 2007/12/08 السبت لاحقا 50% ي حد د All Chapters النهاي ي Websites Phys. 641 s Offical Website http://ctaps.yu.edu.jo/physics/courses/phys641 Phys. 441 s Offical Website 8 http://ctaps.yu.edu.jo/physics/courses/phys441 Phys. 441 s Lecturer notes http://ctaps.yu.edu.jo/physics/courses/phys441 /PDF_Files Lecturer s site on Yarmouk s Portal http://faculty.yu.edu.jo/nershaidat
Lecture 1 Phys. 641: Nuclear Physics 1 Physics Department Yarmouk University 21163 Irbid Jordan Chapter 1: Overview Dr. Nidal Ershaidat http://ctaps.yu.edu.jo/physics/courses/phys641/lec1-1 Brief History - ATOMOS 4 th Century B.C : Democritus (ATOMOS = indivisible) 10 19 th Century A.D : Chemists Dalton, Avogadro, Lavoisier, distinguished many kinds of atoms 1869 : Mendeleev proposes his periodic table of Elements for classifying these atoms according to their masses.
Brief History French Side 1896: Becquerel discovers Radioactivity 11 1897: Jean Baptiste Perrin shows that the cathodic rays are electrons 1898: P. & M. Curie identify two new elements: Polonium and Radium. Rutherford The Nucleus 1903: Rutherford (& Mme Curie) identify three types of the naturally emitted particles. (α, β, γ). 12 1911: Rutherford explains Geiger-Marsden experimental results. He introduces the nucleus
Radioactivity By simply using a deflecting magnetic field Marie Curie demonstrated that a particles are doubly positively charged. b particles have the characteristics of electrons and the g radiation behaves like the X-rays of Roentgen but are more energetic. β γ α 13 Radioactive source Magnetic Field Atoms are Neutral Studies of the hydrogen atom revealed that it contains one electron (electrons are particles seen in a cathodic tube and identified by Perrin in 1897) and a positive particle the proton. 14 The only plausible hypothesis to explain the zero charge of the hydrogen atom was to consider that the charge of the proton is equal to that of the electron*. * Millikan measured in 1911 this charge e and found it equal to 1.6x10-19 C
The Atomic Number (Z) 15 A new classification replaces that of Mendeleev: it is based on the resemblance of chemical behavior of atoms strongly related to their number of electrons. This number of electrons is called the Atomic Number (symbol Z). It also represents the number of protons of an atom. The Geiger-Marsden Experiment 16 In 1911, Geiger and Marsden (students of Lord Rutherford) discovered an amazing result!! When bombarding a thin foil of gold with alpha particles from a radioactive source Geiger and Marsden noticed that a great majority of these α particles were not deviated, a small fraction of them are scattered and 10-6 of them are simply scattered with a scattering angle of 180 0.
The Geiger-Marsden Experiment 17 A beam of alpha particles (charge=2e, mass=m α ) is incident on a thin foil of gold (Gold Au, Z=79). Geiger-Marsden Results 18 Appendix 1 Rutherford Scattering θ
Rutherford The Nucleus 19 Using classical electrodynamics, Rutherford explains the previous findings. Rutherford suggests that an atom is composed of a massive solid core which should have a charge +Ze and should contain all the mass of the atom. This nucleus is surrounded at far distances by Z electrons. The space between the nucleons and the atomic electrons is empty. Lecture 2 Phys. 641: Nuclear Physics 1 Physics Department Yarmouk University 21163 Irbid Jordan The Nucleus Z,N and A Dr. Nidal Ershaidat http://ctaps.yu.edu.jo/physics/courses/phys641/lec1-2
Bohr s Hydrogen Atom Niels Bohr suggests a model to explain experimental results of the measurement of the spectral lines of the hydrogen atom. The atom s electron interacts electrically (now we say electromagnetically) with the proton but within restricted conditions. Quantum mechanics was born. ٢١ ٢٢ The Puzzle of the Atom s Mass See /Chapter-1.pdf k 'U1(UVKDLGDW3K\V&KDSWHU2YHUYLHZ
٢٣ The Puzzle of the Atom s Mass Chemical studies showed that the mass of an atom is much bigger than the sum of masses of its electrons and protons. In order to explain the difference, Rutherford suggested the presence inside the nucleus of particles composed of electrons tightened to protons in a sufficient number to produce the mass of an atom. The He nucleus Do electrons exist inside the nucleus? ٢٤ Such a particle is neutral and has a mass a little higher than the proton s mass Rutherford s suggestion implies the presence of electrons inside the nucleus. Emission of electrons in beta decay of some elements suggests also the presence of electrons in the nucleus But
Electrons cannot exist inside the nucleus ٢٥ أكبر وبالتحديد الربط مع البروتونات يجب أن تكون كبيرة جدا قوة (١! القوة الكولومي ة بين الالكترونات والبروتونات من فا ن h h c p = = x x 200 MeV F c 10 F عام تكون طاقة الالكترونات الناتجة عن نشاط بيتا الاشعاعي وبشكل 1 MeV بحدود c = 20 MeV c : التجارب أن الالكترونات لا ي مكن أن تتواجد داخل النواة أظهرت p x = h : (uncertainty principle) مبدأ هايزنبرغ حسب (٢ النواة داخل للا لكترون الخطي الزخم p هو: ( x = 10 F) ٢٦ (٣ الزخم المغزلي اعتبارات Spin considerations p pe r S tot = r 1 + 2 r r 1 1 + 2 2 الهيدروجين: نظير (Deuterium) نواة الدويتريوم مثال وبروتون (الكترون اعتبرنا هذه النواة مكونة من اتحاد بروتون مع لو الزخم المغزلي الكلي لنواة الدويتريوم يساوي: فا ن ببعضهما) مرتبطين التجربة أن وتثبت S tot = 1 0 1 3 + 1 +, + 2 2 2 z = S 1 0 1 1 tot + 1, + 2 2 2 S deuterium = 1 1 2 or 3 2
Magnetic Dipole considerations ٢٧ (٤ المغناطيسي العزم اعتبارات نعرف أن عزم الثناقطبي المغناطيسي للبروتون يساوي 1/2000 نحن من عزم الثناقطبي المغناطيسي للالكترون تقريبا 1 p µ 2000 e µ عزم يعني بالتالي أن ه لو كان الالكترون موجودا داخل النواة فا ن وهذا المغناطيسي لها يجب أن يكون بكبر عزم الثناقطبي الثناقطبي للالكترون وهذا غير صحيح مخبري ا. المغناطيسي رذرفورد يعلم كل هذا ويفكر بحل افتراض وجود جسيم جديد كان الشحنة وكتلته قريبة من كتلة البروتون. متعادل 1932 A Decisive Year ٢٨ James Chadwick discovers in Cavendish Laboratory (where Rutherford is the director) the neutron. A particle which has the required characteristics predicted by Rutherford, i.e. a neutral particle with a mass of the order of the proton s mass. Rutherford gives an image of the atom which takes into account all the previous considerations and suggests a nucleus with a number Z of protons, a number N of neutrons surrounded at far distances compared to the dimensions of the nucleus, by Z electrons.
Atomic and Atomic Mass Numbers Z and A It was known that the number of the positively charged protons should equal the number of electrons. This number is called the atomic number Z. All elements are classified in the periodic table according to their atomic number The mass of a nucleus would be proportional to the sum Z+N. This number Z+N is called the Atomic Mass Number (symbol A) ٢٩ The Periodic Table ٣٠
The Periodic Table ٣١ Nuclear Spectroscopy Notations A new notation system is invented by the nuclear physicists in order to characterize each nucleus. A nucleus X with numbers Z, N and A is represented A symbolically by: Z XN or A X or simply A Z X Symbol or H or 1 H Nucleus 1 Hydrogen 1H0 4 Helium 2He2 7 Lithium 3Li 4. Uranium 143 235 92U or or or 1 1 4 2He 7 3Li 235 92 U or or or 4 He 7 Li 235 U ٣٢
The Other Two Major Discoveries in 1932 Irène and Fréderic Joliot-Curie discover the Artificial Radioactivity and Urey discovers the deuterium, an isotope of hydrogen. The Isotopes Experiments show that 3 kinds of hydrogen behave similarly when involved in chemical reactions. They should in this case occupy the same place in Mendeleev s table. That s why they are called isotopes (the Latin word isotopos means same (iso) place (topos). Hydrogen(H), deuterium(d) and tritium(t) are the isotopes of hydrogen (symbol H). In the nuclear spectroscopy notation these isotopes are represented by: 1, 2 0 1 H 3 and H 1 2 1H 1
Isotopes, Isotones and Isobars Isotopes of an element have the same Z but different N Examples: Isotopes of the hydrogen are: H, Deuterium and Tritium 1 2 0 1 H 3 1 H 2 235 and 238 92 U143 92U146 1H 1 1H 2He1 are two isotopes of Uranium Isotones are elements which have the same N but different Z 2 3 Examples: and are isotones 1H1 1 Isobars are elements which have the same A 3 3 Examples: 2 and are isobars 2He 112 Elements are known ( 1000 isotopes) ٣٥ Stability vs. Z and N The following curve represents Z vs. N for all known isotopes. Isotopes with Z = N are stable. Instability increases when getting far from the line Z=N ٣٦ = 1
٣٧ Units and Dimensions Energy and Time at the Nuclear Level ٣٨ Time Energy (Mass) Electromagnetic Interactions 10-9 10-10 s Nuclear Interactions 10-20 s Time during which nuclei are within the range of each others nuclear forces γ and β decays : 1 MeV Reactions : 10 MeV A more familiar value in nuclear physics is given using the Einstein energy-mass equivalence.
Mass Energy Equivalence الكتلة والطاقة تكافو تكافو الكتلة والطاقة والتي تعني أن مبدأ E = m c 2 العلاقة تمث ل الكتلة ي مكن أن وأن مادة) إلى ) يأيأيأيأ ي مكن أن تتحول إلى كتلة الطاقة إلى طاقة تتحول 1g mass corresponds to an energy: E = 10-3 kg x (3x10 8 ) 2 =9x10 13 J = 9x10 13 / 1.6 x 10-19 ev = 5.625 x10 32 ev Units الوحدات على المستوى الميكروسكوبي والا بعاد يتناسب نظام لا للتعبير و الوحدات استخدام لذا ن فضل ضل ل ل (... النواة الخ (الذرة الطاقة والمسافات إذ ي صبح من السهل التعبير عن الكميات التي عن في الفيزياء الميكروسكوبية ويكفي لتحويل هذه الوحدة ت صادفنا أن : معرفة نظام إلى في الفيزياء الميكروسكوبي ة للكتلة وحدة استخدام نف نف ف فضلللل الجدول التالي قيمة كتلة السكون في نحتاج فيها إلى النسبية. والتي الجسيمات الا ولي ة بهذه الوحدة. لبعض See: http://ctaps.yu.edu.jo/physics/courses/phys251/pdf/chapter-1.pdf
Rest Energy السكون طاقة Particle Symbol Mass at rest m 0 (Kg) E 0 = Energy corresponding to m 0 Electron e 9.092933 10-31 Proton p 1.672649 10-27 Neutron n 1.674955 10-27 8.181 10-14 J 511003 ev 511 KeV = 0.511 MeV 1.5033168 10-10 J 9.3828 10 8 ev 938.28 MeV 1.5074959 10-10 J 9.4218 10 8 ev 939.573 MeV Alpha α 6.644766 10-27 3727.409 MeV Rest energy for some elementary particles E 0 = m 0 c 2 The ev/c 2 Unit الكتلة: ev/c 2 وحدة Particle Symbol Mass at rest m 0 (Kg) Electron e 9.092933 10-31 Proton p 1.672649 10-27 Neutron n 1.674955 10-27 Alpha α 6.644766 10-27 E 0 0.511 MeV 938.28 MeV 939.573 MeV 3727.409 MeV m 0 (in MeV/c 2 ) 0.511 938.28 939.573 3727.409 Rest mass of some elementary particles m 0 = E 0 /c 2
Units a.m.u The atomic mass unit is defined such that the mass of one mole of (the carbon isotope) = (exactly) 12 g Which gives 1 u or 1 a.m.u = 1.660566 x 10-27 kg Which gives 1 a.m.u = 931.503 MeV/c 2 (order of magnitude = 10 3 MeV/c 2 ) ٤٣ hc (Planck c) h = 6.62 10-34 J.s h c = 6.62 10-34 J.s x 3 x10 8 m.s -1 = 19.98 10-26 J.m = 1.24125 10-7 ev.m h c 12400 ev Å h c 1240 MeV F h h c = c 1970 ev Å = 197 MeV F 2π In HEP (High Energy Physics) h equal to 1 h c 200 MeV F 1 F -1 200MeV (1 ev = 1.6x10-19 J) and c are taken ٤٤
Numerical Examples Consider a photon of wavelength λ = 10 Å The corresponding energy is E = h c / λ = 12400 ev Å / 10 Å = 1240 ev An X-Ray photon has λ= 0.1 Å. Its energy is: 0 hc 12400eV A E = 124000eV = 124 kev λ 0 01. A E 124keV Its momentum is: p = = = 124 kev c c c In γ decay the energies involved are of the order of 1 MeV corresponding to photons of wavelength λ = hc/e = 1240 F = 0.0124 Å ٤٥ De Broglie Wavelength Consider a proton (K = 2 GeV). What is the De Broglie wavelength for the proton? Solution: This proton is relativistic (K/m 0 c 2 = 2.15) λ = = h p = hc pc = E 2 hc 2 ( M c ) 2 p = 1240MeVF 2 2 ( K + M c 2 ) ( M c ) 2 ( 2928.38) 2 ( 938. 28 ) 2 p hc 1240 MeV F 2774 MeV p 0.45 F See: http://ctaps.yu.edu.jo/physics/courses/phys251/pdf/chapter-3.pdf ٤٦
Next Lecture Chapter 2 : Review of Nuclear Structure and Properties (QGRI/HFWXUH