Nuclear Physics PDF - 2007 Lecture Notes

Chapter 39 - Nuclear Physics AA PowerPoint PowerPoint Presentation Presentation by by Paul Paul E. E. Tippens, Tippens, Professor Professor of of Physics...

Chapter 39 - Nuclear Physics AA PowerPoint PowerPoint Presentation Presentation by by Paul Paul E. E. Tippens, Tippens, Professor Professor of of Physics Physics Southern Southern Polytechnic Polytechnic State State University University © 2007 Objectives: After completing this module, you should be able to: Define and apply the concepts of mass number, atomic number, and isotopes. Calculate the mass defect and the binding energy per nucleon for a particular isotope. Define and apply concepts of radioactive decay and nuclear reactions. State the various conservation laws, and discuss their application for nuclear reactions.  Composition of Matter All All of of matter matter isis composed composed ofof at at least least three three fundamental fundamental particles particles (approximations): (approximations): Particle Fig. Sym Mass Charge Size Electron e- 9.11 x 10-31 kg -1.6 x 10-19 C  Proton p 1.673 x 10-27 kg +1.6 x 10-19 C 3 fm Neutron n 1.675 x 10-31 kg 0 3 fm The mass of the proton and neutron are close, but they are about 1840 times the mass of an electron.  The Atomic Nucleus Compacted nucleus: 4 protons 5 neutrons Since atom is electri- cally neutral, there must be 4 electrons. 4 electrons Beryllium Atom  Modern Atomic Theory The Bohr atom, which is sometimes shown with electrons as planetary particles, is no longer a valid representation of an atom, but it is used here to simplify our discussion of energy levels. The uncertain position of an electron is now described as a probability distribution—loosely referred to as an electron cloud.  Definitions A nucleon is a general term to denote a nuclear particle - that is, either a proton or a neutron. The atomic number Z of an element is equal to the number of protons in the nucleus of that element. The mass number A of an element is equal to the total number of nucleons (protons + neutrons). The mass number A of any element is equal to the sum of the atomic number Z and the number of neutrons N : A=N+Z  Symbol Notation AA convenient convenient way way of of describing describing anan element element isis by by giving giving its its mass mass number number and and its its atomic atomic number, number, along along with with the the chemical chemical symbol symbol forfor that that element. element. A Z X Mass number Atomic number  Symbol  9 For example, consider beryllium (Be): 4 Be Example 1: Describe the nucleus of a lithium atom which has a mass number of 7 and an atomic number of 3. A = 7; Z = 3; N = ? N=A–Z= 7-3 neutrons: N = 4 Protons: Z=3 Electrons: Same as Z 7 3 Li Lithium Atom  Isotopes of Elements Isotopes are atoms that have the same number of protons (Z1= Z2), but a different number of neutrons (N). (A1  A2) 3 4 2 He Isotopes 2 He of helium Helium - 3 Helium - 4  Nuclides Because of the existence of so many isotopes, the term element is sometimes confusing. The term nuclide is better. A nuclide is an atom that has a definite mass number A and Z-number. A list of nuclides will include isotopes. The following are best described as nuclides: 3 4 12 13 2 He 2 He 6 C 6 C  Atomic Mass Unit, u One atomic mass unit (1 u) is equal to one- twelfth of the mass of the most abundant form of the carbon atom--carbon-12. Atomic mass unit: 1 u = 1.6606 x 10-27 kg Common atomic masses: Proton: 1.007276 u Neutron: 1.008665 u Electron: 0.00055 u Hydrogen: 1.007825 u  Exampe 2: The average atomic mass of Boron-11 is 11.009305 u. What is the mass of the nucleus of one boron atom in kg? 11 5 B = 11.009305 Electron: 0.00055 u The mass of the nucleus is the atomic mass less the mass of Z = 5 electrons: Mass = 11.009305 u – 5(0.00055 u) 1 boron nucleus = 11.00656 u  1.6606 x 10-27 kg  m  11.00656 u   m m== 1.83 1.83 xx 10 -26 kg 10-26 kg  1 u   Mass and Energy Recall Einstein’s equivalency formula for m and E: E  mc ; c  3 x 10 m/s 2 8 The energy of a mass of 1 u can be found: E = (1 u)c2 = (1.66 x 10-27 kg)(3 x 108 m/s)2 E = 1.49 x 10-10 J Or E = 931.5 MeV When converting amu to energy: c  931.5 2 MeV u Example 3: What is the rest mass energy of a proton (1.007276 u)? E = mc2 = (1.00726 u)(931.5 MeV/u) Proton: EE = Proton: = 938.3 938.3 MeV MeV Similar conversions show other rest mass energies: Neutron: EE = Neutron: = 939.6 939.6 MeV MeV Electron: EE = Electron: = 0.511 0.511 MeV MeV  The Mass Defect The The mass mass defect defect isis the the difference difference between between the the rest rest mass mass of of aa nucleus nucleus and and the the sum sum of of the the rest rest masses masses of of its its constituent constituent nucleons. nucleons. The whole is less than the sum of the parts! Consider the carbon-12 atom (12.00000 u): Nuclear mass = Mass of atom – Electron masses = 12.00000 u – 6(0.00055 u) = 11.996706 u The nucleus of the carbon-12 atom has this mass. (Continued...)  Mass Defect (Continued) Mass of carbon-12 nucleus: 11.996706 Proton: 1.007276 u Neutron: 1.008665 u The nucleus contains 6 protons and 6 neutrons: 6 p = 6(1.007276 u) = 6.043656 u 6 n = 6(1.008665 u) = 6.051990 u Total mass of parts: = 12.095646 u Mass defect mD = 12.095646 u – 11.996706 u m mDD == 0.098940 0.098940 uu  The Binding Energy The The binding energy EEBB of binding energy of aa nucleus nucleus isis the the energy energy required required to to separate separate aa nucleus nucleus intointo its its constituent constituent parts. parts. EB = mDc2 where c2 = 931.5 MeV/u The binding energy for the carbon-12 example is: EB = (0.098940 ( u)(931.5 MeV/u) Binding EB for C-12: EB = 92.2 MeV  Binding Energy per Nucleon An An important important wayway ofof comparing comparing the the nuclei nuclei of of atoms atoms isis finding finding their their binding binding energy energy per per nucleon: nucleon: Binding energy EB  MeV  =   per nucleon A  nucleon  For our C-12 example A = 12 and: EB 92.2 MeV   7.68 nucleon MeV A 12  Formula for Mass Defect The following formula is useful for mass defect: Mass defect mD   ZmH  Nmn   M  mD mH = 1.007825 u; mn = 1.008665 u Z is atomic number; N is neutron number; M is mass of atom (including electrons). By By using using the the mass mass ofof the the hydrogen hydrogen atom, atom, you you avoid avoid the the necessity necessity of of subtracting subtracting electron electron masses. masses. Example 4: Find the mass defect for the He 4 2 nucleus of helium-4. (M = 4.002603 u) Mass defect mD   ZmH  Nmn   M  mD ZmH = (2)(1.007825 u) = 2.015650 u Nmn = (2)(1.008665 u) = 2.017330 u M = 4.002603 u (From nuclide tables) mD = (2.015650 u + 2.017330 u) - 4.002603 u m mDD == 0.030377 0.030377 uu Example 4 (Cont.) Find the binding energy per nucleon for helium-4. (mD = 0.030377 u) EEBB == m c22 where c22 = 931.5 MeV/u mDDc where c = 931.5 MeV/u EB = (0.030377 u)(931.5 MeV/u) = 28.3 MeV AA total total of of 28.3 28.3 MeV MeV isis required required To To tear tear apart apart the the nucleons nucleons from from thethe He-4 He-4 atom. atom. Since there are four nucleons, we find that EB 28.3 MeV   7.07 MeV nucleon A 4  Binding Energy Vs. Mass Number Curve shows that Binding Energy per nucleon EB increases with 8 A and peaks at 6 A = 60. Heavier nuclei are less 4 stable. 2 Green region is for most stable atoms. 50 100 150 200 250 Mass number A For heavier nuclei, energy is released when they break up (fission). For lighter nuclei, energy is released when they fuse together (fusion).  Stability Curve Nuclear Nuclear particles particles are are 140 held held together together by by aa Stable Neutron number N 120 nuclear nuclear strong strong force. force. nuclei 100 A stable nucleus remains 80 forever, but as the ratio 60 of N/Z gets larger, the 40 atoms decay. Z=N 20 Elements Elements with with ZZ > > 82 82 20 40 60 80 100 are are all all unstable. unstable. Atomic number Z  Radioactivity As As the the heavier heavier atoms atoms become become   more more unstable, unstable, particles particles and and photons photons are are emitted emitted from from the the  nucleus nucleus and and itit isis said said to to be be  radioactive. radioactive. All All elements elements with with AA > > 82 82 are are radioactive. radioactive. Examples are: Alpha particles   particles (electrons) Gamma rays   particles (positrons)  The Alpha Particle An alpha particle  is the nucleus of a helium atom consisting of two protons and two neutrons tightly bound. Charge = +2e- = 3.2 x 10-19 C Mass = 4.001506 u Relatively low speeds ( 0.1c ) Not very penetrating  The Beta-minus Particle A beta-minus particle  is simply an electron that has been expelled from the nucleus. - Charge = e- = -1.6 x 10-19 C - Mass = 0.00055 u - High speeds (near c) - Very penetrating  The Positron A beta positive particle  is essentially an electron with positive charge. The mass and speeds are similar. + Charge = +e- = 1.6 x 10-19 C + Mass = 0.00055 u + High speeds (near c) + Very penetrating  The Gamma Photon A gamma ray  has very high electromagnetic radiation carrying energy away from the nucleus.  Charge = Zero (0)  Mass = zero (0)  Speed = c (3 x 108 m/s)  Most penetrating radiation  Radioactive Decay As discussed, when the ratio of N/Z gets very large, the nucleus becomes unstable and often particles and/or photons are emitted. Alpha decay 2  results in the loss of two 4 protons and two neutrons from the nucleus. A Z X Y    energy A 4 Z 2 4 2 X is parent atom and Y is daughter atom The energy is carried away primarily by the K.E. of the alpha particle. Example 5: Write the reaction that occurs when radium-226 decays by alpha emission. A Z X A 4 Z 2 Y    energy 4 2 226 88 Ra 226  4 88  2 Y    energy 4 2 From tables, we find Z and A for nuclides. The daughter atom: Z = 86, A = 222 226 88 Ra 222 86 Rn    energy 4 2 Radium-226 Radium-226 decays decays into into radon-222. radon-222.  Beta-minus Decay Beta-minus  decay results when a neutron decays into a proton and an electron. Thus, the Z-number increases by one. A Z X Y    energy A Z 1 0 1 X is parent atom and Y is daughter atom The energy is carried away primarily - by the K.E. of the electron.  Beta-plus Decay Beta-plus  decay results when a proton decays into a neutron and a positron. Thus, the Z-number decreases by one. A Z X Y    energy A Z 1 0 1 X is parent atom and Y is daughter atom The energy is carried away primarily + by the K.E. of the positron.  Radioactive Materials The rate of decay for radioactive substances is expressed in terms of the activity R, given by: N N = Number of Activity R t undecayed nuclei One One becquerel becquerel ((Bq) Bq) isis an an activity activity equal equal to to one one disintegration per second (1 s disintegration per second (1 s ). 1). --1 One One curie curie ((Ci) Ci) isis the the activity activity of of aa radioactive radioactive material that decays at the rate of 3.7 material that decays at the rate of 3.7 x 1010 Bqx 10 10 Bq or or 3.7 3.7 xx 10 10 disintegrations 1010 disintegrations per per second. second.  The Half-Life The half-life T1/2 of No Number Undecayed Nuclei an isotope is the time in which one- half of its unstable nuclei will decay. N0 2 n 1 N0 N  N0   2 4 Where n is number 1 2 3 4 of half-lives Number of Half-lives  Half-Life (Cont.) The same reasoning will apply to activity R or to amount of material. In general, the following three equations can be applied to radioactivity: Nuclei Remaining Activity R n n 1 1 N  N0   R  R0   2 2 Mass Remaining Number of Half-lives: n 1 n t m  m0   T12 2 Example 6: A sample of iodine-131 has an initial activity of 5 mCi. The half-life of I-131 is 8 days. What is the activity of the sample 32 days later? First we determine the number of half-lives: t 32 d n  n = 4 half-lives T1/ 2 8 d n 4 1 1 R  R 0    5 mCi   RR == 0.313 2 2 0.313 mCi mCi There There would would also also be be 1/16 1/16 remaining remaining of of the the mass mass and and 1/16 1/16 of of the the number number of of nuclei. nuclei.  Nuclear Reactions It is possible to alter the structure of a nucleus by bombarding it with small particles. Such events are called nuclear reactions: General Reaction: x+XY+y For example, if an alpha particle bombards a nitrogen-14 nucleus it produces a hydrogen atom and oxygen-17: 4 2  N H O 14 7 1 1 17 8  Conservation Laws For any For any nuclear nuclear reaction, reaction, there there are are three three conservation laws conservation laws which which must must bebe obeyed: obeyed: Conservation of Charge: The total charge of a system can neither be increased nor decreased. Conservation of Nucleons: The total number of nucleons in a reaction must be unchanged. Conservation of Mass Energy: The total mass- energy of a system must not change in a nuclear reaction. Example 7: Use conservation criteria to determine the unknown element in the following nuclear reaction: 1 1 H  Li He  X  energy 7 3 4 2 A Z Charge before = +1 + 3 = +4 Charge after = +2 + Z = +4 Z=4–2=2 (Helium has Z = 2) Nucleons before = 1 + 7 = 8 Nucleons after = 4 + A = 8 (Thus, A = 4) 1 1 H  Li He  He  energy 7 3 4 2 4 2  Conservation of Mass-Energy There is always mass-energy associated with any nuclear reaction. The energy released or absorbed is called the Q-value and can be found if the atomic masses are known before and after. 1 1 H  37 Li 42 He  42 He  Q Q   11 H  37 Li    24 He  24 He  Q is the energy released in the reaction. If Q is positive, it is exothermic. If Q is negative, it is endothermic. Example 8: Calculate the energy released in the bombardment of lithium-7 with hydrogen-1. 1 1 H  37 Li 42 He  42 He  Q Q   11 H  37 Li    24 He  24 He  1 1 H  1.007825 u 4 2 He  4.002603 u 7 3 Li  7.016003 u 4 2 He  4.002603 u Substitution of these masses gives: Q = 0.018622 u(931.5 MeV/u) QQ =17.3 =17.3 MeV MeV The positive Q means the reaction is exothermic.  Summary Fundamental atomic and nuclear particles Particle Fig. Sym Mass Charge Size Electron e 9.11 x 10-31 kg -1.6 x 10-19 C  Proton p 1.673 x 10-27 kg +1.6 x 10-19 C 3 fm Neutron n 1.675 x 10-31 kg 0 3 fm The mass number A of any element is equal to the sum of the protons (atomic number Z) and the number of neutrons N : A=N+Z  Summary Definitions: A nucleon is a general term to denote a nuclear particle - that is, either a proton or a neutron. The mass number A of an element is equal to the total number of nucleons (protons + neutrons). Isotopes are atoms that have the same number of protons (Z1= Z2), but a different number of neutrons (N). (A1  A2) A nuclide is an atom that has a definite mass number A and Z-number. A list of nuclides will include isotopes.  Summary (Cont.) Symbolic notation for atoms A Z X Mass number Atomic number  Symbol  Mass defect mD mD   ZmH  Nmn   M  Binding energy EB = mDc2 where c2 = 931.5 MeV/u Binding Energy EB  MeV  =   per nucleon A  nucleon   Summary (Decay Particles) An alpha particle  is the nucleus of a helium atom consisting of two protons and two tightly bound neutrons. A beta-minus particle  is simply an electron that has been expelled from the nucleus. A beta positive particle  is essentially an electron with positive charge. The mass and speeds are similar. A gamma ray  has very high electromagnetic radiation carrying energy away from the nucleus.  Summary (Cont.) Alpha Decay: A Z X A 4 Z 2 Y    energy 4 2 Beta-minus Decay: A Z X Y    energy A Z 1 0 1 Beta-plus Decay: A Z X Y    energy A Z 1 0 1  Summary (Radioactivity) The The half-life TT1/2 half-life 1/2 of of an an isotope isotope is is the the time time in in which which one -half of one-half of its its unstable unstable nuclei nuclei will will decay. decay. Nuclei Remaining Activity R n n 1 1 N  N0   R  R0   2 2 Mass Remaining Number of Half-lives: n 1 n t m  m0   T1 2 2  Summary (Cont.) Nuclear Reaction: x + X  Y + y + Q Conservation of Charge: The total charge of a system can neither be increased nor decreased. Conservation of Nucleons: The total number of nucleons in a reaction must be unchanged. Conservation of Mass Energy: The total mass- energy of a system must not change in a nuclear reaction. (Q-value = energy released) CONCLUSION: Chapter 39 Nuclear Physics