Chapter 43 Nuclear Physics 1. Properties of nuclei 2. Nuclear binding and nuclear structure 3. Nuclear stability and radioactivity 4. Activities and half lives 5. Biological effects of radiation 6. Nuclear reactions 7. Nuclear fission 8. Nuclear fusion
Nuclear energy: good & evil Nuclear medicine Proton therapy Little Boy Atomic Bomb Spring 2010 PHYS 53 Eradat SJSU 2
Properties of the nuclei: mass, radius & density Scattering experiments following the Rutherford experiment show that we can model the nucleus as a sphere with radius of R : 1/3 R= R0 A Where A is the nucleon number or mass number and it is the nearest whole number to the mass of the nucleous in atomic mass unit (amu or u for short). 27 1u = 1.66053886(28) 10 kg 15 R0 is an experimentally determined constant R0 = 1.2 10 m= 1.2 fm m p = mass of proton m = mass of n utron 1 u Calculating nuclear density: n m Au Au 3 u ρ = = = = =constant 3 V 4 3 3 1/3 πr 4 π 0 ( R0 A ) 4 π R 3 3 Nuclear density is independent of the type of material. Example: Mass of iron nucleus is 56. Calculate its nuclear radius, approximate mass, and density. Spring 2010 PHYS 53 Eradat SJSU 3
Nuclides & isotopes The basic building blocks of the nucleus: 27 Z : atomic number = number of protons mp = 1.007276u = 1.672622 10 kg = = = 27 N : neutron number number of neutrons mn 1.008665 u 1.674927 10 kg 31 For a neutral atom number of electrons = Z m = 0.000548580u = 9.10938 10 kg A: the nucleun number of mass number A= Z + N e Isotopes: are the atoms with same atomic number Z but different neutron number N Chemical properties of the matter are determined by its electrons which is determined by Z. Isotopes are the same material from chemical point of view. Physical properties of the material such as melting/boiling point, diffusion rate, etc. change with their atomic mass or A We represent the atoms with their chemical symbol with atomic number and atomic mass as follows: Atomic mass A atomic number Symbol Z C A is usually a good approximation for the atomic mass. Spring 2010 PHYS 53 Eradat SJSU 4
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Nuclear binding and nuclear force What do you think of the relationship between E & E 1 +E 2 +E 3? Why the nucleus should exist instead of the individual nucleons? E < E1 + E2 + E3 Lower energy and more stability is the key for any reaction and nuclear reactions are not exceptional. We define the nuclear binding energy: ( ) E = Z m c + m c + Nm c Mc 2 2 2 A 2 B p e n Z E = ZM + Nm M c > ing; < 0 no bonding A 2 B ( H n Z ) 0 bond Where M H = mp + me Is it OK to use M = m + m? Nucleon 1 Nucleon 2 Nucleon 3 E 1 E 2 E 3 Nucleus E H p e Spring 2010 PHYS 53 Eradat SJSU 7
Example: Binding energy/nucleon Find the binding energy per nucleon of the 62 Ni if its neutral atomic mass is 61.928349u ( A ) 2 E = ZM + Nm M c B H n Z A We have Ni so Z = 28; A = 62 & N = A Z = 34; M = 61.928349 u; 8 27 M H = 1.007825 u; mn = 1.008665 u; c = 2.997925 10 m/ s; 1u = 1.660539 10 kg Use the exact value 2 2 931.5MeV uc = 931.5MeV c = u E B B B = ( ZM + Nm A M ) c 2 H n Z ( 28 1.007825 34 1.008665 61.928349 ) 931.5 E = u+ u u E = 545.3 MeV!!! > 0 so this atom can be stable. Keep the energies in units of ev for nuclear physics. E / nucleon= E / A= 545.3 MeV / 62 B B E / nucleon = 8.795 MV MeV / nucleoon?!!! B Compare this to the atomic binding energies. Z MeV u Spring 2010 PHYS 53 Eradat SJSU 8
Binding energy/nucleon A very important graph 8.8 The highest E B /nucleon A big jump in E B Spring 2010 PHYS 53 Eradat SJSU 9
The nuclear force The nuclear force binds the protons to protons despite of the Coulomb repulsion. It belongs to the category of the strong forces Does not depend on the charge and both neutrons and protons are bound the same way. It is a very short range force ~10 15 m (otherwise very large nuclei would be possible to form) There is a saturation property due to the very short range similar to covalent bonding. There is pairing between the nucleons of opposite spin and pairing of pairs like alpha particles (protons and neutrons are spin 1/2 particles or they are fermions). Much stronger than the electrical forces (otherwise the nucleus would fall apart). We don t have a clear picture of inside of the nuclei We gain some insight by studying simple models such as liquid drop or shell model. Spring 2010 PHYS 53 Eradat SJSU 10
Liquid drop model (1928 George Gamow) Based on the observation that all the nuclei have almost the same density. The nuclei are hold together similar to the molecules of a liquid drop with short-range interactions and surface-tension effects. Consider the following contributions to the binding energy: 1) Saturation: an individual nucleon interacts only with few of its nearest neighbors E = CA. S 1 2) Surface effect: the nucleons on the surface are less tightly bound than interior ones. surface 2 2 1/3 2/3 0 2 Reduction Surface E R = C R A = C A 4 π 4π ( ) ( Z ) 3) Coulomb force: every one of Z pronons repels 1 other protons with a force proportional to ( ) Z ( Z ) 2 Z Z 1) 1 1/ R E = Coulomb C C3 1/3 potential 4πε R = A 0 Reduction Spring 2010 PHYS 53 Eradat SJSU 11
Liquid drop model (1928 George Gamow) II 4) The most stable muclei are the ones with balanced number of neutrons and protons. As nuclei depart from this rule a negative energy term is introduced to the Based on experimental data E ( ) = ( ) balance 2 2 4 N Z / A C A 2 Z / A 5) Pairing: nuclear force favors pairing of protons and neutrons. Pairing energy is experimentally determined Epairing =± CA -4/3 5. It is + for both N & Z even and for both N & Z odd and zero otherwise. E B. Spring 2010 PHYS 53 Eradat SJSU 12
Binding energy based on liquid drop model ( -1 ) ( - 2 ) 2 2/3 Z Z A Z -4/3 Putting all together: EB = C 1A C2A C3 C 1/3 4 ± C5A A A Experimentally determined constants: C1 = 15.75 MeV C2 = 17.80MeV C3 C = 23.69MeV C = 39MeV 4 5 Saturation Surface Pairing Coulomb p n balance M M B Z H n Z H n 2 = 0.7100MeV We can use E formula to estimate mass of any neutral atom: B M = ZM + Nm Δm M = ZM + Nm E c A semiempirical formula To separate all of the nucleons of an atom we need to invest energy as much as E EB B which is equivalent to convertion of that energy to mass of Δ m = 2 c M EB ZM H + Nmn = Z M + Example of conservation of mass-energy law. c 2 The model explains the nuclear masses and decay of the unstable nuclei but has no use in explanation of the angular nomentum and excited states. Spring 2010 PHYS 53 Eradat SJSU 13
Example: binding energy and mass 1) CA= 976.5MeV Calculate the five terms in 1 2/3 2) CA 2 = 278.8MeV the binding energy and the total E B for the 62 Z( Z 1) 28Ni 3) C3 = 135.6MeV 1/3 A nucleus. compare it to E B ( A 2Z) 2 calculates based on 4) C4 = 13.8MeV A measured M=61.928349u 4/3 5) + CA 5 = 0.2MeV Find its neutral atomic EB = 548.5MeV mass and compare it to 545.3 MeV, 6% the measured value M = 28 ( 1.007825 u ) + 34 ( 1.008665 u ) Compared to it is larger. = + 548.5MeV 931.5 MeV / u M = 61.925u Δ u% = 61.928349u 61.925u 100 = 0.005% 61.928349 u + 61.925 0.5 Not bad!! ( ) Spring 2010 PHYS 53 Eradat SJSU 14
Undergraduate Research at SETI in Astrobiology Apply online at http://www.seti.org/ursa Projects are all related to the study of life on Earth and the search for life on other planets: Laboratory and computational chemistry Computer simulations of star and planet formation Field and lab investigations in geology and biology URSA students must be: science majors enrolled in a BA/BS degree program at SJSU during AY 2010 11 enrolled in 6 14 units in AY 2010 11 (including one unit for this program) available to work at SETI in Mountain View for 10 hours/week ($15/hour) in good academic standing, and completed Physics 2a/b or 50 series, and Chem 1a US citizens or permanent residents For more information, contact Dr. Kress at mkress@science.sjsu.edu Spring 2010 PHYS 53 Eradat SJSU 15
The shell model Using the central-field ld approximation, we assume each nucleon moving in a averaged-out potential field of all other nucleons. Due to short-range nature of the nuclear forces this may not work but it explaoins many properties of the nuclei surprisingly well. The potential field is approximated by a 3D square potential well. We put that together with the potential for the mutual repulsion of the protons. We can solve the Schrodinger equation and find the energy levels, shells and subshells. Spring 2010 PHYS 53 Eradat SJSU 16
The shell model and the magic numbers The central field approximation in atomic model successfully explained the chamical stability of the inert gasses as those atoms with complete shell. Z = 2,10,16,18,36,54 Nuclear structure exhibits a comparable effect with different atomic numbers called magic numbers: ZorN= 2,8, 20, 28,50,82,126 82 The reason for the difference is the difference in the potential-energy function and also a much stronger spin-orbit interaction in the nuclear structure. Nuclei with both N and Z equal to a magic number are called doubly magic nucleides. Magic and doubly magic nucleides have: a) unusually high number of stable eisotopes, b) zero nuclear spin, c) filled-shell or sub-shell, d) large jump in E B compared to their neighbors. Spring 2010 PHYS 53 Eradat SJSU 17
Nuclear stability and radioactivity Segre chart (1905 1989) 2500 known nuclides 300 stable nuclides Radioactive nuclides: unstable nuclides emit particles and EM radiation to become stable. The process can take from microseconds to billions of years. Segrechartshows the stable nuclides as a function of N and Z. The blue lines are the A=constant lines. Only 4 stable odd odd nuclides exist. Extreme stability of the doubly magic 4 2He prevents A=5 A5 and A=8 nuclides from existence. Stable nuclides are shifted above N=Z line for higher A. Neutrons provide some shielding of the strong proton proton repulsion li Stable nuclides in this area are prevented by strong protonproton repulsion Spring 2010 PHYS 53 Eradat SJSU 18
3 D version of the Segre chart Unstable nuclides with higher A and higher N/Z ratio decay to lighter nuclides with balanced N to Z ration closer to one. In the process neutrons convert to protons and 90% of the known nuclides (2500) are radioactive and they decay to lighter and more stable nuclides through the processes such as processes α decay β decay γ decay Spring 2010 PHYS 53 Eradat SJSU 19
Alpha decay 4 An α particle is He nucleus with two protons and two neutrons with total spin of zero. The nuclei that t are too large to exist can reduce their A by α emission. i The nuclear reaction has to satisfy several conservation laws (can you name them?). A+ B C+ D+ Energy A A A A Z Z Z Z A B C D A B C D Mass-Energy conservation: γ M c + γ M c = γ M c + γ M c ; E = γmc = K + mc 2 2 2 2 2 2 A A B B C C D D total We need to decide if relativistic or classical treatment is needed for a given situation. 226 222 * 4 -decay 88 Ra 86 Rn + Kd + 2 He + Kα Note: M p > M d + Mα to have α possible. Mp M d M α 7 VHe = 1.52 10 m/ s = 0.05c 1 2 2 13 KEnr = mv = 7.7 10 J = 4.8MeV What is the mechanism of escape? Spring 2010 PHYS 53 Eradat SJSU 20
There are three types of β decay: Beta decay Beta-minus β (electron): n p+ β + νe happens whenever N / Z is too large; AA AA Z A ; A Z C+ β + ν A 1 e + Energy M p > M + d + + Beta-plus β (positron): p n+ β + νe happens whenever N / Z is too small; AA AA + Z A C+ + ν + Energy ; M M > 2m A Z β A 1 e p d e Electron capture: p+ β n+ νe happens whenever N / Z is too small; + β is energetically not possible AA AA Z A A Z C+ ν A 1 e + Energy; M p > M d The electron is usually captured from the K shell Important t for β decay: m = 1.008665 u > m = 1.007276 u n The nutrino ν and anti-nutrino ν are particle-antiparticle pairs with little mass e e and no charge. Very difficult to detect (detected by Raines & Cowan in 1953). p Spring 2010 PHYS 53 Eradat SJSU 21
Gamma decay γ -rays are high-energy photons 10KeV-5MeV emitted from the nuclei that are in excited state or undergoing nuclear reactions. This is simmilar to emission from the atoms when they relax from an excited state to the ground state. The difference is in magnitude of the energy of each emitted photon which depands on the spacing of the energy levels. Spring 2010 PHYS 53 Eradat SJSU 22
Natural radioactivity Reactions on the Segre chart We have all kinds of heavy nuclei from the time of formation of stars and planets. All the unstable ones are in the process of decaying into stable ones. Some of these processes are very slow. Example: Segre chart of 238 U decay series terminating with stable nuclide 206 Pb. Note the hlfli half lives (blue fonts) Spring 2010 PHYS 53 Eradat SJSU 23
Activities & half lives Rate of decay of a radioactive nucleon is not affected by any chemical or physical environmental effects. It is a totally random phenomenon. We analyze it with statistical techniques. (): Number (ver (): ( ) () N t : y large) of radioactive nuclei dn t change (negative) in N t during a short time dt dn t dt activity or decay rate that is proportional to the amount of material present in the sample or λ: decay constant or probability of decay/t dn t () λ ln () N ( ) 0 N t ( ) λ () N t λt 0 0 = dt N t = t N t = N e Half-life of the radioactive material: N 2 0 () dn t =λn t dt () large for rapidly decaying small for the slowly decaying Ne λ T 1/2 λt1/2 = e 0 # of remaining nuclei 1 ln 2 0.693 = T1/2 = = 2 λ λ 1 1 T1/2 T1/2 10 10 Lifetime : Tmean = = = = 1Ci = 3.70 10 Bq = 3.70 10 decays/ s decay constant λ ln 2 0.693 Unit of activity Spring 2010 PHYS 53 Eradat SJSU 24
Example: Activity of 57 Co Rdi 57 57 Radioactive isotope of 57 Co decays by 27 Co 26 Fe + +ν e 7 electron capture with a half life of T1/2 = 272days = ( 272days)( 86400 s / day) = 2.35 10 s 272 days. 7 2.35 10 s 7 a) Findthe decay constant and the a ) Lifetime: T mean = T 1/2 / ln 2 = = 3.39 10 s 0.693 lifetime of the process. 8 Decay constant λ = 1/ Tmean = 2.95 10 / s b) If you have a radiation source of 57 Co with activity of 2.00μCi, how dn () t b ) The activity = 2.00 μ Ci many radioactive nuclei does it dt contain? 6 10 4 = ( 2.00 10 )( 3.70 10 / s) = 7.40 10 decays / s c) What will be the activity of the 4 sourceafter one year? dn ( t ) / dt 7.40 10 decays / s # of nuclei: N( t) = = 8 λ 2.95 10 / s 12 N t = 2.51 10 nuclei () t c ) activity after 1 year: N (1 y ) = N λ 0e 8 ( 2.95 10 / s) 365 day/ y( 86400 s/ day) = Ne = 0.394N 0 0 dn ( t ) / dt = λn ( 1y ) = λ ( 0.394N0 ) = λn0 0.394 ( ) ( ) μ ( ) Original activity dn 1 y / dt = (2.00 Ci) 0.394 = 0.788μCi Spring 2010 PHYS 53 Eradat SJSU 25
Radiocarbon dating Spring 2010 PHYS 53 Eradat SJSU 26
Nuclear spins & magnetic moments Spring 2010 PHYS 53 Eradat SJSU 27
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Radioactive dating Spring 2010 PHYS 53 Eradat SJSU 29
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