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Fund Quantum Mechanics Lect & HW Solutions 129

Course: PHY 3604, Fall 2011
School: UNF
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12 Angular Chapter momentum 12.1 Introduction 12.2 The fundamental commutation values relations 12.3 Ladders 12.4 Possible of angular momentum 111

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12 Angular Chapter momentum 12.1 Introduction 12.2 The fundamental commutation values relations 12.3 Ladders 12.4 Possible of angular momentum 111
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UNF - PHY - 3604
112CHAPTER 12. ANGULAR MOMENTUM12.5A warning about angular momentum12.6Triplet and singlet states12.7Clebsch-Gordan coecients12.8Some important results12.9Momentum of partially lled shells12.10Pauli spin matrices12.11General spin matrices
UNF - PHY - 3604
12.12. THE RELATIVISTIC DIRAC EQUATION12.12The Relativistic Dirac Equation113
UNF - PHY - 3604
114CHAPTER 12. ANGULAR MOMENTUM
UNF - PHY - 3604
Chapter 13Electromagnetism13.1The Electromagnetic Hamiltonian13.2Maxwells Equations13.3Example Static Electromagnetic Fields13.3.1Point charge at the origin13.3.2Dipoles115
UNF - PHY - 3604
116CHAPTER 13. ELECTROMAGNETISM13.3.3Arbitrary charge distributions13.3.4Solution of the Poisson equation13.3.5Currents13.3.6Principle of the electric motor13.4Particles in Magnetic Fields13.5Stern-Gerlach Apparatus13.6Nuclear Magnetic Reso
UNF - PHY - 3604
13.6. NUCLEAR MAGNETIC RESONANCE13.6.1Description of the method13.6.2The Hamiltonian13.6.3The unperturbed system13.6.4Eect of the perturbation117
UNF - PHY - 3604
118CHAPTER 13. ELECTROMAGNETISM
UNF - PHY - 3604
Chapter 14Nuclei [Unnished Draft]14.1Fundamental Concepts14.2The Simplest Nuclei14.2.1The proton14.2.2The neutron14.2.3The deuteron119
UNF - PHY - 3604
120CHAPTER 14. NUCLEI [UNFINISHED DRAFT]14.2.4Property summary14.3Modeling the Deuteron14.4Overview of Nuclei14.5Magic numbers14.6Radioactivity14.6.1Decay rate14.6.2Other denitions
UNF - PHY - 3604
14.7. MASS AND ENERGY14.7Mass and energy14.8Binding energy14.9Nucleon separation energies14.10Liquid drop model14.10.1Nuclear radius14.10.2von Weizscker formulaa14.10.3Explanation of the formula121
UNF - PHY - 3604
122CHAPTER 14. NUCLEI [UNFINISHED DRAFT]14.10.4Accuracy of the formula14.11Alpha Decay14.11.1Decay mechanism14.11.2Comparison with data14.11.3Forbidden decays14.11.4Why alpha decay?14.12Shell model
UNF - PHY - 3604
14.13. COLLECTIVE STRUCTURE14.12.1Average potential14.12.2Spin-orbit interaction14.12.3Example occupation levels14.12.4Shell model with pairing14.12.5Conguration mixing14.12.6Shell model failures14.13Collective Structure123
UNF - PHY - 3604
124CHAPTER 14. NUCLEI [UNFINISHED DRAFT]14.13.1Classical liquid drop14.13.2Nuclear vibrations14.13.3Nonspherical nuclei14.13.4Rotational bands14.13.4.1Basic notions in nuclear rotation14.13.4.2Basic rotational bands14.13.4.3Bands with intri
UNF - PHY - 3604
14.14. FISSION14.13.4.5Even-even nuclei14.13.4.6Non-axial nuclei14.14Fission14.14.1Basic concepts14.14.2Some basic features14.15Spin Data14.15.1Even-even nuclei125
UNF - PHY - 3604
126CHAPTER 14. NUCLEI [UNFINISHED DRAFT]14.15.2Odd mass number nuclei14.15.3Odd-odd nuclei14.16Parity Data14.16.1Even-even nuclei14.16.2Odd mass number nuclei14.16.3Odd-odd nuclei14.16.4Parity Summary
UNF - PHY - 3604
14.17. ELECTROMAGNETIC MOMENTS14.17Electromagnetic Moments14.17.1Classical description14.17.1.1Magnetic dipole moment14.17.1.2Electric quadrupole moment14.17.2Quantum description14.17.2.1Magnetic dipole moment14.17.2.2Electric quadrupole mom
UNF - PHY - 3604
128CHAPTER 14. NUCLEI [UNFINISHED DRAFT]14.17.2.4Values for deformed nuclei14.17.3Magnetic moment data14.17.4Quadrupole moment data14.18Isospin14.18.1Basic ideas14.18.2Heavier nuclei14.18.3Additional points
UNF - PHY - 3604
14.19. BETA DECAY14.18.4Why does this work?14.19Beta decay14.19.1Energetics Data14.19.2Von Weizscker approximationa14.19.3Kinetic Energies14.19.4Forbidden decays14.19.4.1Allowed decays14.19.4.2Forbidden decays allowed129
UNF - PHY - 3604
130CHAPTER 14. NUCLEI [UNFINISHED DRAFT]14.19.4.3The energy eect14.19.5Data and Fermi theory14.19.6Parity violation14.20Gamma Decay14.20.1Energetics14.20.2Forbidden decays14.20.3Isomers
UNF - PHY - 3604
14.20. GAMMA DECAY14.20.4Weisskopf estimates14.20.5Comparison with data14.20.6Cage-of-Faraday proposal14.20.7Internal conversion131
UNF - PHY - 3604
132CHAPTER 14. NUCLEI [UNFINISHED DRAFT]
UNF - PHY - 3604
Appendix AAddendaA.1Lagrangian mechanicsA.1.1IntroductionA.1.2Generalized coordinatesA.1.3Lagrangian equations of motionA.1.4Hamiltonian dynamics133
Syracuse - IST - 444
Ashley StephensIST 444Laurie JervaLogo AssignmentWhen designing the logo for my groups company, Syracuse Miracle Works, there were afew factors that attributed to the overall design. Being a Hair Salon and Spa with the ultimategoal of providing rela
Syracuse - MAS - 361
A1. Analyzing DataObjective: To analyze each of the nine variables associated with keeping it connected withfamily members by using graphical and numerical descriptive techniques.A2. Our team constructed a questionnaire containing nine questions. These
Boise State - BIOL - 191
Pho synthe istos11/07/11Quick Revision :Where does photosynthesis occur Leaf X-section :PalisademesophyllChloroplast under electron microscopeChloroplast 11/07/11Warm-up QuestionsWarm-upThe following microphotograph is atransverse section
Boise State - BIOL - 191
Boise State - BIOL - 191
HydrogeLightenergy nOxygenChlorophyllChemicalATPenergywaterADP + PiHydrogenCarbonDioxideATPADP and PiGlucoseThis powerpoint was kindly donated towww.worldofteaching.comhttp:/www.worldofteaching.com is hometo over a thousand powerpoints
Boise State - BIOL - 191
Day makingPlants length food 1.Joe lives for basketball.He plays basketball every morning.He eats breakfast after playing, then goes to school.His mum says he needs the energy the food provides.At school he wrote down the food he had for breakfast.
Boise State - BIOL - 191
PlantsPlantsStructure LeavesLeavesstemrootsrootsLeavesLeavesFunction of leavesFunction Trap light energy for photosynthesisProducing sugar from photosynthesis Exchange of gases Exchangeoxygen and carbon dioxideoxygenStructureStructureW
Boise State - BIOL - 191
LightDependentReactionObjectivesUnderstandingtheprocessesofphotoionisation,photophosphorylation&amp;photolysis.Knowledgeofwheretheseprocessestakeplace.Knowledgeofthedifferencebetweencyclic&amp;noncyclicphosphorylation.OverviewofphotosynthesisOverviewofpho
Boise State - BIOL - 191
AbsorptionofLightbyChlorophyllHigherBiologyLightabsorptionbychlorophyllMeasuringlightabsorption Aspectrometer isusedtomeasuretheamountofabsorptionateachwavelength Anaction spectrumshowstherateofphotosynthesisatdifferentlightintensitiesChlorophyllA
Boise State - BIOL - 191
Slide 1StructureofPlantsSlide 21.2.3.A. Functions of RootsAnchor &amp; supportplant in the groundAbsorb water &amp;mineralsHold soil in placeRoot HairsFibrous RootsSlide 3B. Root TypesTap Root1. Fibrous Roots:2. Tap Roots larger centralbranch
Boise State - BIOL - 191
Boise State - BIOL - 191
PlantsEvolutionaryPastCyanobacteriaBluegreenalgaePhotosyntheticResponsibleforthemarkedincreaseinOxygen2000millionyearsagoCyanobacteriaEvidenceDNA shows the firstgreen plants evolvedfrom prokaryotesbetween 2500 and1000 million yearsago.Same
UNF - PHYSICS - 4360C
(b) T =T=1211 mg 1 m 2 g2mv = mvt = m =222 c2 2 .22 D 2(.145kg )2m 9.8 2 skg( 2 ) (.22 ) ( 2 ) (.0366 )m2= 87 J3 t 23 Fdx = cv dx = c v dt = c vt tanh dt t t t 31= cvt tanh 2 + tanh d 2t t 31= cvt tanh 2 + ln cosh
UNF - PHYSICS - 4360C
F2F1 2F2t1 + t1 ( t t1 ) +( t t1 )m2m2mF 2 F 2 F 2 5F 2At t = 2t1 : x =t1 + t1 + t1 =t1mm2m2mdv dv dxdvc3a== = v = v2dt dx dtdxm1cv 2 dv = dxm1vxmaxcv v 2 dv = 0 mdx1c2v 2 = xmaxmx=2.111xmax2.122mv 2=cGoing up:
UNF - PHYSICS - 4360C
v=vt v(vt2.142+v122), vt =gmg=kc2Going up: Fx = mg c2 v 2dvc22a = v = g kv , k =dxmvxvdvv g kv 2 = 0 dxv1ln ( g kv 2 ) = xv2kg + kv 2= e 2 kx2g + kvgg2v 2 = + v e 2 kx kk2Going down: Fx = mg + c2 vdvv = g + kv
UNF - PHYSICS - 4360C
22c2 v c1 c1 + 4mgc2t1=ln22mc1 + 4mgc2 2c2 v c1 + c1 + 4mgc2(t2c1 + 4mgc2m)12( 2c v + c += ln( 2c v + c 2v0) (c + 4mgc ) ( c +21c1 + 4mgc221c122)+ 4mgc )112c1 + 4mgc2c1222as t , 2c2 vt + c1 c1 + 4mgc2 = 0c1 c1
UNF - PHYSICS - 4360C
2.17dvdv= mv = f ( x )i g ( v )dtdxmvdv= f ( x ) dxg (v)mBy integration, get v = v ( x ) =If F ( x, t ) = f ( x )i g ( t ) :dxdtd 2xd dx = m = f ( x )i g ( t )2dtdt dt This cannot, in general, be solved by integration.If F ( v , t ) =
UNF - PHYSICS - 4360C
Integrating11A = tuu mand substituting e v = uAln 1 + e v t m(b) t = T @ v = 0A v = ln 1 + e v T mA vm1 e v e v = 1 + e TT=AmdvAvdvA= dx(c) v = v = e vvdxmem(a) v = v 1again, let u = e vdu = udvordv =duuv=1ln u1 d
UNF - PHYSICS - 4360C
3r 2= 1gr4Using Mathcad, solve the above non-linear d.e. lettingwhich reduces to: r +1 103 and R 0.01mm (small raindrop). Graphsshow thatv r t and r t 210
UNF - PHYSICS - 4360C
CHAPTER 3OSCILLATIONS3.1x = 0.002 sin 2 ( 512 s 1 ) t [ m ]mmxmax = ( 0.002 ) ( 2 ) ( 512 ) = 6.43 ss22 mmxmax = ( 0.002 ) ( 2 ) ( 512 ) 2 = 2.07 104 2 s s 3.2x = 0.1sin t [m]When t = 0, x = 0and = 5 s 13.3mx = 0.1 cos t smx = 0.5
UNF - PHYSICS - 4360C
3.63.711l=T =s 2.5 s9.82g6For springs tied in parallel:Fs ( x ) = k1 x k2 x = ( k1 + k2 ) x1(k + k ) 2= 1 2 mFor springs tied in series:The upward force m is keq x .Therefore, the downward force on spring k2 is keq x .The upward force
UNF - PHYSICS - 4360C
Fr = mg +mkmkdkcosxm = mg +tM +mM +mM +mFor the block to just begin to leave the bottom of the box atthe top of the vertical oscillations, Fr = 0 at xm = d :mkd0 = mg M +mg ( M + m)d=k3.9x = e t A cos ( d t )dx= e t A d sin ( d t ) e
UNF - PHYSICS - 4360C
3.1117 2 mx = 02317 = and 2 = 2222222 r = 2 = 4 mx + 3 mx +(a)Amax =(b)=2d = 2 2 =25 24 d =522A15 2e Td =3.12F2m d r = 2 =121ln 2 = f d ln 2Td1 d = ( 2 2 ) 2(a)122So, = ( + )2d1111 2 2 2 ln 2 2 2f = fd
UNF - PHYSICS - 4360C
21Td d 1 2=== 1 + 2 2 2 d 4 n TFor large n,Td1 1+ 2 28 nT13.14((a) r = 22122)22 = 2 2 = 0.70721 1 22 1 4(b) Q = d === 0.86622 2 2 2 ( 2 )222(c) tan = 2= 2 2 =2 43122() 2 = tan 1 = 146.331222
UNF - PHYSICS - 4360C
3.16(b) Q =2 2d=222 =1,LC=R2L2 1 R 2 LC 4 L L 1Q== 2 R R C 42 2L L R(c) Q = = C =RR 2 3.17Fext = F sin t = Im F ei t and x ( t ) is the imaginary part of the solution to:mx + cx + kx = F ei ti.e.i t x ( t ) = Im Ae
UNF - PHYSICS - 4360C
= tan 1 ( c 2m )m ( 2 2 ) c + kUsing sin 2 + cos 2 = 1 ,2F22= m ( 2 2 ) c + k + 2 ( c 2m )2AFA=cfw_m ( ) c + k + ( c 2m ) 2222122and x ( t ) = Ae t cos ( t ) + the transient term.3.19l A2 T 21 g8(a)for A =(b)412l1.041g,
UNF - PHYSICS - 4360C
1 Tf ( t ) = c + 2T f ( t ) cos ( n t ) dt cos n tn T2T1+ 2T f ( t ) sin ( n t ) dt sin n t ,n T21Tn = 1, 2, . . ., and c = 2T f ( t ) dtT 2Now, due to the equality of terms in n :2 Tf ( t ) = c + 2T f ( t ) cos ( n t ) dt cos n tn T2T2
UNF - PHYSICS - 4360C
3.22In steady state, x ( t ) = An e (i n t n )nAn =Fnm1( 2 n 2 2 )2 + 4 2 n 2 2 24F,n = 1, 3, 5, . . .andNow Fn =n9 2Q = 100 so 2 40, 00024F1A1 =1m22229 22( 9 ) + 4 40000 FA1 2m 24F1A3 =13m2 2 22 9 ( 9 2 9 2 ) +
UNF - PHYSICS - 4360C
3.24The equation of motion is F ( x ) = x x 3 = mx . For simplicity, let m=1. Then(a)(b)x = x x 3 . This is equivalent to the two first order equations x = y andy = x x3The equilibrium points are defined byx x 3 = x (1 x ) (1 + x ) = 0Thus, the p
UNF - PHYSICS - 4360C
3.25 + sin = 0d 2 cos = 0dt 22Integrating:2= cos 0)0T = 4 2 = 2 ( cos cosord1 2 ( cos cos ) 2Time for pendulum to swing from = 0 to = isNowsubstitute sin =sinsin2so =and after some algebra 2at = 2cos = 1 2 sinand use the iden
UNF - PHYSICS - 4360C
CHAPTER 4GENERAL MOTION OF A PARTICLEIN THREE DIMENSIONS-Note to instructors there is a typo in equation 4.3.14. The range of the projectile is v 2 sin 2v 2 sin 2 2 NOT . 0R=x= 0gg-4.1V V VjkxyzF = c iyz + xz + kxyj(a) F = V = i()(b)
UNF - PHYSICS - 4360C
er1 F = 2r sin r kr n4.3e r0e r sin = 0 conservative0(a)i F =xxyjycx 2k= k ( 2cx x )zz3jycxzy2kzxy2cx x = 01c=2(b)i F =xzya l so4.4(a) x cx = i 2 2 +yyx cx 2 2 =0yycz z+=0y2 y2 1 1 + k cz + z j2y
UNF - PHYSICS - 4360C
v2 = v2 2( + + )m122v = v 2 ( + + ) m(b)v2 2( + + ) = 0m122v = ( + + ) m(c)mx = Fx = Vxmx = Vmy = = 2 yyVmz = = 3 z 2z4.5(a)jF = ix + ydr = idx + dyjon the path x = y :(1,1)(0,0 )11110000F dr = Fx dx + Fy
UNF - PHYSICS - 4360C
(1,1)(and, with x = 1on this path1,0 )(1,1)(0,0 )1F dr = dy = 10F dr = 0 + 1 = 1F is not conservative.4.6From Example 2.3.2, V ( z ) = mgre2( re + z )zV ( z ) = mgre 1 + re From Appendix D,(1 + x )11= 1 x + x2 + z z2V ( z ) = mg
UNF - PHYSICS - 4360C
h=re 1 2 2re v 2zre g22h=re re2v 21 z22greFrom Appendix D,( h, z1(1 + x ) 2 = 1 +h=x x2 +28re re v 2zv4++ z +2 2 2 g 4 grehre )v 2z v 2z 1 +2 g 2 gre 1v2 v2 From Example 2.3.2, h =1 2 g 2 gre v2 v2 1And with (1 x ) 1
UNF - PHYSICS - 4360C
sin = gbv2cos 2 = 1 sin 2 =v 4 g 2b 2v4gb 2 v 4 g 2b 2 gb 2 v 2+= 2+v22 gv 22v2gMeasured from the ground,gb 2 v 2hmax = b + 2 +2v2ghmax = gb The mud leaves the wheel at = sin 1 2 v4.8x = R cos so t =andx = v x t = ( v cos ) tR
UNF - PHYSICS - 4360C
Now sin = cos = cos + 4 2 4 2 2 4 2 2v 2 Rmax =cos 2 + 2g cos 4 2Again using Appendix B, cos 2 = cos 2 sin 2 = 2 cos 2 1v2 2v 2 1 1Rmax =cos + + =cos 2 + + 1222g cos 2 2 g cos Using cos + = sin ,22vRmax =(1 sin )g (1
UNF - PHYSICS - 4360C
2sin 2 =2gh= 1 22csc v01gh 1 + 2 = v0 1 + ghv0 2gh Finally csc 2 = 2 1 + 2 v0 (b)Solving for Rmax Rmax =hhh==2sin 1 2 sin 1 2 csc2 Substituting for csc 2 and solving v2gh Rmax = 0 1 + 2 g v0 4.10 We can again use the results
UNF - PHYSICS - 4360C
zzmaxv0h0h1x0xx1RWe have reversed the trajectory so that h0 (= 9.8 ft), and x0 , the height and range withinwhich Mickey can catch the ball represent the starting point of the trajectory. h1 (=3.28ft) is the height of the ball when Mickey strik