#### L25-E1112 solution

Delaware, PHYSICS 424
Excerpt: ... Class exercise #11 Quarks carry spin . Two quarks (or actually a quark and an antiquark) bind together to make a meson (such as pion or kaon). Three quarks bind together to make a barion (such as proton or neutron). Assume all quarks are in the ground state so the orbital angular momentum is zero). (1) What spins are possible for mesons? (2) What spins are possible for baryons? Solution: Now add the third spin: Lecture 25 Page 1 Class exercise #12 The electron in a hydrogen atom occupies the combined spin an position state: Note that in both cases , what values might (a) If you measure the orbital angular momentum squared your get and what is the probability of each? You get with probability P=1 (100%). . (b) Same for z component of the orbital angular momentum Possible values of m : (c) Same for the spin angular momentum squared . (d) Same for z component of the spin angular momentum . Lecture 25 Page 2 E12 (e) If you measure the total angular momentum squared your get and what is the proba ...

#### homework12

BU, PHYSICS 451
Excerpt: ... PY 451: Quantum Physics I Problem Set 12 Due date: Thursday, April 30 2009 by 5pm in the graders' box 1. (Griffiths 4.35) Quarks carry spin 1/2. Three quarks bind together to make a baryon (such as the proton or neutron); two quarks (actually a quark and anti-quark) bind to make a meson (such as the pion or kaon). Assume that the quarks are in their ground states so that the orbital angular momentum is zero. (a) What spin quantum numbers are possible for baryons? (b) What spin quantum numbers are possible for mesons? 2. By following the steps outlined in lecture, determine by detailed calculation the spin matrices Sx , Sy , and Sz for a spin-1 particle. For the ambitious students, note that this same calculation can be extended to a particle with arbitrary spin quantum number s. You are invited to try this latter calculation; the answer is given in problem 4.53 of Griffiths. 3. (part of Griffiths 4.36) Two particles with spin quantum numbers 1 and 2 are at rest in a configuration such that the total spin qu ...

#### 362molecularterms

Illinois State, CHE 362
Excerpt: ... Chemistry 362 Fall 2007 Dr. Jean M. Standard November 28, 2007 Diatomic Molecular Term Symbols and Electronic Spectroscopy Molecular Term Symbols For diatomic molecules, molecular term symbols can be constructed much like those for atoms. These term symbols label the electronic states of the molecule. For atoms, the term symbols 2S+1 2S+1 LJ . For molecules, the term symbols have the form have the form , where is the total molecular orbital angular momentum . We will not be considering the total angular momentum J in determining the molecular term symbol. In many cases, the parity of the state (g or u) is listed as a subscript in the molecular term symbol, 2S+1 g or 2S+1 g , where g or u corresponds to either gerade or ungerade parity, respectively. For atoms, the letters S, P, D, and F are used in the term symbol to represent L=0, 1, 2, and 3, respectively. For diatomic molecules, the letters , , , and are used in the term symbol to represent =0, 1, 2, and 3, respectively. To determine the total molecular or ...

#### Lecture 25

Delaware, PHYSICS 424
Excerpt: ... Lecture 25 Addition of angular momenta Ground state of hydrogen: it has one proton with spin and one electron with spin ( orbital angular momentum is zero). What is the total angular momentum of the hydrogen atom? Total spin Electron's spin, acts only on electron's spin states Proton's spin, acts only on proton's spin states The z components just add together and quantum number m for the composite system is simply Combination of two spin particles can carry a total spin of s =1 or s = 0, depending on whether they occupy the triplet or singlet configuration. Three states with spin s = 1, m = 1, 0, -1 This is called a triplet configuration. and one state with spin s = 0, m = 0: This is called a singlet configuration. How do we prove that it is true? Lecture 25 Page 1 L25. P2 To prove the above, we need to show that for triplet states since and for singlet states since This can be shown by direct calculation (see page 186 of the textbook for the proof). In general, it you combine any angular momentum va ...

#### lecture21

Dartmouth, PHYS 13
Excerpt: ... on). EXAMPLE: point particle 1.2 Multi-particle system Recall for a multi-particle system psys = p1 + p2 + . + pN We do the same thing to get the angular momentum of a multi-particle system (add up angular momentum of all the pieces) Lsys = L1 + L2 + . + LN write out. r cross p for each However, it is often useful to divide the angular momentum up into two pieces: angular momentum of the center of mass ( orbital angular momentum ) and angular momentum of particles in system relative to CM. 13 DRAW L = (rCM + r1 ) p1 + . = (rCM psys ) + (r1 p1 + .) EXAMPLE: Earth orbits the Sun. The CM of Earth is rotating around the Sun. But, the Earth also spins on its axis. Individual points on Earth rotate with respect to the CM of earth. Notice: the orbital angular momentum depends on our choice of origin. It is relative to some point (book uses subscript A). 2 Conservation of Angular Momentum Angular Momentum Principle: dL = ext dt if some component of torque aroun ...

#### section19

East Los Angeles College, PHYS 3002
Excerpt: ... 19 19.1 Parity, Charge Conjugation and CP Intrinsic Parity In the same way that nuclear states have parity, so hadrons, which are bound states of quarks (and antiquarks) have parity. This is called intrinsic parity, and under a parity inversion the wavefunction for a hadron acquires a factor P {P } (r) = = {P } (-r) = {P } {P } (r). can take the values 1 noting that applying the parity operator twice must bring us back to the original state. The lighter baryons (for which there is zero orbital angular momentum ) have positive intrinsic parity. On the other hand, antiquarks have the opposite parity from quarks. This means that the light antibaryons have negative parity. It also means that the light mesons, such as pions and kaons, which are bound states of a quark and an antiquark have negative intrinsic parity. The lightest spin-one mesons, such as the -meson, also have zero orbital angular momentum and thus they too have negative intrinsic parity - they have spin-one because of the alignment of the spin ...

#### Lecture17

CUNY Baruch, PHYSICS 330
Excerpt: ... filled The Periodic Table Transition Metals: Three rows of elements in which the 3d, 4d, and 5d are being filled Properties primarily determined by the s electrons, rather than by the d subshell being filled Have d-shell electrons with unpaired spins As the d subshell is filled, the magnetic moments, and the tendency for neighboring atoms to align spins are reduced Lanthanides (rare earths): Have the outside 6s2 subshell completed As occurs in the 3d subshell, the electrons in the 4f subshell have unpaired electrons that align themselves The large orbital angular momentum contributes to the large ferromagnetic effects Actinides: Inner subshells are being filled while the 7s2 subshell is complete Difficult to obtain chemical data because they are all radioactive Have longer half-lives The Periodic Table Total Angular Momentum Orbital angular momentum Spin angular momentum Total angular momentum L, Lz, S, Sz, J, and Jz are quantized. Total Angular Momentum If j and mj are quantum numbers for the si ...

#### spin1

University of Florida , PHY 3101
Excerpt: ... other words, the energy degeneracy with respect to magnetic quantum number ml of the H atom is broken with the application of an external magnetic field. It is possible to see the effect of the quantization of angular momentum, because the energy levels will be split into 2l + 1 different levels, and the line spectra, therefore, will be split also. This is known as the normal Zeeman effect. It gives fine structure to spectral lines. The energy splitting between two levels is given by: E eh E = B Bml = B with m l = 1 2m E For a 1 Tesla field, this splitting is 5.79 10 5 eV, which is very small. n=2 ml = 1 ml = 0 ml = 1 l =1 n=1 D. Acosta Page 3 ml = 0 1/3/2001 PHY3101 Modern Physics Lecture Notes Spin 1 Spin and the Stern-Gerlach Experiment The normal Zeeman effect is not always seen. It turns out that orbital angular momentum is not the complete story, and that there exists intrinsic angular momentum. Lets consider the Stern-Gerlach experiment of 1922. In that experimen ...

#### ClassEx1

Excerpt: ... Class exercise #-1 1) The electron In a hydrogen atom occupies the combined spin and position state Note that (a) If you measure the orbital angular momentum squared L2, What values might you get and what is the probability of each (b) Same for the Z component of the orbital angular momentum , LZ (c) Same for the spin angular momentum S2 (d) Same for the z component of the spin angular momentum Sz Let J=L+S be the total angular momentum (e) If you measure the total angular momentum squared J2, what values might you get and what is the probability of each GlebschGordanCoefficientsJ1=1,J2=1/2 m=3/2 m1,m2 m=1/2 m1,m2 1,1/2 0,1/2 J= 3/2 1/3 1/2 2/3 1,1/2 J= 3/2 1 2/3 1/3 2 ) A p-electron in an atom is one that has orbital angular momentum quantum number l = 1. What are the possible angular momentum states of this electron in ...

#### notes_Lecture19_slides

Washington, CHEM 152
Excerpt: ... tum number, and ranges from 1 to infinity. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). ml, the "z component" of orbital angular momentum . Ranges in value from -l to 0 to l. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). m, the "z component" of orbital angular momentum . Ranges in value from -l to 0 to l. We can then characterize the wavefunctions based on the quantum numbers (n, l, m). Orbital Shapes Let's take a look at the lowest energy orbital, the "1s" orbital (n = 1, l = 0, m = 0) 1 1s = p Z e a o 3 2 - Z r a0 3 1 = p -s Z 2 e a o a0 is referred to as the Bohr radius, and = 0.529 1 2 Z - 18 E n = - 2.178x10 J 2 - 2.178x10- 18 J = n 1 Orbital Shapes (cont.) Note that the "1s" wavefunction ...

#### notes_Lecture18_powerpoint

Washington, CHEM 152
Excerpt: ... ntum number, and ranges from 1 to infinity. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). ml, the "z component" of orbital angular momentum . Ranges in value from -l to 0 to l. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). m, the "z component" of orbital angular momentum . Ranges in value from -l to 0 to l. We can then characterize the wavefunctions based on the quantum numbers (n, l, m). Orbital Shapes Let's take a look at the lowest energy orbital, the "1s" orbital (n = 1, l = 0, m = 0) 1 Z 1s = e ao 3 2 -Z r a0 3 1 Z 2 - = e ao a0 is referred to as the Bohr radius, and = 0.529 1 Z2 -18 E n = -2.178x10 J 2 = -2.178x10-18 J n 1 Orbital Shapes (cont.) Note that the "1s" wavefunction has no a ...

#### p1250410

Caltech, PH 0304
Excerpt: ... in the expansion of the eikx expansion). Consider transitions between states of specified initial and final orbital angular momentum , i and f , and initial and final projections of the orbital angular momentum on the z-axis, mi and mf . In the electric dipole approximation, what transitions are permitted (e.g., what are the permitted values of f and mf if i and mi are given)? 19 ...

#### classes_winter09_113AID28_P_26_113A_W09

UCLA, CHEM 113A
Excerpt: ... Preview #26 (Chem113A W09, due Fri. 3/06/09 at 12:05pm in class) Attendance record: I am [ ] present at or [ Assigned reading: None ] absent from this class meeting (see date and time above). Name:_ Forming study groups is permissible, but you must construct your solutions independently. By writing down my name, I confirm that I strictly obey the academic ethic code when doing this preview and my statement on attendance (above) is correct. Please write down the names of everyone who you worked with on this preview in the space above. [1] Orbital angular momentum QM originates from CM. So lets see what CM says about orbital angular momentum . In classical mechanics (CM), angular momentum L (a vector) is defined as L=rp where vector L = Lx i + Ly j + Lz k, vector r = x i + y j + z k, and vector p = px i + py j + pz k. [1](a) Please show by algebra that the x, y, z components of L are: Lx= ypz - zpy; Ly = zpx - xpz ; Lz = xpy ypx. [1](b) Commutator relations. Given the form ...

#### 10-11-99

University of Iowa, C 004132
Excerpt: ... Coupling quantum mechanics yield these energy levels 1 E l,s,j = hcA{ j(j + 1) - l(l + 1) - s(s + 1)} 2 Aisthespin - orbitcouplingconstant Term Symbols Term symbols - method for labeling electronic states in a multi-electron system Contains three pieces of information Central capital letter (S, P, D, .) indicates total orbital angular momentum , quantum # (Q.N.) = L. Left superscript gives the multiplicity of the term. Right superscript is the value of the total angular momentum, quantum # = J. 1 Rules for Term Symbols- part.1 Total orbital angular momentum Q.N., L, gives the magniture of the orbital angular momentum from {L(L+1)}1/2h. It has 2L+1 orientations distinguished by the mL Q.N. where mL= -L, -L+1, . , +L. L is obtained by coupling individual orbital angular momenta using a Clebsch-Gordon Series from total orbital angular momentum eigenfunctions L = l1+l2, l1+l2-1, ., |l1-l2|. L = 0, 1, 2, 3, 4, . (written S, P, D, F, G, .) Total Orbital Angular Momentum of a p and d e ...

#### 362angmomvectors

Illinois State, CHE 362
Excerpt: ... Chemistry 362 Fall 2007 Dr. Jean M. Standard November 2, 2007 Orbital Angular Momentum Vectors Two examples of quantum mechanical orbital angular momentum vectors are shown below for l = 1 and l = 2. l =1 l=2 ...

#### ProblemClassI_2007

East Los Angeles College, MAB 503
Excerpt: ... Nuclear Physics II PROBLEM CLASS 1 1. Show that, in addition to the static electric multipole moments discussed so far, the next highest order multipole moment expected in nuclei would be described by an expression of the form 1 2 2 2 4 e (r , ' ) 5z 7z 6r + 3r dV 8 (you may assume that P3 ( x ) = ( ( ) ) 1 1 (5 x 3 3 x ) and P4 ( x ) = (35 x 4 30 x 2 + 3) ) 2 8 2. Show that the size of the spin-orbit splitting in nuclei is proportional to (2l+1) where l is the orbital angular momentum quantum number 3. Show, using the method of counting m-states, that it is not possible to generate an odd-spin state with two identical particles in the same shell-model level. Do this by following this procedure separately for two different shell model levels, and analysing the result. 4. Electric quadrupole moments: Comparison of equation I.10 with the previous equation in the lecture notes shows that the effective quadrupole moment measured for a single proton in a level with spin j is differen ...

#### ps13

Cornell, PHYS 316
Excerpt: ... Homework for Physics 316, Modern Physics I (Hostaetter/Drasco/Thibault) Due Date: Friday, 05/06/03 - 9:55 in 132 Rockefeller Hall Exercise 1: Compute the radial part Rnl (r) of the Hydrogen wave function for n < 3 for all possible l. Show by direct integration that the hydrogen state |n = 1, l = 0, m = 0 is orthogonal to the state |n = 2, l = 0, m = 0 . Exercise 2: For Hydrogen, an eigenstate of the total angular momentum J 2 and its z-component Jz is expanded as a linear combination of orbital angular momentum eigenstates: 1 1 1 1 |nljsmj = A|n, l, ml = mj , s, ms = + B|n, l, ml = mj + , s, ms = . 2 2 2 2 Using J = L + S, argue that the linear combination cannot contain any other states. Exercise 3: (a) Use the Schroedinger Equation for the radial part of the wave function for hydrogen, and verify that a solution of the form Rnl = (c0 c1 r)rl+1 ebr (2) (1) corresponds to the energy eigenvalue 1/(l + 2)2 in dimensionless units. (b) Putting l = 1, sketch (i) the form of the w ...

#### handout_Lecture18

Washington, CHEM 152
Excerpt: ... E n = - 2 2 2 = -2.178x10-18 J 2 n 8e0 h n n is the principle quantum number, and ranges from 1 to infinity. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). ml, the z component of orbital angular momentum . Ranges in value from -l to 0 to l. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). m, the z component of orbital angular momentum . Ranges in value from -l to 0 to l. We can then characterize the wavefunctions based on the quantum numbers (n, l, m). Orbital Shapes Lets take a look at the lowest energy orbital, the 1s orbital (n = 1, l = 0, m = 0) 1 y1s = p Z e ao 3 2 -Z r a0 1 = p Z 2 -s e ao 3 a0 is referred to as ...

#### 260 2008 Lecture 14

Sveriges lantbruksuniversitet, CHEM 260
Excerpt: ... or Quantisation of Angular Momentum: Stern-Gerlach Experiment Pass a beam of silver atoms through an inhomogeneous magnetic field to study the directions of each particles magnetic moment How they are the atoms directed by the field ? CLASSICAL z component of angular momentum can take any value expect to see broad band of atoms QUANTUM L is quantized sharp bands corresponding to discrete orientations of z. Proves quantization of angular momentum? BUT. 1. Electron configuration of Ag0 = [Kr].4d10.5s1 orbital angular momentum = 0 no splitting 2. L quantized by integral quantum numbers 0, 1, 2,so why only split into 2? (degeneracy = 2l + 1 need l = ) 5 Stern-Gerlach Experiment I was then the equivalent of an assistant professor. My salary was too low to afford good cigars, so I smoked bad cigars. These had a lot of sulfur in them, so my breath on the plate turned the silver into silver sulfide, which is jet black, so easily visible. It was like developing a photog ...

#### lecture_4

UNC Charlotte, CHEM 2141
Excerpt: ... Answer: 4d, since 0=s, 1=p, 2=d and the values of m range from -l to +l =-2,-1,0,1,2 S -orbitals S orbitas (n = 1, l = 0, ml =0) have forrm: -r a0 = e a 0 3 There is no angular dependence thus function is spherically symmetric. 2 Volume element between r and r+ r spheres is: 4r r=V 2 2 2 The probability of finding the electron between r and r+dr is V =4r r 4r2 2 = radial distribution fn. Shell Probability: What's probability of finding electron between a0 and (a0+ 1 pm) Use this for illustration 13.3 in which r = 1 pm, a0 = 52.9 pm, and is given by the above expression , thus the probability of an S-orbital electron between a0 and a0+1pm: Note that max value of is 0 when r = 0 a0 =P(r) the radial distribution function occurs at r = a0 for 1S orbital 3 = 1 however maximum value of 4r22 P & d orbitals P orbitals have l = 1 Existence of nodal plane for all orbitals with l>0 (p,d,f.) is due to nonzero orbital angular momentum being interpreted as the electron circulating about the nucleus, ...

#### lecture18

Rutgers, PHYSICS 361
Excerpt: ... Quantum Mechanics and Atomic Physics Lecture 18: Angular Momentum Raising and Lowering http:/www.physics.rutgers.edu/ugrad/361 Prof. Eva Halkiadakis Hydrogen Atom Summary Coulomb Potential: is a product of which is oscillatory which is a polynomial in cos R which is a product of a decaying exponential and a polynomial in r depends on 3 quantum numbers Principal quantum number n=1, 2, 3, Orbital angular momentum quantum number l = 0, 1, 2, (n-1) Today we will concentrate on this Magnetic quantum number ml = 0, 1, . l or ml =-l,-l+1, l-1, l Next time we will focus on this Review of Orbital Angular Momentum Recall that the orbital angular momentum in spherical coordinates is: Is the H atom an eigenstate of Lx, Ly, Lz? You should already be able to tell me the answer to this question! Lets start with Lz: Yes, it is an eigenstate of Lz with eigenvalue mlhbar! Now lets try Lx and Ly So, no, the H ...