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Atomic Physics

Quantum Physics 2002. Atomic Physics. Recommended Reading: Harris Chapter 7. L. Current. r. m. v. . Angular Momentum and Magnetic Dipole Moment.

solomon-robertson
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Atomic Physics

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  1. Quantum Physics 2002 Atomic Physics Recommended Reading: Harris Chapter 7

  2. L Current r m v  Angular Momentum and Magnetic Dipole Moment. Imagine an electron orbiting a proton counter-clockwise in an orbit of radius r. Being negatively charged, it represents a conventional clockwise current of magnitude I = e/T, where e is the fundamental charge and T is the period of revolution. If we have a current I flowing around a loop of area A then the magnetic dipole moment  is defined as  = IA. Thus (1) where me is the mass of the electron, v is its velocity and we have used the definition of angular momentum L = mevr. Both the angular momentum and magnetic moment are vectors so (2) Result: any orbiting charged particle will have a magnetic dipole moment

  3. We can exploit this relationship to reveal angular momentum quantisation In the absence of an external force to align them the magnetic moments  of atoms point in random directions in space. However if we place the atoms in an external magnetic field B then they gain potential energy (3) And they will experience a force given by (4) If the magnetic field is constant everywhere in space then the derivatives w.r.t x, y and z are all zero and there is no force acting on the atoms  they have a constant potential energy.  if we apply a nonuniform magnetic field then we can exert a force on an atom if it has a magnetic dipole moment. In 1922 O. Stern and W. Gerlach carried out an experiment to see if they could detect this force by applying a nonuniform magnetic field to a beam of atoms

  4. Stern - Gerlach Experiment.

  5. Stern - Gerlach Experiment. The pole pieces of the magnetic are situated such that there is a channel in which there is a non-uniform field in the z-direction. The magnetic field in the x and y (horizontal) is uniform so there is no force in the x and y direction. • Atoms produced from a metal (Ag) heated in an oven are sent down the channel in the magnet. Because of the non-uniform field they will experience a force that will deflect them in the vertical direction. • If the angular momentum is quantised then it can only point in certain directions relative to the z-component of the magnetic field (space quantisation) and the amount of deflection will depend on the z-component of the angular momentum • However. if the angular momentum behaves classically then it can point in any direction in space, it will still experience a force but this force will be in a random direction and therefore the magnitude of the deflection will also be random. • The pattern observed on the screen will then distinguish between ‘classical’ and ‘quantum ‘ behaviour of the atoms in the magnetic field.

  6. Since only the z-component of the force remains in equation (4) (5) We can relate this force to the angular momentum using equation (2) (6) l = 0 If the angular momentum L is zero then it will have no z-component and there will be no force acting on the atom  it should be undeflected by the magnetic field classical, l  0 If L  0, then the classical expectation is that Lz could be any of the continuous range of values from -L to +L leading to a continuous band of deflected atoms on the screen

  7. If the angular momentum L, is quantised then its z-component will be quantised and therefore the force exerted on the atom will be quantised. Using the quantisation condition Lz =mL Suppose we passed hydrogen atoms in the l = 1 state through the magnet, in this case the quantum number mL can take on three values -1, 0 and +1 and we would expect to see three lines on the screen corresponding to to an upward force (mL = -1), a downward force (mL = +1) and no force (mL = 0). In general we would expect to see 2l+1 lines on the screen l = 1 mL = -1 mL = 0 mL = +1 Ag: l = 0 Stern and Gerlach used Ag atoms which has l = 0 and therefore they should have seen only one undeflected line on the screen. However when they did the experiment they observed two deflected lines and no undeflected line in the centre of the screen

  8. This result was baffling, since there should be 2l + 1 lines on the screen and this result implies that the angular momentum quantum number l should be 1/2, but we know that l can only be an integer. Clearly a problem here. What is going on? in the early 1920’sW. Pauli was the first to suggest that a fourth quantum number, ( in addition to n,l,mL) assigned to the electron was needed to solve this problem. In 1925 S. Goudsmit and G. Uhlenbeck, proposed that the electron must have an intrinsic (built in) angular momentum. This proposal runs into problems if it is considered classically, so we must regard this additional angular momentum an a purely quantum mechanical effect To explain the experimental data Goudsmit and Uhlenbeck proposed that the electron has an intrinsic spin quantum number s = 1/2 and therefore it also has an associated magnetic moment.

  9. We can treat the spin quantum number, in exactly the same was as we treated the orbital angular momentum. By analogy, there should be 2s+1 = 2(1/2)+1 = 2 components of the spin angular momentum vector s. The magnetic spin quantum number ms (I.e. the z-component of the spin vector) should have have only two values +1/2 and -1/2, so the electron’s spin components will be either ‘up’ or ‘down’ in an applied magnetic field. The magnitude of the spin vector is The wavefunction of the hydrogen atom now has four quantum numbers to describe each state: z and the following eigenvalue equations apply.

  10. For each state described by the quantum numbers (n, l, mL) discussed previously there are now two distinct states one with ms=+1/2(spin up) and a second with ms= -1/2(spin-down) . These states will be degenerate unless the atom is placed in a magnetic field. In a magnetic field these states will have different energies and this degeneracy is removed. In a magnetic field these states will have different energies and this degeneracy is removed. The degeneracy of a state with principle quantum number n is now where the factor of 2 comes from the spin quantum number.

  11. we now have two magnetic moments associated with the electron. The first is due to the orbital motion of the electron about the nucleus. The second is associated with the spin of the electron Where B is the fundamental unit of magnetic moment, called the Bohr Magneton We can now explain the result of the S-G experiment. If the atom was in a state with l = 0, there would be now splitting due to mL. However there is still space quantisation due to the intrinsic spin of the electron and the magnetic moment associated with this spin can interact with the inhomogeneous magnetic field to produce two lines, one due to the +1/2 state and the other due to the -1/2 state.

  12. Total Angular Momentum. Each electron had both orbital angular momentum (its L quantum number) and spin angular momentum (its S quantum number). These are vector quantities and they can combine vectorially to produce a Total Angular momentum, denoted by the quantum number J. If we consider one-electron atoms, for example Hydrogen or the hydrogen like ions, He+, Li2+ ..., or atoms that have a single electron outside of a filled core for example the alkali metals such as Li: 1s2 2s1 = [He] 2s1. Na: 1s2 2s2 2p6 3s1 = [Ne] 3s1. K: 1s2 2s2 2p6 3s23p6 4s1 = [Ar] 4s1 Then for an electron with orbital angular momentum L and spin angular momentum S, the total angular momentum J is given by Since L, Lz, S and Sz are quantised, the total angular momentum J and its z-component Jz are also quantised. If j and mj are the corresponding quantum numbers for the single electron then

  13. S S L L J J The value of mj ranges from -j to +j in integer steps. For example if j = 3 then mj = -3, -2, -1, 0, 1, 2, 3. The total angular momentum quantum number for a single electron has the values and because ms = ½ for a single electron  j = l + ½ or l - ½ . Example: for an s-state l = 0 and  j = +½ or -½ for an p-state l = 1 and  j = 3/2 or 1/2 for an d-state l = 2 and  j = 5/2 or 3/2 etc The notation commonly used to describe these states is Where n is the principal quantum number, j is the total angular momentum and L is an uppercase letter representing the orbital angular momentum Example: an s-state, l = 0, j = +½ or -½  nS1/2 or nS-1/2 a p-state, l = 1, j = 3/2 or 1/2  nP3/2 or nP1/2 a d-state, l = 2, j = 5/2 or 3/2  nD5/2 or nD3/2

  14. Russell Saunders Coupling (LS coupling) The situation becomes more complicated atoms with more than one electron outside an inert core. In this case we obtain the total angular momentum J as follows: 1) we couple the orbital angular momenta li of all the electrons to form a resultant orbital angular momentum L. For example if we had a p- electron (l1 = 1) and a d-electron (l2 = 2) then the total orbital angular momentum would be L = l1 + l2 = 3. Note that this is the maximum value L can have. 2) we couple the spin angular momenta si of all the electrons to form a resultant spin angular momentum S. For example if we had a p- electron (s1 = 1/2) and a d-electron (s2 = 1/2) then the total spin angular momentum would be S = s1 + s2 = 1. Note that this is the maximum value L can have. 3) The total L and S vectors then couple to form the total J.

  15. L, S, J, Lz, Sz and Jz are quantised in the same manner as the single electron case with quantum numbers L, S, J, mL, ms and mj. We define the multiplicity of a state as 2S+1. For example if a multielectron state has a total spin S = 1 then it has a multiplicity of 3 (a triplet state). For a total spin S = 3/2 the multiplicity is 4 (a quartet state). Example:Find the possible states under LS coupling for a two electron atom whose electrons are in p and d states. electron 1: p-statel1 = 1 and s1 = 1/2. electron 2: d-state l2 = 2 and s2 = 1/2 Since S = 0, 1multiplicities of 1 (singlet states) and 3 (triplet states) J can have values from L - S  to  L + S in integral steps  J = 0, 1, 2, 3, 4

  16. Notation The state of a multi-electron atom is denoted by Example: n = 3,L =1, S = 0, J = 1  31P1 state n = 5,L = 3, S = 1, J = 4  53F4 state etc... LS coupling of the p and d electrons leads to all of the following possible states of the atom: Singlets:S = 0 then J = L and L = 1, 2, 3 so the singlet states are 1P1 , 1D2 , 1F3 . Triplets:S = 1, L =1  J = 0, 1, 2 3P0 , 3P1 , 3P2 . S = 1, L =2  J = 1, 2, 33D1 , 3D2 , 3D3. S = 1, L =3  J = 2, 3, 43F2, 3F3 , 3F4 . Note that there are 12 possible angular momentum states of a atom in which a p and d electron couple together. In general these states will not be degenerate. So how do we know which state has the lowest energy???  Pauli Exclusion Principle,Afbau principle and Hund’s Rules

  17. Pauli Exclusion Principle Early in the development of the quantum mechanics Wolfgang Pauli (1900 - 1958) recognized that no two electrons, in the same atom, could have the same set of four quantum numbers. Two electrons may have the same n, l and ml quantum numbers, placing them in the same atomic orbital, but then they must have different spins. Thus each orbital can accommodate two electrons assuming that their spins are "paired”, I.e one +1/2 and the other -1/2. This rule is called the Pauli Exclusion Principle. Starting with hydrogen, the lowest energy state has one electron in the 1s state, 1s1 (n,l, mL, ms ) = (1,0,0,+1/2) or (1,0,0,-1/2). Helium 1s2 1st electron = (1,0,0,+1/2), 2nd electron = (1,0,0,-1/2). Litium 1s2 2s1 = He + (2,0,0,+1/2) or (2,0,0,-1/2). Beryilium 1s2 2s2 = He + (2,0,0,+1/2) or (2,0,0,-1/2) Boron 1s2 2s2 2p1= Be + (2, 1, 0,  1/2) or (2, 1, 0, 1/2) or (2, 1, -1, 1/2) Boron has a single electron in the 2p level, but where does the next electron go when we move to carbon? Let's consider the several possibilities here:

  18. Possible electronic configurations of carbon (a) (c) (b) We can establish a simple rule that electrons will tend to singly occupyorbitals until forced to begin pairing up. This favours (b) and (c) over (a) Unpaired electrons in sublevels with degenerate orbitals (p, d, f, etc) will tend to have their spins parallel. From an energy standpoint, (c) < (b) < (a).Thus (c) is the correct configuration.

  19. Afbau (building up) Principle Fluorine: 1s2 2s2 2p5 Nitrogen: 1s2 2s2 2p3 Oxygen: 1s2 2s2 2p4 Neon: 1s2 2s2 2p6 See http://www.chem.uidaho.edu/~honors/aufbau.html

  20. Hund’s Rules 1) The term with maximum multiplicity lies lowest in energy  ( 3P, 3D and 3F ) are lower in energy than the (1P, 1D and 1F ). 2)For a given multiplicity, the term with the largest value of L lies lowest in in energy ( 3F < 3D < 3P ) and ( 1F < 1D < 1P). 3)For atoms with less than half-filled shells, the level with the lowest value of J lies lowest in energy. Otherwise the level with the largest J lies lowest in energy. Since we have only one electron in the p and one in the d shell they are less than half full so (3F2 < 3F3 < 3F4) Conclusion: if we have an atomic state with two electrons, one in a p-state and one in a d-state. Then the atomic state with lowest energy is a 3F2 state Hund’s rules only apply when L-S coupling holds and this breaks down for electrons in heavy elements (then we have to use a different coupling, j-j coupling See: http://hyperphysics.phy-astr.gsu.edu/hbase/atomic/hund.html

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