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W.N. Catford/P.H. Regan 1AMQ 102 Optical Spectra of Atoms. Example: Sodium, Z=11, thus the electron configuration in the atomic ground state is 1s 2 2s 2 2p 6 3s 1 The first excited state has 1s 2 2s 2 2p 6 3p 1 The lowest energy levels of the atom are due to excitations of electrons between outer levels (where the smallest energy gaps and the first vacancies exist). For electrons in the n=3 electron orbitals, the nuclear charge (Z=+11e) is screened by the inner 10 electrons from the n=1 and n=2 shell (Q=-10e). Thus the energy is similar to that of the n=3 Bohr orbit for hydrogen. Screening effects give an energy shift of the levels, which depends on the l of the orbital.

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W.N. Catford/P.H. Regan 1AMQ 104 Optical photons are emitted when valence (outer) electrons make transitions between energy levels. e.g. 3p  3s,  E=2.10 eV (3.37 x J), thus  hc/  E) = 589 nm (i.e. yellow street lamps).

W.N. Catford/P.H. Regan 1AMQ 105 Not all of the energetically possible transitions are allowed. The selection rules are (as in x-rays) The transition 3p  3s shows fine structure giving the sodium D lines, i.e. D 2,  nm and D 1, =589.6 nm. If an external B field is applied, the Zeeman effect gives a further splitting of the levels. The Zeeman splitting is generally smaller (for typical laboratory size eternal fields) than the fine structure (which increases as approx. Z 4 ). A further selection rule is observed, namely, Summary: Optical spectra arise from transitions of valence e - s. Optical photon energies are approx. several eV. Inner e - s are left undisturbed by optical transitions. When more energy is injected, and inner electrons are removed, x-rays (with energies ~keV) are emitted when these vacancies are filled

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W.N. Catford/P.H. Regan 1AMQ 107 Addition of Angular Momenta-(1) The properties of Alkali atoms such as Na are determined by The single electron outside the closed system. The whole atom behaves as if it had the quantum numbers of that single electron. L1L1 L2L2 ● Take an atom with two electrons L = L 1 + L 2 Where L =  l(l + 1). h Thus we get the total ang. Momentum by adding the vectors but they have to obey the rules of quantum mechanics. ● The maximum value of the total orbital ang. Momentum is L max. = l 1 + l 2 ● The minimum value of total ang. Momentum is L min = l 1 - l 2 ● The allowed values of L range from the max. to min. values in integer steps. ● The z-component of L is the sum of the z components of the individual vectors L Z = L Z1 + L Z2

W.N. Catford/P.H. Regan 1AMQ 108 Addition of Angular Momenta-(2) ●The case of Carbon (Z = 6) Here we have the 1s and 2s filled and two electrons in 2p They can combine to give L = 0,1 or 2 S = 0 or 1 ● How do we know which combination gives the ground state? The rules for finding the answer are called Hund’s rules. 1)First find the maximum value of M S consistent with the Pauli Principle. Then S = M S,MAX 2)For this M S find the maximum value of M L consistent with the Pauli Principle. Thus for Carbon the max. value of MS is +1.Thus S = 1. The max value of m l for electron1 is +1 The second electron cannot have the same value since then they would have the same quantum numbers. So the second electron must have m l = 0. ● Thus M L,MAX = +1 and L = 1 and the ground state of carbon has S = 1 and L = 1

W.N. Catford/P.H. Regan 1AMQ 109 The Helium Atom: 2 Active Electrons While Sodium has, for low excitations, only one active electron for optical transitions, helium and other elements will have more. L2L2 L1L1 Selection Rules:-  L = 0,+/- 1,  S = 0  l = +/- 1

W.N. Catford/P.H. Regan 1AMQ 110 Ground State of Helium: two electrons, both with l=0 (same (1s) orbital), gives configuration of 1s 2 n l m l m s / /2 Using the coupling rules for l and s: i) L max = 0+0 = 0, Smax=1/2+1/2=1 ii) L min = |0-0| = 0, Smin=|1/2-1/2|=0 iii) L=0 (and thus M L =0), S=1 or 0 iv) M L =0, M S =-1,0,+1 are the only possibilities NB. Use capital `L’ and `S’ for > 1 electron. The Pauli Principle means that S=1 is not allowed, since would need m s =+1/2 for both e - s for S=1. i.e. for 1s 2, can only have (L=0, S=0), i.e. ground state of helium has L=0, S=0, hence, J=0

W.N. Catford/P.H. Regan 1AMQ 111 Excited State of the Helium Atom. The first excited state (above the ground state) has configuration 1s 1 2s 1. n l m l m s /2, -1/ /2,-1/2 As for the ground state, L=0, S=0 or 1 from coupling of L 1 and L 2 vectors etc. Allowed states are (L=0, S=0) and (L=0, S=1). For S=0, only one state (singlet state) For S=1, three states (i.e. M S = -1,0,+1) (triplet state) Hund’s rules lead to E triplet <E singlet. (This is related to the PEP ; aligned electrons tend to “repel” each other).

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W.N. Catford/P.H. Regan 1AMQ 113 Transitions in Helium Experimentally, the selection rules for He are  L=0,+-1 and  S=0 (i.e. no “spin-flips”). Singlet (S=0) and triplet (S=1) states CANNOT BE CONNECTED by transitions. 1s 2 (singlet) 1s 1 2s 1, singlet but not triplet The triplet 1s 1 2s 1 state is rather long lived or metastable (= “isomeric”). This is due to the fact that no decay can occur to the ground state via single photon emission (  S=1 is forbidden). It subsequently decays by transferring kinetic energy in collisions. The singlet 1s 1 2s 1 state is also isomeric since a transition between 2s and 1s states would violate the  l =  1 selection rule for individual l values.

W.N. Catford/P.H. Regan 1AMQ 114 1AMQ, Part V Molecules Ionic Bonding-NaCl. The Molecular Hydrogen Ion Hydrogen Molecule and Covalent Bonding K.Krane, Modern Physics, Chapter 9 Eisberg and Resnick, Quantum Physics, Chapter 12.

W.N. Catford/P.H. Regan 1AMQ 115 Molecules-Some general Points ● Molecules are combinations of atoms. They can be simple or very complex in structure from NaCl to DNA ● Various experiments suggest that the spacing of atoms in molecules is about 0.1 nm. In addition the binding energies are of the order of eV. ● All of this sounds familiar. These are the spacings and energies we associate with electronic orbits in atoms. It suggests that the forces binding molecules together are due to the electrons. Somehow the electrons(-vely charged) lie between the +vely charged nuclei and bind them together. ● Why do the negative electrons not just repel each other? ● One way we can think of this is in terms of slowly bringing two atoms together. Then the atomic states become the molecular states and these states are filled in order of increasing energy. ● As for atoms it is the probability densities of the occupied states that determine the nature of the molecular bonds and hence the structure and properties of molecules.

W.N. Catford/P.H. Regan 1AMQ 116 Molecular Bonding For a molecule, we need at least two centres of positive charge i.e, two atomic nuclei. The Time Independent Schrödinger Equation (TISE) must be solved for a two- centred potential, with the electrons occupying the allowed levels. The H 2 + Molecular Ion: this comprises of a single electron and two protons. It turns out that a particular value for the separation of the two nuclei is favoured (i.e. corresponds to a minimum energy) compared to the unbound (free= infinite separation) system. The electron is in effect “shared” between the two centres of positive charge. Most cases are too complicated to solve so we start with a simple case.The H 2 + mol. Ion.The H molecule with an electron Removed.

W.N. Catford/P.H. Regan 1AMQ 117 Molecular Bonding ● We might imagine it is a H atom with an extra proton attached. WRONG:-Atom is neutral so why should a proton be attached to it. ● It seems that the electron must spend a significant amount of time sandwiched between the two protons. It must have a large prob. density in this region.

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W.N. Catford/P.H. Regan 1AMQ 119 (a) Consider when the protons are very far apart, then if   and  2 are the two separate hydrogen w.functions, and the electron is equally likely to occupy either   or    then the total w.function is Recalling that for hydrogen, in the 1s state,

W.N. Catford/P.H. Regan 1AMQ 120 (b) For smaller separations the   and   wavefunctions overlap and then,     and      have different probability distributions.

W.N. Catford/P.H. Regan 1AMQ 121 Since the wavefunctions have different probability density distributions, the energies E + total and E - total are also different. For (     ) it is more likely to find the electron between the protons, which reduces the repulsion and hence a lower energy solution results, i.e. to take the electron from the state described by (     ) and take it to infinite distance costs energy. It also follows that E + total < E - total The total energy is given by E + total = E + + U, where E + is the energy of the electron in (     ) and the energy U represents the repulsion between the two protons. Note, that there is a bound state which has a minimum energy at -16.3eV (corresponding to a separation of 0.11nm). This is 2.7eV less total energy than if we had a single H atom plus one extra separate proton at infinity (-13.6 eV)

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W.N. Catford/P.H. Regan 1AMQ 123 The H 2 Molecule. This consists of two electrons and two protons. At infinite separation of the four particles, the total potential energy = 0. (a) At infinite separation of the two protons, the lowest energy corresponds to having two hydrogen atoms, Energy=2 × (-13.6eV)=27.2 eV (b) At small separations, suppose we add one more electron to the (     ) energy level in H 2 + (labelled U+E + in figure on previous page). This is allowed by the Pauli principle (m s = ±1/2). This approximately doubles the overall energy (  -16.3eV × 2  eV at approx 0.11nm). More accurately, the electron interaction is more complex (  31.7eV at 0.07nm separation). We see that the extra electron binds the 2 protons closer together. This COVALENT BONDING is produced by the symmetric wavefunction, (     ). Because eV < eV the H 2 molecule exists!

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