§ Spectroscopic Terms : L-S and j- Couplings 1. Terminology We define some terminology associated with atomic energy levels State : The ‘state’ of an atom is the condition of motion of all the electrons. It specified by listing four quantum numbers for each electron. If several states have sam, energy, they are said to be degenerate. The state with lowest energy is the ground state. Energy Level : A collection of states having the same energy in the absence o external magnetic or electric field constitutes an “energy level’, An energy level i characterised by a quantum number J, that is, by a particular value of the total angula momentum, The level with the lowest energy is the ground level. Sublevel : An external field splits an energy level into several ‘sublevels', cac! characterised by one or more magnetic quantum numbers. Term : A collection of levels characterised by an orbital angular momentum an multiplicity (that is, spin) comprises a ‘spectroscopic term’. For example a ‘D tem means the weighted average energy of the ’Ds, *D; and *D, levels. Configuration : The specification of the quantum numbers nm and 1 for the orbital of all the electrons of an atom is called the ‘electron configuration’ of the atom. Fo example, the electron configuration of the °C atom is Is? 2s? 2p. Equivalent Orbitals : Orbitals with same 7 and 7 values are said to be ‘equivalent The elecrons in equivalent orbitals are called “equivalent electrons’ Statistical Weight : The number of distinct states in a specified collection is th ‘statistical weight’. The statistical weight of a level is 2J + 1; for a term tt (28 + 1) QL + 1); fora single electron itis 2°. Spectral lines are categorised according to the following nomenclature : Component : A transition between two sublevels is called a ‘component’. Line : A transition between two levels is a ‘line’. Thus, a line 1s @ blend © components. Multiplet : A collection of transitions between two terms | a multiplet consists of a number of lines. Resonance Line : Among the lines arising from transitions between the ground lew and higher levels, the line of lowest frequency is called the ‘resonance line’ 2. Atoms with Several Optically Active Electrons ‘a yes O In the last chapter we determined the electron configuration (n and ! vit weve individual electrons) of the atoms by the application of Pauli’s principle. This: ae does not tell us the term Lypes associated with the ground and the excited electron’ called a ‘multiplet’ ‘Thus aopic Terms : L-S and j-j Couplings 9 fon ee im type is obtained by adding together the angular yum vectors I and $ of the individual electrons. For this Purpose, we have to first the interactions (couplings) between the angular momentum vectors of the Let us consider a multielectron atom (or ion) having a nuclear char; ge +Ze, punded by NV electrons (NV < Z). The electrons constitute a ‘core’ of completely subshells surrounding the nucleus, and a few of them remain in a partially-filled 7 ‘subshell and are optically active. The Hamiltonian for such an atom must consist of lowing terms : (i) The kinetic energy of the electrons, given by N 2 : Si. abe F izt 2m re p; is the momentum of the / th electron (mass m). (ii) The electrostatic energy of the electrons due to their interaction with the nucleus, is given by N 1 ze z my i=1 4%& 7 is the distance of the i th electron from the nucleus. The mutual (residual) electrostatic energy of the electrons, given by N W-1 7 QB = —— i=t ja 408 My is the distance between the ith and the j th electrons. iv) The spin-spin corelation energy. The spin-spin corelation makes electrons with spins repel each other, and electrons with antiparallel spins attract each other. This rises because of the exchange-symmetry postulate which states that the acceptable ions for a system of electrons are antisymmetric with respect to an exchange space and the spin coordinates of any two electrons. The spin-orbit magnetic interaction energy of the electrons. ich, iw, above, there are a number of smaller interaction terms incuding relativistic the Hartree central field approximation, each electron is treated as if it were moving “ntly in a spherically symmetrical net potential that describes the average of = lan interactions with the nucleus and with the other electrons. That is, th includes from above only the terms (i), (ii) and a pane ~~ ee we SY of each i i tically-active electron is determin: ai alee 7 the constant total energy of n and 1. The total energy of the atom is By 0 optical electrons. Consequently, in rmined by the configuration (n, 1 h configuration are a number of ber of different quantum Plus the sum of the total energies of the Proximation, the energy of the atom is dete the optical electrons. Now, associated with eacl Of quantum numbers m, and m,, that is, a num140 Atomic and Molecular Spectra Lay, states, Since the energy does not depend on m and m,, the different quantum States a degenerate. Also, there is an exchange degeneracy because the energy docs not depeng , which of the electrons has a particular set of quantum numbers. Thus, in the Hart approximation there are a number of degenerate energy levels associated with ac configuration. Many of these degeneracies are removed when the remaining of the aboy terms are included as perturbations. The relative magnitudes of the remaining terms a different for different atoms. In general, we can divide the atoms into two main Classes | this respect : I. For most atoms, mainly the lighter ones, the residual (non-spherical) electrostay interaction effect of the term (iii) and the spin-spin corelation effect of the term (iv) are th largest of the remaining terms; the spin-orbit magnetic interaction term (v) is considerab] smaller, while the rest are negligible. Such atoms are governed by L-S coupling. IL. For some atoms, mainly the heavier ones, the spin-orbit term (v) predominates oy, others. Such atoms are governed by j-j coupling. 3. L-S Coupling The L-S coupling is also known as ‘Russell-Saunders’ coupling after the tw astronomers who first used it in studying atomic spectra emitted by stars. In atoms whic obey this coupling we introduce the various perturbations in the order : (a) spin-spi corclation, (b) residual electrostatic interaction, (c) spin-orbit interaction. (a) As a result of spin-spin corelation effect, the individual spin angular momeniur vectors of the ‘optical’ electrons are strongly coupled with one another to form resultant spin angular momentum vector 5’ of magnitude VS(S + 1) # which is constant of motion. The quantum number S can take the values : >, 2 >, 2,2 S=|H+ 2+ 3+ wat [Rt RH HH. +1, min ans (5, + 52 + 53 4 eee The states with different values of S$ have considerable energy difference, the state 0 highest S- being of lowest energy. It means that due to spin-spin corclation, # (unperturbed) energy level of the central field model is splitted into a number ¢ well-separated levels, equal to the number of different values of S that can be forme from the individual spins of the optical electrons of the atom. The different levels a designated by their multiplicity, (25 + 1). Thus : For one electron : Sess t (25 + 1) = 2 (doublet level). For two electrons : n= 4, ns } S= | -]. |s- 9 Hy (+) = 01 (25 + 1) = 1, 3 (singlet and triplet levels) Forthree electrons: sy = 4, s, 2 4 «i 2°" 9_ spectroscopic Terms : L-S and :j Couplings 141 ‘To combine three spins, we first combine two of them to obtain S$’ = 0, 1, and ther i wath combine the third 53 = 3 (© cach of them, Thus, to 5? = 0; we if we couple 54 = / ese and if we couple a=} to S’ =] | qectrons, we get » We get S = Rie vie 4 ‘ Thus, for three ssiil “2° 2° 2 (twosets of doublets, and one set of quartets), ‘the following branching diagram (Fig. 1) illustrates the sumbers which can be obtained by combining several indeper Possible total spin quantum ‘ndent electron spins. ELECTRON S=4= (DOUBLET) IMO ELECTRONS S=0 (SINGLET) Say a (TRIPLET) THEE ELECTRONS =} oN : (DOUBLET) s=Zt (wougier) (QUARTET) AIR ELECTRONS s<0 $=) SINGLET) (TRIPLET) S=0 $57 ssi s=2 ‘ (SINGLET) (TRIPLET) (TRIPLET) (QUINTET Fig. 1) Inthe general case of NV electrons, the possible values of S are $001, 2, cnn © foreven WN, 1 3 5 N im iy 2, for odd N. than it is in the singlet state (antiparallel spins). (Positive) electrostatic repulsion energy is smaller in the triplet state than it Slate. Hence, triplet level lies deeper. “result of the residual electrostatic interaction, the individual orbital angular "eclors of the ‘optical’ electrons are strongly coupled with one another to h orbital angular momentum vector of magnitude VE (L + 1) an Constant of motion. The quantum number L. can take the values : "lis PLP > > W+ B+ Pe. las | H+ P+ Be. wi min vowel Be BP ivecceee ).‘Atomic and Molecular Spectra : Lase 142 ‘The states with different values of L_ have fairly large energy difference, the state g largest L being of lowest energy. it means that each of the levels splitted by spin-pi cere ion effect is further splitted by the residual electrosial® interaction into a number g Jess-separated levels, equal to the number ‘of different values of L that can be formed fron | orbital angular momenta of the optical electrons of the atom. The differen according as L = 0,1,2,3,4, the individual levels are designated as S,P,D,FsGy.-- Thus : For 3p 34 electrons : hel h=2 i t= |h-]- ja] tl ee +b) = 1,2,3 (P,D,F states). two p electrons which ar For 2p 3p 4d electrons : Let us first combine the more tightly bound to the atom. For this hele. . L'=0,1,2. 2. which is less tightly bound, with each of thes Now, combining the d electron, [5 = gives L’=0,h=2. . L = 2 (Dstates), L'=\h=2 L = 1,2,3,(P,D,F, states), and Bis 2 L=0,1,2,3,4,(8,P,D,F,G, states) ‘Thus, in all we obtain one S, two P's, three D's, two F's and one G states. ‘That the state of largest L is of lowest energy can be understood by considering two electrons in a Bohr atom. Because of the coulomb repulsion between the electrons, the electrostatic energy will be a minimum when the electrons stay at the opposite ends of a diameter, that is, at a maximum distance apart (Fig. 2). In this state, the two electrons would be revolving together “in the same direction” about the nucleus, that is, with their individal orbital angular momentum vectors parallel. The magnitude of the total orbital angular momentum vector, L, would clearly be a maximum in this state of lowest 9.2) C tion havi inclu static interac energy. pin corelation and the residual electro i interaction is ™ (c) The dominant spin-s} been taken into account as a first perturbation, the smaller spin-orbit in L-S coupling as an additional perturbation. ange! ‘Asa result of smaller spin-orbit magnetic interaction, the resultant orbital if momentum vector L’ and the resultant spin angular momentum vector Sr yf strongly coupled with each other to form a total angular momentum vector atom : Patek, if the magnitudes of J, Land S? remaining constant, The magnitude © WJ + 1) where the quantum number J takes the values :level characterised by gi futher broken up into comparative closer QS +1) arncterised by-a different J value*. The Or (2L + 1) levels, each if J Sroup of these J-levels forms a ‘fine-structure wate’ ‘The relative spacing of the fine-structure levels within a multiplet is governed tyLande interval rule. 4, Lande Interval Rule Under L-S coupling, the spin-orbit interaction e 481-0 ( 7.5), nergy is of the form ‘where @ isan interaction constant, Let us write PeDee Taking the scalar self product, we have PeP=P.viziy 2 : Mu G04 0 -La+ ys n] A, P= Wen Hand so on. We can write ita ; Abs =ALIU + D-La+ s+ a), 4 is another constant, ‘ The various fine-structure levels of a Russell-Saunders multiplet have the same values 5; and S$, and differ only in the value of J, Hence, the energy difference between two. ct levels corresponding to J and J+ 1 is Bu-B=Ale+ig4n-sus »] = 2A + 1). : nergy interval (spacing) between consecutive levels Jand J+1 ofa 4s proportional to J +1, that is, to the larger of the two {mvolved. This is ‘Lande interval rule,’ WS lake few examples. According to Lande interval ule, the fine-structure levels Pex Py have separations in the ratio 1:2, the levels 3D, Ds, °D; inthe ratio | levels “D,., "Din. “Dans ‘Din in the ratio 3 ; 5 ; 7; and so on. The = agreement between the experimentally observed and the theoretically Predicted lighter atoms Provides evidence of L-S coupling in these atoms, Deviations from tule occur with increasing deviation from L-S coupling. a” “evel nas st @ degeneracy of (2 + 1) which can be— iad Atomic and Molecular Spectra: 1a, , 5. Norm: id Inverted Multiplets Within a given multiplet, usually the level with smallest J value lies lowest multiplet is a ‘normal’ multiplet. There are, however, multiplets in which the lar level lies lowest. Such multiplets are “inverted” multiplets. The reason why in a normal multiplet the smallest J level Understand. The magnetic field B” produced by the orbital motion clecric field of the nucleus isin the same direction as the orbital angular Momentum 2 In this field the most stable state, that is, the state of lowest energy must be One in which the spin magnetic moment $i. of the electron lines up in the direction of BP (Fig. 3.) We a f | Such a Rest is lowest is easy 4 OF the electron in ty (a) (by 's directed opposite to §” because inthe lowest energy state, L° and 5 know that ji? ‘ ‘ the electron is negatively charged. Thus, are in opposite directions and so the value of J is lowest. When 2” and 5 are inthe i p. same direction (Fig. 3 . ‘ Jy then ji? is opposite to BP and the state is ices oe Desspos ps The inverted multiplets arise due to some Perturbing influences, 6. i ‘ma am Spectral Terms for L-s Coupling ©XCiled states of atoms Tor whe io terms associated with the ground state and the ‘the angular momentum of oPute Holds. Since the core electrons do tt electro the completed sibahepy the atom, we must consider only the (optical iC Configuration we i: Computing the terms, The terms if com fi odd; but if computed from an even elect » tern mee are denoted by a superscript ° flown ising from gn configuration 2s 2p is eno ™ aN even configuration 2p 3p is written simp * Sections, 3) Configuration i 252 Sty yon on to ontiguration 2 Is even. For example, the 2p ap ‘Sot even parity, “ °Spectroscopic Terms : L-S.and J Couplings 145 (1) Atoms with One Optical Electron ; configuration is Fora hydrogen-like atom, the ground state Is. For this, we have set, 1=0, ce S= 5 = muliplcity 25 4 1 « 2, L=1= 0(Sstate), a J IL-1, son ht) ad, ‘Thus, the ground state term of “oceans ko son 172 The ground-state configuration of an alkali atom, say Li, . Hence, as for hydrogen, the ground state term of Li is 6 - This is so for all the alkalies, (2) Atoms with Two or More Non-Equivalent Optical Electrons : Let us consider ‘@alom having two optical electrons with outer configuration 4p 4d*. Forthis, we have aes sez mats heat, m2, ‘Mhepossible values of 5 and L are’: reg Sly 2h, I-14 1,. = 0, 1 ; multiplicity (25+ 1) = 1,3 Lath = Bll = bl t bce Ch +h) = 1,2,3,(P,D, F, states), ete have in all six ee singlet terms and three triplet terms. All these terms Ee ise the configuration 4p 4d is odd (1 = 1 + 2 = 3). Wecan write these a+ (3) + 53) t Ip) lpe 3p Ine ape Pi Pha kPa account spin-orbit interaction, let us combine L and S to form J. Now JIL = SU, ceecceseeseseee “+S. terms, we have S=0. pe, La pit=li Pr $20 5.9, Ip L=2i7 SY hve, ces aes eas ie pond Ua cela146 Atomic and Molecular Spectra . Laie, $< oe $=0 523; 'Fy. L=3 For triplet terms, we have Sel. peoaas Ty. Ry. Py. L=1 Sel 70 1,2,3; 2D, Ds, Dy, L=2 32234; JARS Fe. Thus, a single degenerate level of configuration 4p 4d is splitted into 12 levels as shown in Fig. 4. The spin-spin corelation splits this level into two levels, the single t leve) unperturbed spin-spin residual spin-orbit 5 tevel corelation + electrostatic + magnetic energy energy energy (Fig. 4) (S = 0) and the triplet level (S = 1); the triplet (higher multiplicity) level being lower. The residual electrostatic interaction splits each of these levels into three; the P, D and F levels, the level of largest L (that is, F level) being lowest. Now each of the triplet levels is further splitted by the spin-orbit interaction into three fine-structure levels, each characterised by a J value. The spacings between each set of fine-structure levels is according to Lande rule, and the level with smallest J lies lowest (normal triplet). Each of the J level is still (27 + 1)-fold degenerate and this degeneracy can be removed by an external magnetic field. The total degeneracy of 4p 4d level is 60 (verify). For practice, certain other electron configurations and their terms are given in the following table. Try these ! Terms of Non-Equivalent Electrons Electron Configuration 7 | "So. 5, sp 1 sd 1 ppt ddSpectroscopic Terms : L-S and j-j Couplings 147 valent electrons the terms are ey; spins and the three orbital angular momenta. The spin cor combining the spins of two of the electrons and then combining the third spin with each of n jomenta combinations are obtained by first combining the orbital angular momenta of two More-tightly bound of the electrons and then couple the less-tightly bound electron with each of them, Fér example, let us consider the Configuration 3d 45 Sp. aluated by combining the three mbinations are obtained by first Spin Combinations : 5; = On combining 5; and s; we obtain S$’ = 0,1, Then, on combining s; with each of be, ’ 1 1 these S’ values, we obtain § = 2°32: 3 Thus, the multiplicities (25 + 1) are2, 2.and 4 which correspond to two Sets of ‘doublet’ states and One set of ‘quartet’ states. Orbital Angular Momenta Combinations : We first combine 3d and 45 electrons. For them, we have h=2b=0. 6 Vad We now add 5p-electron momentum to it, which is 4, = 1. Thus, L = 1,2,3(P,D, F states) 2p 458, 2p 2p 2p, ar ‘p ‘D, “p that is, (2), °D' (2), FQ), PY Apr, te, Including spin-orbit interaction, we can write the various levels as : *PiraraQh "Dasa Fain Pian ‘Dinsasnias “Peasarmoni Thus, we get a total of 23 distinct levels. We note that one of the quartet terms, namely “P, has three (not four) fine-structure components. This is because for this term £1) is tess than § (= 3/2) and so the number of components is 2L + 1, and not +1, (3) Atoms with Two or More Equivalent Electrons : For two equivalent electrons n and J values) the values of at least one of the remaining quantum numbers (my & ™) must differ to satisfy Pauli’s exclusion principle. Hence certain terms which were Possible for two non-equivalent electrons are now not allowed. For instance, two * valent p-electrons, such as 2p and 3p, give tise to the terms :'P.'D.3S,3P 30 tu if the two electrons are equivalent, say 2p", then the terms ‘ yo ' id “P. Let us now see + Sand 'p do not exist and we have only the terms 'S, 'D an bow 6 obtain terms from a configuration involving equivalent electrons. Before we do so NE must mention two im; facts : portant facts : ' A closed sub-shell, such as ?, p', d®, su... always forms a 'Sy term only. The Sub-shell consists of maximum number, 2 (2! + 1), of equivalent electrons in “iparaliel pairs so thatM, = 0, My = 0 L= 0 (S-Sta), “ve S=0,25+1=1 (singlet) ot J=0. That is, the only possible term is 'Sy. Hence, we conclude that when « ,, Senwleily filed, he only allowed state is one in which the tot momentum, total orbital angular momentum and total angular ‘This also means thatthe subshel has no net magnetic Pin enaae Moment, Rule dipole moment, * all cere, easing peketon 2 2 NO% Al Fob combination of my a ae too 4 mel 1 1 “i j wm 4) 2 2 7} ca fae 0401.9 © king al pons POE sates ~»') in Which a single p-electron’ Checaase by ombittions of the ape” “AUiValemt electrons can be obtained 8 alike by sce et St Sates aken woe corey cn ™ and vale “S300 Te willbe Sac, . ™s cannot be same for two equi ~Spectroscopic Terms : L-S and j-j Couplings —————— oo lt ab, ac, ad, ae, af ; be, bd, be, bf; + and two values of m, to form Ms (Em, = M; and Em, = Ms }. This leads to the following tabulation : [| ab | we | ad [ae [of | be [bd be] d aft fofa [i Po Tli fo Pipe fat-ali poten welt ti fofolots fofofofelofof=f-ta] The highest value of M; is 2 which indicates a D term (L = 2), Since this value of M, occurs only with Ms =0, the tem is 'D (5 = 0). Apart from M, = 2; M= 1,0,~1,~2 also belong to this term, each having Ms = 0. Thus, out of the above AS combinations, the following combinations form 'D term, awa ff of M=2 1 0 -1 -2}'p, Ms=.0 0 0 0 0 Ofthe remaining M, and Ms values, the highest M; is 1 and the highest M; is 1. These values must belong to a *P term: (L = 1, S = 1), because only for such a term can the highest values of M, and Mz be 1, But L = 1 corresponds to M, = 1,0,-1; and S= 1 corresponds to Ms = 1,0,-1. Hence all the following nine combinations belong tothe *P term: ab ac be bd cd co de df of M= 1 0 -1 1 0 =-1 1 0 -1} 3p, Ms=1 1 1 0 0 0 -1 -1 -1 Only one combination be is left for whith M, = 0 and Ms = 0. Hence, it can give only 'S term (L = 0, S = 0). be medi Ms = 0 Thus, two equivalent p-electrons give rise to 'D, *P ad 's terms; and no others. The fine-structure levels P 'Da,Pa.1,2 and "Ss. same ily calculated from Breit's ene, In this scheme we cote in s ube all the M values of Mz which can be formed by the bination of m, and m,, of the two electrons. For Ns ve write the values of my, and my, ina row and in 5 Tespectively. The sums My are written below (two equivalent ) "Md 0 the left of m,,. These nine values of M eli halular Spectra : Laser _ These sets are ! re aivided by the shaped (does) Hines" (sed) form tree sets ANS Fj ool (Used) 1 0 ~ (ill set) 1 and 0 respectively, that is, to one D, 0 22, These sets of Mr-values correspond to L form either S = 0 (singlets) or ne p, andone S term an or ‘The spins of the (WO ae eke he fom in quant enone sa ar their values We cannot, ae any of the and hence they ™ . eee tne wpm Se ct ‘m, ). Also, w¢ can use only the My-values from corresponds to equ me i a merel comreipiiid wa ‘itieen one ve ote eect (otherwise they are identical with, and are @ mirror image of sinie nth first side). Thus, with S= 1 (triplets), we are limited to the following Mp-values. +o -Latseo - 3 which are the components of aterm with L = 1. This corresponds to a “P term ora 3pp,,2 multiplet. When 5 = 0, the e restriction on the values of My setof Mz, values has already been and III sets to combine with S = with L = 2 and L = 0 respectively. Hence, Thus, two equivalent p-electrons give 3Po, 1,2 multiplets. ‘These will also be the terms for p* configuration. Let us now consider two equivalent d-clectrons, that is, Breit's scheme for the possible M,-values is : Jlectrons differ in their spin quantum numbers and there is no which may be combined with this value of S. As the I used to form the °P term, we have only the remaining I 0 (singlets). These sets are the components of terms they correspond to 'D and 's terms. 16 1p and 3P terms or ‘Sp, ‘Dz and (nd) configuration. The (two equivalent electrons) h=2;h=2 Se -spectroscopic Terms : L-S and j-j Couplings 151 ae ‘There are 5 sets of M,-values ; 43:2 10 -1 -2 _3 4 (set) 5 Pro -1 2 13 (Il set) 2 10 -1 ~2 (III set) 10 -1 AV set) Q (V set) These sets correspond to L = 4, 3,2, 1,0 respectively, that is, to G, F, DP, § terms respectively. ‘The spins of the two electrons can be combined to form either § = 0 (singlets) or $= 1 (wiples). For S = 1, we are limited to the Mz values from one side of the diagonal, that is, to the following sets ; 3.2 10 -1-2-3 (I set) 1 0 -1 (IV set) These sets correspond to L = 3 and L = 1 and give °F and °P terms or *F,,s,4and 4pq ,2 multiplets, The remaining I, Ill, and V sets of M;-values are to be combined with $ = 0 (singlets). They yield 'G, 'D and 'S terms. Thus, two equivalent d-electrons give "So. "Dr. 'G., *Pé.1a + Fes s.. ‘These will also be the terms for d* configuration. As a final example; we now calculate the spectral terms arising from P configuration. The six possible states for a single p-electron in a very strong field are m=1 0. =1 1 0 -1 =i 1 i at 2 2 8 2 2 2 2 @ © (©) @) (©) © The possible states for three (equivalent) electrons can be obtained by taking all possible Combinations of the above six states taken three at a time, with no two alike. There will be 2 nations (°c, = ——®! _ = 29). such combinations ( C= Te Fi 20) They are abe abd abe abf acd ace acf ade adf aef bed bee bef bde bdf bef cde cdf cef def Foreach of these 20 combinations we obtain M; (= Em) and Ms (= E m,) . This leads to the following tabulation : abe abd* abe* abf* acd** ace** acf* ade* adf* aef* Ms 9 2 1 0 1 0 -1 2 1 0 Bais ot 4 Peo fh ood 4 ghd la 2 2 2 2 2 2 2 2 2 2 bed bce** bef* bde** bdf** bef* cde cdf** cef* def Me gy Ya i 0 -1 O -1 -2 0 Ms 1 1 Nim vie Nin t Rin 1 Nie I vie 1 1 \ t I ' NIwW152 Atomic and Molecular Spectra : — ‘The highest values of M, are 2 which indicate a D-term (L = 2). Since they occur with . 1 Ms = 4 and Ms = = i which are the magnetic field components of S = 2 the tery is 7D. Apart from M, = 2; M,=1,0,-1,-2 and each with Ms = 3 imal Ms = - i also belong to this term. Thus, out of the above 20 combinations those Marked by a single star (*) go to form the *D term. Of the remaining combinations, the highest M, are 1, and again they occur with My = 3 and Ms=- 4 They indicate, therefore, a 2P term(L = 4), Si 3] Apa from My = 1; My =0,-1 and each with Ms= 4 and Ms =~} also belong oth term, Hence the combinations marked by a double star (**) belong to the *P term. ‘The remaining four combinations are : M=0 0 0 0 aan i] Mey 2072 7 iw These M, and Mg values are the components of L = 0 and S = 3 which correspond toa ‘S term. Thus, the terms of p* are ap D, ’ or Pasi Drs ‘Syn. Some important results are as follows : Terms of Equivalent Electrons aa Is, z Pinar he. $0 .'D2, Po, 1,2 2 2 4, Pins D'yrs1. S30 2 Dy sa 1 1 1 S01 Dry 'G47Po,1,2.7F 2.3.4 cons, then" {fa given configuration contain i i ' 5 both equivalent ang, non-equivalent clectr rey derive the spectral ae by starting with the equivalent electron terms and er Other electrons, one , wit nfig ays, diieene one, with them. Thus, Suppose we have a CO! The terms for 4 p* are 's."D and 2pspectroscopic Terms : L-S and j-j Couplings 153 Let us add the 5s electron to each of these : 's + s = 3 = "SA W=0,8'=0) (1=92=3) (4-05-35) 'D + 5 = *D = Diasn ; L i W=2, 8 =0) (isa 2-3) (e255) 3p ia s = 2p 4p = Pass Piasasn ’ - a 1 = 13 Wes =1) (! as=3) (4-4 5=5.3) 7. Order of Terms and Fine-structure Levels The relative energies of the various terms and levels which arise from a given (equivalent) electron configuration may be deduced from a sct of rules given by Hund, These rules are : (1) Of the terms arising from equivalent electrons, those with largest multiplicity lie lowest. (2) OF the terms with given multiplicity, and arising from equivalent electrons, that with largest L value lies lowest, (3a) In the multiplets formed from equivalent electrons in a less than half-filled sub-shell, the level with lowest J lies lowest (“normal order"), (3b) In the multiplets formed from equivalent electrons in a more than half-filled sub-shell, the level with highest J lies lowest (“inverted order”), (3c) Terms arising from half-filled sub-shells show only very slight fine-structure splitting. (Gd) The lowest terms arising from the half-filled sub-shells are the S-termis and are Specially stable. These terms are °S;/. for ahalf-filled s sub-shell, “Sua for a half-filled P sub-shell, °55o fora half-filled d sub-shell, and *S,-> fora half-filled Ff sub-shell. Hund was quite successful in Predicting the lowest terms and levels of many atoms, ‘ver, exceptions do occur. Let us apply these rules to some examples. The normal Sonfiguration of the carbon atom is 1s* 2s? 2p? which gives rise to the following terms : "So, "Das *Po. ‘Pi, "Pa. Applying Hund’s set of rules and remembering that these terms arise from a less than hal-illed sub-shell (p?), we find that these terms are in the following order of increasing energy ; 3p, Spy, *p,, "Dy, "So. The oxygen atom, on the other hand, has the configuration 1s°2s? 2p, The terms are ain 's,, "D,,Po.1,25 the ground state being a 'P term, However, since the 2p ‘ub-shell is more than half-filled, the multiplet will be inverted and the state of lowest etBy will now be the "P, term. _‘Atomic and Molecular Spectra: Lay ae Fata, 1225%2p>, which gi ip, which gives tise 4g 154 i configuration P oy s Te ° i the eal state, the next above itis D” and then 1 g, Normal Electron Configuration and Spectral Terms of Certain Atoms We give below for certain important atoms the normal electron configurations and the terms (or levels) arising from them. Atom. Configuration Terms y Is a "He is 15, Ai 1s?25 25.70 ‘Be eae 1s, 3 1s 28 2p Pp (ground), 2s, ‘6 1525? 2p" 159, "Da 'Po ground),"Ps "Ps N weaeay Pin. Psp, Dias D's, 45%52 (ground) ie) 1522s? 2p* 15, Dy 3Po. 21» Pacem *F 128° 9p° tpn ya wound) "Ne 1s 25" 9p° Is, yy —_|is?25° 29°38 25,9 "Mg 1s? 29° 2p° 38° Is, "ol 1222 299° 38° 3p bp a ?P salen io 15? 25? 2p° 3s? 3p 4e. 's, cu 15228? ap? 3¢ ap 3494s 59 “Zn 1s as 2p 32? 3p ad" ae Is, "Hg 15? 2s? ap 3s? 3° 3d!94s? ap? 4d af! 5325) 6 54° 692 Is, 9, Selection Rules for Multi-electron Atoms in L-S Coupling jectron atoms a The ction » mi rules forthe elecrc-dipole transitions in multi-el a to the selection rules for the one-electron atom. ds i) Most of the transitions occur in whi j a that its ur in which only one electron jumps 8 0"" its /-value changes by one unit, that is, , m Abst. +f «the pari 4 This is a special special case of the more general Laporte rule thal ‘odd > even) *The pay | ‘odd or even according as £1 of the coniiguration is odd oF °¥°" espesecrscopic Terms: L-S and -j Couplings Iss In ease more'thanyone electron jumps, the Laporte rule requites that the sum of the individuall’s must change by an odd number of units, For example, if we have a transition involving two electrons simultaneously, then A! must be even for one electron and odd for the other. Since we have Al = +1 for a one-electron transition, for a two-electron tanstion we would have Ah =+1; 4h =0, +2, Thus, a double electron jump may take place from the configuration 3d 4d (even) to 4s4p (odd). There are two possibilities. 4d may go to 4p (Al = - 1) and 3d to 4s (4h = - 2); or 3d to 4p (Ah =- 1) and 4d to 4s (Ah = - 2), (ii) There is no restriction on the total quantum number 1 of either electron. ii) For the atom as a whole, the quantum numbers L, S and J must change as follows : AL = 0, £1 (Inone-electron atoms A L = 0 is not allowed) AS=0 4J=0,+1 but J=0)=0. Therules for L and S hold only when the spin-orbit (magnetic) coupling is weak, that is, when the splittings between the various fine-structure levels of a multiplet are very small compared with the separations between the various multiplets themselves, We know that the spin-orbit interaction increases rapidly as we go to heavier atoms. Hence the above rules for L and S hold quite accurately for light atoms but not well for heavy atoms. This is why there are no intercombination lines (singlet-triplet transitions) in the spectrum of helium, but they are quite strong in the spectrum of. mercury. As a matter of fact, for heavy uoms the whole concept of L-S coupling breaks down and one approaches to j-j coupling. 10. }{ Coupling In its ideal form, the: j-j coupling is an opposite extreme to the ideal L-S coupling ind is approached by the heavier atoms, for which the spin-orbit (magnetic) interaction ‘em in the Hamiltonian predominates over the residual electrostatic interaction and the ‘in-spin corelation. This means that the interaction between the orbital and the spin Tomenta of a single electron is much greater than the interaction between the orbital Momenta of different electrons or between the spin momenta of different electrons. "efore, in this case the splitting of the unperturbed energy level due to the introduction Various perturbations takes place in the order : (a) spin-orbit interaction, (b) residual ic interaction and spin-spin corelation. (4) Asa result of the stronger spin-orbit interaction, the orbital and the spin angular lum vectors of each individual electron are strongly coupled together to form a rT h a angular momentum vector § of magnitude WO+D QR’ where ; 1 : 2 2 padi+ i, that is, j takes half-integral values only. This means that due sin orbit interaction, the unperturbed energy level is splitted into a number of Paced levels, each corresponding to a different combination of the possible j-values x ‘ndividual optical electrons; the level corresponding tol the electrons having their F¥alue (i ya > Jing lowestAtomic and Molecular Spectr, 156 " th rong stom, "aks ty ctrostatic interaction and Spin-spin Corelay (0) As a result of the residual J’ of the individual electrons are tet Tern = oni tar tel olpilarlnomextiinVatin,- P of the lel ot ons malt rm Har momentum quantum numbe, on WO eT A. The total angular J 2n magnitude values : i | a Be one is | Re Bt oun Las HLS essere ting = 1 This s that each of the above levels is further splitted into @ Dumber of ikea i acid ji if J, — eeuiee wa ane the terms for icone confi Se N ing this configurati, j it ll that under L-S coupling under jj coupling, (We recal Prs'Das Fy 912:°Diasi Faxed , For the p-clectron: ly = 1,5) = ziand so jy = 1 : For the d-lectron: |, = 2.5, = 2iandso j, = This gives four (j, +42) combinations of the possible Ai and jy val lues. These are 1 3).f1 s) (3 3 3 s\ (3:3 ; ri): a*3| me Buration pay ‘OR BIVES the tem 13 es" a5 arg Thus, the spin-orbit interaction’ splits Unpertu: energy leve! which. 3 is west and 3 into four levels, o ji highest, is ‘trostatic interact @ number of J-levels, equal to the number of integral Spaced values of J that can t of the two J -values. above four (j, oh ‘Combinations Bive J values as under : 1 2 3 2 i) 23 Bes Ja O1 23 5 2 Bes Jaa ag The ee ening is shown in Fig. 5 (see Next page), : \ . tha the J vatucg et levels is the same (12) as forthe L- coun Ji coupl ively ett rem se fatively Seldom, In fact, there is a Sradual shift nal feu cm T atoms towa, +4 couplin for the heavier atom elements NSition ig Seen in ¥ sramgC SUG yg the levels of the first excited state of the My fovea an clo: have practically pure 1.8 courl i coupii jo “upling whch ed Olin For the ground state of these 3spectroscopic Terms : L-S and. Jj Couplings —<———!}2 157 J (3) 59) ace ab , S=loe, a7 (Sey Ya) a Cie a NN Wanag 4p 4d 2» S}2) <“aeeee ae Sa) \ < 2} aShs9 5 \ \_ Ole, 32) ot ————_, Oe ted,» unperturbed level + spin-orbit energy + electrostatic and spin-spin energy (Fig. 8) 47 =0, t1butJ=0e sa, The selection rules AS = Oan longer good quantum numbers. id AE= 0, *1 no longer hold, since L and $ are no SOLVED PROBLEMS 1 Inthe 34 3d configuration of the 2 mp, Py. The measured energy ini 167 x 1974 Ca atom there is a normal triplet of levels : terval between the °P, and *Py levels is eV and that between the °P; and 2p, levels is 33-3 x 10°*eY, Show these values verify the Lande interval rule, Solution, By Lande interval Tule, the interval between two consecutive fine-stricture corresponding to Jand J +1 is proportional to J + 1, thatis, Eis. - Ey= 2A +1), here A is some Constant. Therefore, the energy interval between °P, and *Py levels is E, ~ E=2A(1) “Sthat between 3p, and °P, levels is Ey - E,=2A(2). Ths, the Lande interval rule predicts that xa |. osm, E,- "2 Neoing ‘oexperiment, we have £, - & 7 ~ Sx I eV L959, f.- B 333 x 10 ev _ eyAtomic and Molecular 5, Pectra Late 158 the experimental and the predicted ratios verip, Mes th, ‘The excellent agreement between atom. This, in turn, provides evic, lence foray Lande’s interval rule is applicable in presence of L-S coupling in Ca. 2, Measurements of the line spectrum of a certain atom show 4 levels of increasing energy in a mobile th are separations between adjacent energy the ratio 3 : 5. Use the Lande interval rule to assign the quantum numbers 5 : Shay these levels. Solution. Let J,J+1 and J+2 be the quantum numbers of the lowest successively higher energy levels of the (normal) multiplet, as shown. If E be ‘eee th 78 nd J+1 levels, then the interval between J a and J+2 levels would be J+2 interval between J+1 E (5/3) E, as given. a Now, according to Lande interval rule, the J+1 energy interval between adjacent levels of a ) te fnultiplet is proportional 10 the J value of the olor upper level. Thus, E=2AU +1) and G/3)E=2A 0 + 2). er 3. J+) Dividing : 57342: Solving : = + ‘Thus, the J values of the levels, in the order of increasing energy, are 3 vege 22° To determine the values of S and L for the multiplet, we use JsIL— St, IL —SV4 Vy eee (L + S). ‘The minimum andihe maximum values of J are 3 and 5. Thus, 1 Ib+ Sl= > 2 and (L+S= 3. Let us suppose that L > S$, Then 1 L-S=> § 2 and L+S= 3. Addition and subtraction give L=3 and $=. annot be half-in But, the total orbital angular momentum quantum number Le Hence, our supposition L > S is wrong. Obviously, L < S, Then S-L=as Lig‘spectroscopic Terms : L-S and j-j Couplings 159 5 oa Sting L+s=2, S= 1, sothat 254123 L=1(P state), case L< gs would give the same result* ~ Thus, the levels of the lov, sr state alues of the Upper state levels, in the order of J = 1,253. J=IL~SiIL-~S| +1‘eo Atomic and Molecular Spectra But S = 1 (for triplet) and so ' L = 2(D state). The levels of the upper state are 3D), "Dz. "Dy. ‘The energy level diagram has been drawn in Fig. 6. %, 3, D. 55) Ih tt * 3, *, ITT I] COMPOUND TRIPLET Fig. 6) Applying the selection rule AJ = 0,41 butJ = 0¢4J=0, we obtain six transitions. The spectrum is a compound triplet. 4. In a quartet transition for C* observed by Fowler and Selwyn, the separation between the upper state multiplet levels are found to be 14-72, 25-07 and 36:30 em! while the separations between the lower state multiplet levels are found to be 23 and 45-0 em’ ', Obtain the term designation of the particular transition involved am draw the energy level diagram also. Which of the components would be most strong: (Meerut Solution, Let J, J + 1,J+2,J +3 be the J-values of the lowest, and th successive higher levels of the upper state multiplet. By Lande rule, the energy separate between two successive levels J and J + 1 of a multiplet is proportional to J * Thus, Z+1 _ 1472 _ 2944 ~ J+2 ~ 2507 ~ 5014 ~ and L+2 _ 2507 _ 5.014 ~ ; J+3 ~ 3630 ~ 7.26 ~ These give J = 2 Thus, the J-values of the levels are 3 5 3 2. 13 5 7 des ,2 2 7 2° 2*2°2‘Spectroscopic Terms ; L-S ang ij Couplings Todetermine the values of 5 and L for the multiplet, We write Jrll-Siit- stay 161 ‘The minimum and maximum values of J are 3 and i. Thus, L~sjol S 2 ant t+ =. Let us suppose that L > S. Then 1 L-S= 2 and beset. 3 Solving : S= 5) sothat 28 +1 = 4 (quartet) and L = 2(D-state), ‘The levels of the upper state are ‘Dia. ‘Dar. ‘Ds, ‘Dy. Mwesuppose that L < S, then, we shall write 1 S-tal tro 1 und L+s=3 These would give ! = a2, S22, L=3 But this is not possible because the total orbital angular momentum quantum number L cannot be hall-imegral. Hence, the assumtion L,< S is wrong. Proceeding as above for the lower state multiplet levels, we have J+t _ 2384 | 265. 3. J+2 > 450 “5 5 Thsgives J = 1. The J-values ofthe levels are thus 3 3 J=5, 2°2 Again, J=IL-SIIL-Si+1,.. Ts, IL-si=4 ng “+923. letus first suppose L > S. Then ud—— 162 Atomic and Molecular Spectra : Laser These would give L = 3 which isnot possible. Hence, let us now suppose L < S$. Then we shall write i S-L=5 5 and L+S=3- i These give = 3, sothar2s + 1 = 4 (quartet) | and L= 1 (P-state). The levels of the lower state are ‘Pay “Pan, "Psp. We note that a “P term has only three components (because L < S, hence the number of components is 2L + 1 and not 2S + 1). ‘The energy level diagram has been drawn in Fig. 7. J Te 40) Fa 32 Ya | 1 ott TTbl Sk 45 * Ya (Fig.7) Applying the selection rule AJ=0, 410 #04 J = 0) we obtain eight transitions, ai, : those transitions for which J and L alter in the same sense a® "p, ace ; of these the most intense is that with greatest J. Thus, the transitio® 2 ~ “Psy is the most intense,‘Spectroscopic Terms ; LS and j-j Couplings 163 5. The quantum imbers of the’ ‘atom are : SNe'0 Opti estrone in «two-valnce casos "6h a3 01 m=Sh=ts 01 ‘@) Assuming L-S coupling, find th, Possibl () Assuming jj coupling fn the pons ne, ae Of Band hence of J Solution. (a) Given that : (Meerut 2008, 2000 special paper) h=3h=1, “Leth bhih- big = 2,3,4, : g saline 2 * S=ls, ~ alls st 1, =0,1. The J-values are : =IL-Sl,. For$=0 and b= 2,34, wehave J=1235 2,34 and 3,45, 1 4 =3,5,=1, * A= shy th = 44, (ty + 54) =i 7 ~2°2 ) 1 b= 2 pet 3. fon 3i 9 n % give four j,, j2 combinations : [2EE-9.0-9)-8-2) gives J = 1,2,3,4. i 2 (3.3) gives J = 2,345. 2 number) are the same in both cases.164 Atomic and Molecular Spectra Lan Ser 6. Compute the possible terms and energy levels for a configuration with opticaliy active electrons 2p 3p 4d- three Solution. The configuration 2p 3p 4d is even. Let us first comput combinations. We have = 1 1 nz 2g BET On combining s and 52, we obtain S’ = 0,1. Then, on combining sy with each of these S’ values, we obtain 1 3 1 S909" 2 so that 28+1=2,2,4, which correspond to two sets of doublet terms and one set of quartet terms. Let us now compute orbital angular momenta combinations. We have hel helb=2 Oncombining /, and /;, we obtain LU =0,1,2. = 2 toeach of these L’ values, we obtain L=2; 1,23; 0123,45 Combining 45 which correspond to D; P,D,F:S,P,D,F,G states. The possible terms are Ip °p, *D: *P, 25. 25, 4s, 2p, p, 4p, *D, 7D, P, 2D, 2D, *D, °F, °F. “Fs ip, 2F, 2F, 4F,7G, 7G, “G that is, 25 (2), P (4), D6), 2F (4). GQ). ‘S(1), “P). ‘D@), *FQ, “GU. where the numbers in parentheses indicate the number of the corresponding terms. Including spin-orbit splitting, we can write 35.02) ?Pinsa (4s *Dsr.52 Oh *Fs2.72 (4 Grron Qs {S39 (0s PraaasaQ) ‘Din32,52.72 8) “Fra sarnsaQh ‘“Gsa1n.92,.20)s a total of 65 levels. 7. Write down the normal electronic configuration of carbon atom Z obtain the spectral terms arising from equivalent electrons. (Meerut special paper 2004, 03, 01, 0° 5, 96, ‘Also write down its first excited configuration and obtain the spectral term Indicate the allowed transitions. (Meerut 2000 Solution. The normal (ground-state) electronic configuration of °c is 2p. We have to find spectral terms for two equivalent computed in § 6) e p electrons. ‘These terms 4g; 1D, 5B. The energy levels areBy Hund’s rule, the components of the triplet 3 ae 2. Pi- "Py The energy levels for the normal state 2p") of °C atom are drawn in Fig, 8 1 FIRST r is EXCITED, t STATE ae 2p36 ape i VA i ! ape fe Pitti) i ' - 's, NORMAL 4 STATE eee 2p? RB SINGLET. TRIPL 8 er TRANSITIONS | TRANGITIGNe ? ' H 4L= 41; asst) {4L=0;AJ=0, +1, Js0 +fes=o ' (Fig. 8) “The first excited configuration i 2p 3s, is odd. This configuration (ps) would give the terms 1 Pp . 3, ‘p*, levels would be PP. PL Pa, *ttiplet levels being deeper. The rst) excited-state (2p 3s) energy levels are also drawn in Fig. & _ Tatsitions are allowed under the following selection rules : tions can occur only between configurations which differ in the nm and / ‘Sumbers of a single electron. This means that two. or more electrons cannot 'Y make transitions between energy levels. tion can occur only between configuration in which the change in / of that ‘Atisfies the same restriction as for one-electron atoms, that is, eS ly mat 2Atomic and Molecular Spectra : Laser ued even ¢-—+ odd. these configurations for which - between states in these confi eh the 3. Transitions can O=°ur On ity the following restrictions: AL changes in £, 5, J quantum nu AS =O sy=0, thbus = 0 > J = 0. Franko drawn according (o the above rules are 8 in number, a QUESTIONS 1. Describe and explain the different types of couplings in atoms. Give illustrative . (Meerut 94) 2 wn Les coupling ? Deduce the various interaction energy terms for L.5 coupling, / (Meerut 2001 sp. paper) 3. Distinguish between L-S and Jj couplings for two-valence electron systems Show that the same number ‘of terms are obtained for the configuration ps in the two coupling schemes. (Meerut 2004 sp. paper, 2000) 4. Prove that the number of terms for pd configuration is the same for L-S coupling as that for j-j coupling. __ (Meerut 96 sp. paper, 93) 5. Distinguish between L-S and jj coupling schemes in the case of two-valence clectgon systems. Under what conditions can a transition from L-S to j-j coupling scheme has been observed ? Illustrate your answer with examples. (Meerut 99 sp. paper) 6. Give the selection rules for the allowed transitions in an atom for two valence ‘electrons. What selection rules operate for a transition between 'D and 'S terms of the oxygen atom. (Meerut 2004 sp. paper) 7. State and explain Lande interval rule for Russell-Saunders’ coupling. (Meerut 2003 sp. paper, 03, 92) PROBLEMS L. Show that an atom having filled subshells has 'Sp ground state. (Meerut 96 sp. paper) 2 Find the values of /,5,j and the corresponding values of L,5,J for an atom heaving electron configuration 15° 2s? 2p! . (Meerut 89 sp. paper) pmtoisi,saSaut, jest, 2. A. The sluminivm atom has two 3s electrons and one 3p electron outside filled sence shells. Find the term symbol of its ground state. Sit: Al: 19° 25° 2p° 39 Sp. Ans. ?Pia- < The fctisam stom has one 2s electron outside a filled inner shell. Its ground state 6 “Sug What ace the term symbols of the other allowed states, if any ? (2. +6), as shown, ‘hes. There are no other allowed states. , ee 3s electrons outside filled inner thee FaSpectroscopic Terms : L-S and j-j Co lings 6. Express the followi States 7 4 1 Notation : =LSal, ? @L=1,5 2? ML=25=2, OL=3,5203 167 82 sp. paper) S6- spectral levels : (Meerut 92 sp. paper) Pra sarasaun ; ©) "F, S.J and the ae ‘ OLang (0) "50 5) Des Gey Spee ites of he following Pectral terms for (i) two configuration), and (ii) two equivalent electrons (p? of Pauli’s exclusion Principe: 7 . (Meerut. Ans. (i) 'S), 'P), "Dy, 35. Sp 5, Dias: Gi) '5y, "Dy, Poa. ic ing two-electron systems ? non-equivalent p-electrons op configuration) on the basis 2006, 04) @) ap n’d, (ii) np)’, Gti) (nd)? Ans. (i) 'P, 'D, 'F, *P, 'b, °F. iy 's, 1p, 3p, Git) 'S, 'D,'G, 2p, Ip Find out the spectral terms for two equivalent d electrons, ‘Ams. ‘So. 'D2, Ga, Paras Fase. Work out the different electronic terms obtainable from the following electronic configurations and state with reasons, which of these terms represents the ground State of the corresponding atoms : @)1s72s?, 0) 15° 257297, (6) 1s? 25° 2p CIs? 2s? 278 35? aps Ans.(a) "Sy, (b) 'Sy, "Dz, *Po (ground), °P, , °P, ©) Py. Pr2, Dar, Dea. ‘Sur (ground) (@) "50, 'D2,3Py.°P,.°Py (ground) ©) P12, *Prp (ground). > Spectral terms should be present in the odd configuration 2p* 3p ? | 08. S12 sa). Divas Qs Fearn: Sya. f Pinsasa, ‘Diasasaia. + Deduce terms of 3p 4d configuration system in L-S and j-j couplings. Show _ them in a diagram, ‘uantum numbers of two electrons in a two-valence electron atom are : ” (A) 1s? 25? 2p 35? 3p, aly 1 x my =5,4 =0,5 = 5, Soe 1 ma4ib=ls=z16. 17. 18. 19 21. Atomic and Molecular Spectra : Las ‘Laser (a) Assuming L-S coupling, find possible values of L and hence of J. (by Assuming jj coupling, {ind possible values of J (Meerut 90 Assuming Fre 150, 1,2 (0) = 0.1 1% )) Write down the normal electronic configurations of helium and lithium atoms and determine the states (0 which these configurations give rise. ‘Ans. "He: 157, 'Sq3 “Li: 18°25, "Sia Show that the configuration 2 p? in carbon atom gives 'So. °Poi.2 and 'b, ‘tates, Draw energy level diagram and denote the ground state. (Meerut 2002 sp. paper) Hint : *Po is ground state. The atomic number of carbon is 6. (i) State its electronic configuration (ii) Calculate the spectroscopic terms for this configuration. (iii) If one of the 2p electrons gets excited to the ‘M-shell what other spectroscopic terms will be possible ? Carbon is governed by L-S coupling. "Ans, (i) 152 25° 2p" (i) 'S, 'D. Sp, (ii)2p 38 : "Ps *P, ap3p : 'S, 'P, ‘Ds °S, *P. *D, 2p3d:'P,'D, 'F 3 °P.°D, *F, Explain equivalent and non-equivalent electrons. Write down normal electronic configuration of neutral nitrogen atom and spectral terms arising from it, and show that 7D* is next state to 4s" ground state. (Meerut 97) ‘Ans. 18°28? 2p's Prva» 2D sa and “S's. * and obtain the spectral terms. Write down the electronic configuration of N ‘Ans. 157257 2p"; 'So, 'Dz» "Po.1.2- Obtain the terms for the ground state of neutral oxygen atom. (Meerut 98 sp. paper) ‘Ans. 'Sp, "Da. *Po. °P1. °Pz (ground). Write down the normal electronic configuration of O* ion and determine the spectral terms obtained from this configuration. State with reasons the ground term of O° ion. Ans. 1s?2s? 2p’; *P, °D, “5 (ground). Write electronic configuration of Ca (Z = 20). What are various spectral terms? (Meerut 92) Ans. 15°25? 2p 35° 3p 48°; 'Sp-10 Spectra of Alkalj Elements {Brod Features of akan ogg ets Affer hydrogen, the simplest atoms aoms’; “Li, "Na, "°K, "Ry ) series in the infee gog ngien » and a ‘fundamental’ (or Taking clue from the Balmer's formula for hydrogen spectral series, y= y_ — n Rydberg represented the alkali spectral series by similar f Principal Ya = Ve ~ —a_ (m + py * R Sharp Vig = Ven — ‘ . * (m + sy Diffuse va = vi - — (m + dy “Fondamental vi, = vy, — —K " (m +f) Yu 'S are the wave numbers of the convergence limits of the corresponding series, ae called ‘fixed terms’. In the so-called ‘running terms’ ; P»s,d and f represent ‘corrections for the corresponding series. Rydberg noticed the following relations among different series of the same alkali The sharp and the diffuse series have a common convergence limit (v! = vf). {The common convergence limit is equal tothe frst running term (with m =2) Principal series : ave i. 2 +p) The convergence limit of the principal series is equal to a running term (with Vol the sharp series : we ; (l+s) The convergence limit of the fundamental series is equal to the first running ™ = 3)af the diffuse series : es‘Atomic and Molecular Spectra : Lase, - 170 In view of these relations - the above formulae may be written in the following form ; Principal a - 223.4, ce << mas (mt py 5 pier at et m=2,3.4, Seer Vm = 4 pyr (mt sy * iffuse dj gee = _ Rk a= pf ‘n= Ga pe meay % R -—-,--— m=4,5,6, 7 Fundamental Vm men 3 +d) The following laws were found to hold in the alkali series © ‘The wave number difference between the principal series Rydberg-Schuster Law + limit and the sharp (or diffuse) series limit is equal to the wave number of the first line of the principal series = ve vied = _rk_ __® 5s a aq+y @+Pr ce between the diffuse series limit and the The wave number differen ber of the first line of the diffuse series : Runge’s Law : fundamental series limit is equal to the wave num R vi- w= —_ - = vi. @+tp G+dy , 2. Ritz Combination Principle Ritz pointed out the possibility of occurrence fixed term in the formula for the chief series. Such observed in many spectra including that of hydrogen. ‘Asan example, the (chief) principal and sharp series of alkalies are represented in the abbreviated notation by of other series obtained by changing the additional series have actually been +2, = 1S -— mP, pase 2,34), cveeevee oo. i) vi, = 2P — mS, m = 2,3,45 creer oo. (ii) The series predicted by Ritz are obtained by changing the fixed terms 1S and 2P to VS, 3S, ee and 3P.4P,......- Thus, we obtain the combination principal series represented by 25 - mP, m= 3.4.5, Le ti) 3S - mP, m=4,5,6, oo and the combination sharp series represented by 3P - mS, m=4, . Bie i) 4P — mS m=5,6,7, 1 Oe i) are included in the running terms of (ii) sand We note that all fixed terms occurring i the fixed terms occurring in (iv) are included in the running terms of (i) Thus predicted series are simply combinations (sum or difference) of the terms of the chit series. The resulting series are therefore called ‘combination series’ + and the possibility q their occurrence is known as ‘Ritz combination principle’. 3. Explanation of the Broad Features of Alkali Spectra gas core compos An alkali atom (of atomic number Z ) consists of an inert: ingle nucleus and a few completed subshells having Z — electrons, plus @ * —