Lecture 21: Hyperfine Structure of Spectral Lines:
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In this lecture, we will go through the hyperfine structure of atoms.
Various origins of the hyperfine structure are discussed
The coupling of nuclear and electronic total angular momentum is explained.
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When individual multiplet ( JJ→ transitions) components are examined with spectral apparatus of the highest possible resolution, it is found that in many atomic spectra each of these components is still further split into a number of components lying extremely close together.
This splitting is called hyperfine structure.
− The magnitude of the splitting is ~2cm 1 .
Hyperfine structure is caused by properties of the atomic nucleus.
Isotopic effect:
Heavier isotopes present 1 in 5000 in ordinary hydrogen.
Therefore, different isotopes of same element have slightly different spectral lines.
Consider 1H (hydrogen) and 2H (deuterium):
1 71− RRH = ∞ =1.09677 × 10 meter 1+ m M H
1 71− RRD = ∞ =1.097074 × 10 meter 1+ m M D The wavelength difference is therefore:
λD ∆=λλHDH − λ = λ1 − λH R =λ 1 − H H R Figure shows the Balmer D line of H & D This is known as isotope shift.
observed calculated −−11 ()cm ()cm H 1.79 1.787 α Hβ 1.33 1.323
Hγ 1.19 1.182
Hδ 1.12 1.117
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A quantitative explanation of the isotope effect is not simple, exception H atom. For the heavier elements the effect is traced back to the change of nucleus radius with mass.
Sm150− Sm 152 is double that of Sm152− Sm 154 .
Usual increase is not from Sm150→ Sm 152 .
In many cases the isotope effect is not sufficient to explain the hyperfine structure.
The number of hyperfine structure components is often considerably greater than the number of isotopes.
In particular, elements which have only one isotope in appreciable amount also show hyperfine structure splitting.
Likewise, the number of components of different lines is frequently quite different for one and the same element.
These hyperfine structures can be quantitatively explained, when it is assumed that the “atomic nucleus possess an intrinsic angular momentum with which is associated a magnetic moment”.
This angular momentum can have different magnitudes for different nuclei and also of course, for different isotope of the same element.
This is known as Nuclear spin
.
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Magnetic moment & Angular momentum of the nucleus.
Nucleus consists of Proton & Neutron.
Proton:
1 (i) Possesses angular momentum I described by the spin quantum number ; this angular P 2 momentum obeys the general rules of quantization.
1 Component along oz-axis (I ) = ± P z 2
11 Magnitude IP = +1 22 (ii) Possesses a magnetic moment µP parallel and in the same sense as its angular
momentum IP .
e Magnetic momentum of proton µµ=( ) =2.79 = 2.79 µN PPz 2MK B
em 1 µ N = nuclear magneton =≈=µµ B 2MKBB M 1836
µ So, µ N = B B 1836
Neutron: also possesses magnetic moment
µµ= −1.913 N Neu B
But µµNeu+≠ P total magnetic moment of the nucleus.
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The structure of the nucleus is complex.
Inter nuclear forces are non-central forces involving angles between the magnetic moments and the radius vector joining the nucleus.
Further more, within the nucleus the nucleus possess an orbital angular momentum which can be zero for certain nuclei.
The nuclear magnetic moment µN is related to nuclear angular momentum I, µµN= gI IB 1 = gNNµ I So, g= g N IB II1836 ggor N is called nuclear Lande’ factor. I I
The general adopted sign for gI :
The Lande’ factor is positive when the nuclear magnetic moment and angular momentum are in the same direction, and is considered negative when in the opposite direction.
I I
µN
µN
gII<<0,µ 0 gII>>0,µ 0
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Nuclear spin and Magnetic moment:
(1) All isotopes having an even mass no. A and an even atomic number z , have zero nuclear spin and zero nuclear magnetic moment.
4 16 20 Example: 2He; 8 O ; 10 Ne ; .....
(2) All isotopes having an even mass number A and an odd atomic number z , have an integral nuclear spin.
2 6 10 135D, I= 1 ; Li , I = 1 ; B , I = 3 ; .....
(3) All isotopes having an odd mass number A , have a half integral nuclear spin.
1 113 39 3 1H,;,;, I= 2 He I= 19 K I = 222 1s 12 ss [ _4] s
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Magnetic field due the orbital motion of electron:
A point charge is qe= − is moving in a classical orbit with a velocity V . At a given instant, the field it creates at the nucleus, is µµid×− r r µ q B =00 =qV ×= 0m r × V 44ππr3 rr 33 4 πm µ q l = 0 4π m r3
r is directed from the nucleus towards the charge q .
Hence the magnetic field due to the orbital motion of the electron is
µµ00q 11 BllB= = − 2µ l 44ππm rr33
The interaction energy between the nuclear magnetic moment and the orbital motion of electron
µ0 2 1 El=−⋅=µµ N B l2 g IB lI ⋅ 4π r3
Using special case of Wigner-Eckart theorem, ( Lande formula) Jm|| A⋅ J Jm Jm'| A | Jm = Jm'| J | Jm JJ( +1)
jm|| l⋅ j jm jm'| l | jm = jm'| j | jm jj( +1) lj. So we can substitute, lj= and we get jj(+ 1) µµ0022 1 lj.1 El=2 g IBµµ l ⋅= I 2. gIB (Ij) 4ππrr334 jj(+ 1) ……………………..(21.1)
Page-8 The interaction energy between two magnetic dipole moments µN and µS separated by r is given by µ µµ. 3(µµ⋅⋅rr)( ) =0 NS + NS ESpin 35 …………………………………(21.2) 4π rr Substituting the value of µN and µS in this equation, we get 2 µµgg Is. 3(Ir⋅⋅)( sr) E =0 ISB −+ ……………………………..(21.3) Spin 4π rr35 Is. Let us take the first term r3 jm|| s⋅ j jm Using the relation jm'| s | jm = jm'| j | jm jj( +1)
sj. So we can substitute, sj= and we get jj(+ 1)
2 I. s js .1. ls+ s 1 = (.)Ij = (.)Ij r3 jj(++ 1) r 33 jj ( 1) r
3(Ir⋅⋅)( sr) Now let us take the second term r5 rj. we can substitute, rj= and we get jj(+ 1)
33(Ir⋅⋅)( sr) ( I ⋅⋅ j)( sr) rj. = r55r jj(+ 1)
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jr.= ( l + s ). r = + Now, lr.. sr =0. + sr = sr.
2 33(Ir⋅⋅)( sr) ( I ⋅⋅ j)( sr) So, = r55 jj(+ 1) r
Is. 3(Ir⋅⋅)( sr) Substituting the values in equation-21.3 of and , we get the interaction r3 r5 energy for spin 2 µµgg Is. 3(Ir⋅⋅)( sr) E =0 ISB −+ Spin 4π rr35 2 2 2 µµgg 3(I⋅⋅ j)( sr) ls.1+ s = 0 ISB− (.)Ij …………………..(21.4) 4π jj(++ 1) r53 jj ( 1) r 2 µµgg 2 (Ij⋅ ) 3(sr⋅ ) ls. + s2 = 0 ISB − π + 53 4 jj ( 1) r r
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Now we will calculate the total interaction energy due to electron orbital (equation 21.1) and spin (equation 21.4)
EHF= EE l + Spin 2 µµl.1 j 2g µ 2 (Ij⋅ ) 3(sr⋅ ) ls.+ s2 =+−002.gµ 2 Ij IB ππIB ++3( ) 53 4jj ( 1) r 4jj ( 1) r r
Here we have substitutes gS = 2 for the electron.
So
2 µ 2g µ 2 (Ij⋅ ) l.. j 3(sr⋅ ) ls+ s2 E =0 IB +− HF π + 35 3 4 jj ( 1) r r r 2 222 µ 2g µ (Ij⋅ ) l.. s++ l 3(sr⋅ ) ls s =0 IB +− π + 3 53 4 jj ( 1) r r r 2 µ 2g µ 2 (Ij⋅ ) ls223(sr⋅ ) =0 IB +− π + 3 53 4 jj ( 1) r r r
1 2 11 3 Substituting sr. = rand s= ss( += 1) + 1 =, we get 2 22 4
2 µ 2g µ 2 (Ij⋅ ) ll(+ 1)3(r) 3 E =0 IB +− HF π + 3 53 4 jj ( 1) r 4 r 4 r µ 2g µ 2 (Ij⋅ ) ll(+ 1) = 0 IB …………………….(21.5) 3 4π jj (+ 1) r µ 2g µ 2 ll(+ 1) 1 = 0 IB ⋅ 3 (Ij) 4π jj (+ 1) r
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So the Hamiltonian including the hyperfine interaction for one electron system is
HH=++0 HSpin− Orbit H Hyperfine
Here the hyperfine interaction is coupling the total angular momentum of the electron j and the nuclear angular momentum I. So we need the new angular momentum F which will be the good quantum number for the total Hamiltonian.
So we define, F= jI + and the eigenfunction is FmF which will be the coupled state arising from the
uncoupled states of jmjI Im The interaction energy EHF= A HF′ Ij ⋅
AHF′ → constant, characteristic of the level jl and .
Note that the value of AHF = 0 for l = 0, i.e. for the s-states.
2 However, experimentally splitting is observed for the S 1 state of hydrogen. This can not be 2 explained by this classical explanation.
However, starting from the Dirac equation, if one evaluates the Hamiltonian for the hyperfine interaction including the vector potential (Reference: Atoms and Molecules by M. Weissbluth), it becomes
µ 2g µ 2 ( +1) 18 π = 0 IB ⋅+δ ⋅ Hh 3 I j( r)( Is) 4π jj( + 13) r
( +1) =A I⋅+ j AIs ⋅= A′ I ⋅+ j AIs ⋅……………………(21.6) HF jj( +1) F HF F
The first term is the dipole-dipole interaction with corresponding to classical expression as we derived earlier. The last term is known as Fermi Contact Interaction term, it has no classical analog and contributes only for s-states. Since ψ (0) at r = 0 for non-s states is zero, Fermi contact term goes to zero for non-s states.
2 1 δψ(r) =(0) = 3 π a0 Where a0 is the Bohr radius.
Page-12 For = 0; i.e. s states first term zero; second term is important. For penetrating orbit contact term is important. FIs= + F2=++ I 22 s2 Is ⋅ FIs222−− ⇒⋅=Is 2
AF EHF= FmAIsFm F F. F = FF( +−111) II( +−) ss( +) 2
2 11 For hydrogen, S1 state, Is= , = , F = 0,1 2 22
AF 11 11 EFHF (==+−+−+ 1) 1( 1 1) 1 1 2 22 22 So AA3 =FF2 −= 2 24
AF 11 11 EFHF (== 0) 0( 0 +−+−+ 1) 1 1 2 22 22
A 33 =F 0 −=−A 224 F
The hyperfine splitting = ∆=EHF EF HF ( =− 1) EFHF ( == 0) AF
-1 The calculated values of AF = 0.047 cm .
F =1 n =1 AF 4 2 −1 S 1 0.047 cm 2 3A F 4 F = 0
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For non s states, We have
( +1) E= A Ij⋅ HF HF jj( +1) F222=++⋅ I j2 Ij FI222−− j ⇒⋅=Ij 2 + AHF ( 1) EHF = FF( +−111) II( +−) jj( +) 21jj( + )
2 1 3 For hydrogen, P3 state, lI= 1, = , j = , F = 2,1 2 22
+ AHF 11( 1) 11 33 EFHF (== 2) 2( 2+−+−+ 1) 1 1 233+1 22 22 So, 22( ) A 8 18 2 =F 6 −=A 2 15 4 5 HF
+ AHF 11( 1) 11 33 EFHF (== 1) 1( 1+−+−+ 1) 1 1 233+1 22 22 And 22( ) A 8 18 2 =F 2 −=−A 2 15 4 3 HF
16 The hyperfine splitting = ∆=E EF( =− 2) EF ( == 1) A HF HF HF 15 HF
F = 2 2
2 AHF P3 5 2 2 A 3 HF F =1
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2 11 For hydrogen, P1 state, lI= 1, = , j = , F = 0,1 2 22
+ AHF 11( 1) 11 11 EFHF (== 1) 1( 1+−+−+ 1) 1 1 2 11+1 22 22 So, 22( ) A 8 32 =F 2 −=A 23 2 3 HF
And
+ AHF 11( 1) 11 11 EFHF (== 0) 0( 0+−+−+ 1) 1 1 2 11+1 22 22 22( ) A 83 =F 02 −=−A 23 2 HF
8 The hyperfine splitting = ∆=E EF( =− 1) EF ( == 0) A HF HF HF 3 HF
F =1 2
2 AHF P1 3 2 2AHF
= F 0
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For multielectron atom The interaction energy is E= AI' ⋅ J HF
A' → hyperfine constant, characteristic of the level J and L F222=++⋅ I J2 IJ FIJ22−− 2 ⇒⋅=IJ 2 A' EHF ( F) = FF( +−11) II( +−) JJ( + 1) 2
EFHF ( +−1) EFHF ( ) = hyperfine splitting or hyperfine structure.
Hyperfine splitting is very small – measurements can be made to a high degree of precision.
The general conference of weights and measures (1964) defined “atomic second” from the transition between the hyperfine energy levels Fm=4,F = 0 and Fm=3,F = 0 of the 6 133 S 1 , Ground state of Cs atom 55 Cs . 2
These two sublevels correspond to parallel and anti parallel orientations of the spins S = 1/2 of the valence electron and I = 7/2 of the nucleus of the Cs atom
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Various Corrections:
Since the hyperfine splitting is very small, a lot of small corrections are needed.
(1) Polarization of the inner shells: For atoms with many electrons, the resultant of the electron spins in the completed inner subshells cannot be regarded as zero, statistically each spin has a slight tendency to align parallel to the spins of the valence electrons. In
evaluating the field B0 , it is necessary to take account of this magnetization of the inner shells 30%. (2) Relativistic effect: As a result of high electrostatic charge of the nucleus with high atomic number z , the velocity of the electrons is high in the neighborhood of the nucleus and corrections are necessary. These corrections can modify the result with far heavy atoms for levels with small J , by a factor of the order of two. (3) Volume effects: With increasing t , the approximation of a point nucleus cannot be preserved.
Electric Quadrupole Effects
The distribution of the charge qN within the nucleus is not spherically symmetric.
In classical theory, if the origin is taken as the center of gravity of the electric charges. Within the nucleus, the corresponding electric dipole moment is zero, there then remains the problem as to relative positions of the center of gravity of the electric charges and of the center of gravity of the masses within the nucleus.
In quantum theory, symmetry rules result in zero dipole moment for the nucleus. The first term in the multipole moment expansion corresponds to the interaction of electric quadrupole moment with electric field gradient create by the electrons in the region of the nucleus.
Let’s assume that the nucleus has a cylindrical charge distribution around its own Oz axis, I is also Oz axis. The electron cloud has cylindrical symmetry around Oz axis (direction of J ). The electric field gradient,
2 ∂Ez ∂ V φzz =−= ∂ z ∂ z2
Q → quadrupole moment of the nucleus
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The additional energy ∆EQ resulting from quadrupole coupling will be,
eQφzz 312 ∆=EQ cos θ − 42 2
Where θ is the angle between Oz( I ) and Oz( J ).
Define a quadrupole coupling constant =D = eQφzz
D 312 ∆=EQ cos θ − 42 2
From vector model,
FF( +−11) II( +−) JJ( + 1) cosθ = 2II( ++ 11) JJ( )
Or, using quantum mechanics,
3 CC+−12 II + 1 J J + 1 D ( ) ( ) ( ) ∆=E 2 Q 4II( 21−−) JJ( 2 1)
Where, C= FF( +−11) II( +−) JJ( + 1)
F = 5 2 D 4 3A′ J =1 2
I = 3 A′ 2 F = 3 D 2 5A′ 2 5D 4 F = 1 2
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Recap
In this lecture we came to know the origin of hyperfine structure such as isotope effect, hyperfine interaction etc.
The hyperfine structure is very small and can only be observed with a very high resolution.
We have understood the interaction of interaction of nuclear magnetic moment and the total electronic angular moment.
We now know that the ground state hyperfine splitting of hydrogen can not be described by classical concept.
The Fermi contact term is important to describe this splitting and quite accurately predict the experimental observation.
We have also understood the various corrections due to quadrupole effect, volume effect and relativistic effect.