Laser spectroscopy of diatomic metal-containing molecules

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Laser spectroscopy of diatomic metal-containing molecules Powered By Docstoc
					                                   Imperial lectures, April 2007
                                       Timothy C. Steimle,
                                              Lecture #3
         Summary from Lecture #2
a) “Effective” Hamiltonian for 2P state
  Fine structure
                                                                        NJ-S


  Mag.Hyf. terms
                                                                           “+” raising op.

  2S +   state drop the spin-orbit (A), -dbl (p&q). and a & d terms

b) Expressions for Hund’s case (a) basis set matrix elements



                                                                                        Eq. 1



                                                                                        Eq. 2
  Summary from Lecture #2
Near term goal: Parameters  spectrum & 

Example: Model energy levels and spectra for OH X2P

 Parameters (MHz) for X2P:               Energy levels X2P:
Rotation B=555660.97
Rotation D=57.175                              Proton mag. Hyf. splitting
Spin=orbit: A=-4168639.13
                                                  F=2
Spin=rotation: g=-3574.88                                   Parity:+
-doubling: p=7053.09846                          F=1     Does not look
                                                          like simple model
-doubling: q=-4168639.13        Cm-1                       J=3/2
                                                          prediction. i.e.
Mag. hyf: a=86.1116                         -doubling    E=BJ(J+1)

Mag. hyf: bF=-73.2537
                                                F=2        Parity:-
Mag. hyf: c=130.641
                                                F=1
Mag. hyf: d=56.6838

                                                          Next
  Summary from Lecture #2
What we have calculated:
                        Parity                             F DE(MHz)
                         +                                2       70.15
                                                          1
                            -doubling                                Notice large diff.
       J=3/2                                                          between -doublets. Due
                         7799.101 MHz
                                                                      to the “d” hyperfine
       Rotational                                                     parameter
       spacing; 3.3B       _                             2      11.99
                                                          1
       J=1/2
                            _
                                                          1     90.53
                                                          0
                             4733.123 MHz
        Spin-orb. A
                            +                             1     14.69
                                                          0


       J=3/2                +                             2       55.16
                                                          1
                             1667.145 MHz
                            _                             2
                                                          1      55.06
                                     This transition is used to detect OH in the interstellar
                                     media. We will look at it in more detail below.
Summary from Lecture #2
How good is the calculation ? Compare the transition freq. for lowest -doublet
        A B C   D
                    2
                    1



J=3/2

                    2
                    1

                    F
                        Outline for Lecture 3
 A) Interpretation of spin-orbit & hyperfine parameters of OH.
    i) Spin-orbit of OH from O information
    ii) hyperfine parameters of OH
 B) Energies and wave functions for 2S+ states

 C) Prediction Relative Transition Intensities
    i) OH -dbl. transitions
    ii) 174YbF N=0 N=1
 A) Interpretation of the parameter in X2P OH i) spin-orbit,A
“Interpretation” means obtaining of a molecular orbital level understanding of
the magnitude and sign of the parameters used to model the energy levels. I.E
is our m.o. correlation diagram and our knowledge of atomic properties
consistent with the molecular parameters B,p, q and A?
                     (2p0-1s)s

                                           This requires that the “ macroscopic”
                     2pp+1 2pp-1
   2p+1    2p0 2p-                         operators used in Heff be written in
                 1
                                   H(1s)   “microscopic” form. For spin-orbit
                     (2p0+1s)s
                                           interaction the two operators are: .
                                           The Macroscopic “fitting” spin-orbit Heff=ALzSz
          O 2s         2ss
                       OH

 The way to obtain this molecular orbital level understanding is to:
  a) select any non-zero matrix element of Heff .
  b) write down the el as anti-symmeterized product of one-electron
      function which is an eigen-function of Sz & Lz. (i.e a linear
     combination of Slater determinants)
 c) Evaluate the selected matrix elements using the one-elec. functions.
 A) Interpretation of the parameter in X2P OH i) spin-orbit,A
  a) A matrix element of the Macroscopic effective operator used to model the
  data for J=3/2 =3/2 X2P of OH.

        Focus on this matrix element
        for a comparison between
        Micro and Macro description.




                                               = -69.52 cm-1

The same matrix element using the Microscopic operator form: (Note only diagonal
terms of operator required):



                                            = ??? cm-1

b) re-write the basis function in terms of Slater determinant of
         molecular orbitals: p3 Config. with electrons arranged to have =1, S=1/2
                                                     Short-hand notation used by
                                                     “L-B&F” and “B&C”

                                                              This wave function is an
                                                              eigenfunction of Sz and Lz
                                                              with eigenvalues ½ and 1.
A) Interpretation of the parameter in X2P OH i) spin-orbit,A

  Remember that              Note that lzi & szi are one-electron
                             operators. Use rules for evaluation of matrix
                             elements of one elec. op. for Slater
                             determinaental representations of el. L-B&F


                                            Where because the 3 electrons are
                                            all in the same p type orbital we
                                            have ai(r)  ap. .



                                       Note that p orbitals are non-
                                       bonding and look like O-atom
                                       p-orbitals.
                                   Therefore
                                           Atomic spin-orbit par for O

                                 ap                              =153cm-1
                                             = -153/2 =-75.8 cm-1
 Our m.o. picture does a great job of predicting spin-orbit (-75.7
 vs. -69.52 cm-1)!
A) Interpretation of the parameter in X2P OH ii) hyperfine parameters


Hyperfine parameters (MHz) for X2P: a=86.1116 bF=-73.2537
                                      c=130.641     d=56.6838




                                                         Slater Determinant




                                    Need a value for <r-3> to proceed. Should
                                   be  bond length because unpaired elec. is
                                   on O whereas the spin is on H.  let r 
                                   1.0Å <r-3> 1.010-30m-3
              a 97 MHz     Simple model works well for a !
A) Interpretation of the parameter in X2P OH ii) hyperfine parameters


    The microscopic form of parameter:




                                             Spin-density at nucleus. Only 
                                             for electrons in orbitals that
                                             have some “s-orbital” character.




          The p-orbitals have no “s-orbital” character.  to
         the first approx. it is predicted that bF =0

                 Exp. Value is bF=-73.2537 MHz

     Need to modify the simple m.o. picture for X2P of OH.
A) Interpretation of the parameter in X2P OH ii) hyperfine parameters
           Current Model for bonding in OH

                          (2p0-1s)s       3s*
                                                                                                       Proton Spin
                          2pp+1   2pp-1   1p
    2p+1    2p0 2p-1

                           (2p0+1s)s      3s      H(1s)




                                       2s                                   This interaction is
           O 2s                                                             different from interaction
                            2ss
                                                                            of up-down pair. i.e. Spin-
                                                                            polarization. This is treated
                            OH
                                                                            by mixing in a bit of a
   Config: .. (2s)2 ((3)s)2(1p)3                                            higher energy configuration.
                                           X2Pi

  Config2: .. (2s)2 ((3s)1 (1p)3 ((3s*)1        2P        2P       & 4P r
                                                     i,        r

   It only takes a very small admixture of Config2 because aF(H-atom) =1470 MHz)

  Need the wavefunctions for the two 2P states from 3 open shell configuration:
A) Interpretation of the parameter in X2P OH ii) hyperfine parameters
 Nesbit’s table:




                                                               Eq. 1


                                                               Eq. 2

                                                         Eq. 3

                                                       Eq. 4


 Eq. 4 will produce a small negative bF value.   bF  -100 MHz
B) Energy levels for    174YbF   X2S+
                                              174Yb   (I=0)   19F   (I=1/2)
 YbF X2S+        =0, S=1/2, S= 1/2

 Basis set
  Y(case(abJ))=|>|SS>|(JI)FMF>                             FJ        I
        S   S      J    I                                                   R
 1=|0;1/2,1/2;F-1/2,1/2,F,0.5>                               S
 2=|0;1/2,1/2;F+1/2,1/2,F,0.5>                    S
 3=|0;1/2,-1/2;F-1/2,1/2,F,-0.5>
 4=|0;1/2,-1/2;F+1/2,1/2,F,-0.5>           4x4 Matrices

Although , we will use a Hund’s “case a” basis set, S states have an energy level
pattern that is close to that of a Hund’s “case b” because the electron spin is not
tied to the internuclear axis by spin-orbit interaction. In a Hund’s case (b) Hrot
and Hspin-rot have only diagonal matrix elements:

                                     HSpin.R. small splitting between J= N+1/2
                                     and J=N-1/2 levels which increases approx.
                                     linearly with N.


Therefore we expect that the predicted energy level pattern will look like:
B) Energy levels for   174YbF   X2S+ (cont)
     3/2                   2
 1                         1
  1/2                      1       We want to model these energy levels exactly
                           0
                                   using first a Hund’s case a basis set. We neen
                                   matrices for F=0,1,2. There are only 2 levels
                                   with F=0 but 4 levels for F=1 and 2 (same for
                                   higher J values.
                           1
 0 1/2
                           0
N J                        F

We will only calc. these energy levels.  need matrix rep. for F=0,1 &2
 F=0.0                                         F=1
                                              1=|0;1/2,1/2; 1/2,1/2,1,0.5>
2=|0;1/2,1/2; 1/2,1/2,0, 1/2>
                                              2=|0;1/2,1/2;3/2,1/2,1,0.5
4=|0;1/2,-1/2; 1/2,1/2,0,-1/2>               3=|0;1/2,-1/2; 1/2,1/2,1,-0.5>
                                              >
                                              4=|0;1/2,-1/2; 3/2,1/2,1,-0.5>

  F=2                                  The spectroscopic parameters (this group)
 1=|0;1/2,1/2; 3/2,1/2,2,0.5>                     B =7233.8271     MHz
 2=|0;1/2,1/2;5/2,1/2,2,0.5>                      D=0.00715        MHz
 3=|0;1/2,-1/2; 3/2,1/2,2,-0.5>                   g=13.41679       MHz
 4=|0;1/2,-1/2; 5/2,1/2,2,-0.5>                   bF= 170.26374    MHz
                                                   c=85.4028        MHz
 B) Energy levels for        174YbF   X2S+ (cont)
      Need Eq. 1 &2 first slide +




  The results( Fortran program provided)
  F=0
ROT AND CENT COR. HAM                   THE SPIN ROTATION HAMILTONIAN
0.00000 0.00000   0.00000 0.00000       0.00000 0.00000       0.00000 0.00000
0.00000 0.24129   0.00000 -0.24129      0.00000 0.22377D-03 0.00000 -0.22377D-03
0.00000 0.00000   0.00000 0.00000       0.00000 0.00000       0.00000 0.00000
0.00000-0.24129   0.00000 0.24129       0.00000 -0.22377D-03 0.00000 0.22377D-03
B) Energy levels for         174YbF     X2S+ (cont)
THE HYPERFINE HAMILTONIAN                         THE COMPLETE HAM
0.00000D+00 0.00000D+00 0.00000D+00 0.00000D+00   0.00000D+00 0.00000D+00 0.00000D+00 0.00000D+00
0.00000D+00-0.18946D-02 0.00000D+00-0.23649D-02   0.00000D+00 0.23962D+00 0.00000D+00-0.24388D+00
0.00000D+00 0.00000D+00 0.00000D+00 0.00000D+00   0.00000D+00 0.00000D+00 0.00000D+00 0.00000D+00
0.00000D+00-0.23649D-02 0.00000D+00-0.18946D-02   0.00000D+00-0.24388D+00 0.00000D+00 0.23962D+00


                              THE EIGENVALUES AND           EIGEN VECTORS
                                 0.483506 cm-1
                                                                                      “-” parity
  F=0                          0.0000 0.7071 0.0000         -0.7071
                                 0.000000
                               1.0000 0.0000 0.0000          0.0000
                                 0.000000
                               0.0000 0.0000 1.0000          0.0000
                                -0.004260 cm-1                                    “+” parity
                               0.0000 0.7071 0.0000          0.7071




         THE EIGENVALUES AND EIGEN VECTORS                  THE EIGENVALUES AND EIGEN VECTORS
           1.449375                                            1.449375
        -0.0003 0.7071    -0.0003 -0.7071                   -0.0003 0.7071    -0.0003 -0.7071
 F=1       0.484722
         0.5968 -0.3792   -0.5968 -0.3792         F=2
                                                               0.484722
                                                             0.5968 -0.3792   -0.5968 -0.3792
           0.478313                                            0.478313
         0.3792 0.5968    -0.3792 0.5968                     0.3792 0.5968    -0.3792 0.5968
           0.001420                                            0.001420
         0.7071 0.0003     0.7071 -0.0003                    0.7071 0.0003    0.7071 -0.0003
B) Energy levels for                    174YbF        X2S+ (cont)
Many matrix elements to calculate. Best to write program. Here is sample of code provided
               SUBROUTINE HSROT12(FL,FSI,FOM,FJ,HW,PAR,S)
                                                                           Rotation terms; Eq. 1
               IMPLICIT REAL*8(A-H,O-Z)
               DIMENSION FL(4),FOM(4),FSI(4),FJ(4),HW(4,4),PAR(6)
               DIMENSION HCD(4,4)
               DO 500 I=1,4
               DO 500 J=I,4
               HW(I,J)=0.0
               IF(FJ(I).NE.FJ(J))GOTO 499
               IF(FJ(I).LT.0.0)GOTO 499
               IF(FJ(J).LT.0.0)GOTO 499
               PAR4=0.0
               DO 400 K=1,2
               IF(K.EQ.1)Q=-1.0
               IF(K.EQ.2)Q=1.0
               X=FJ(J)-FOM(I)+S-FSI(I)+0.1
               IP=IDINT(X)
               PHASE=(-1.0D0)**IP
               PAR1=THREEJ(FJ(I),1.0D0,FJ(I),-1.0*FOM(I),Q,FOM(J))
               PAR2=THREEJ(S,1.0D0,S,-1.0*FSI(I),Q,FSI(J))
               PAR3=(FJ(I)*(FJ(I)+1.0)*(2.0*FJ(I)+1.0)*S*(S+1.0)*(2.0*S+1.0))
               PAR4=PAR4+PAR1*PAR2*DSQRT(PAR3)*PHASE
400            CONTINUE
               PAR5=0.0
               IF(FSI(I).NE.FSI(J))GOTO 480
               IF(FOM(I).NE.FOM(J))GO TO 480
               PAR5=FJ(I)*(FJ(I)+1.0)+S*(S+1.0)
 480            CONTINUE
C      NOW CALC THE N MAT
               HW(I,J)=PAR5-2.0*PAR4
499             CONTINUE
               HW(J,I)=HW(I,J)
 500            CONTINUE
C      NOW SQUARE N MAT. TO GET CENT. COR
               DO 610 I=1,4
               DO 610 J=1,4
               HCD(I,J)=0.0
               DO 610 K=1,4
               HCD(I,J)=HCD(I,J)+HW(I,K)*HW(K,J)
 610           CONTINUE                                             DO 630 J=1,4
C      NOW CALC. HROT AND HCD                                630     HCD(I,J)=HCD(I,J)*PAR(2)
               DO 620 I=1,4                                          DO 640 I=1,4
               DO 620 J=1,4                                         DO 640 J=1,4
620             HW(I,J)=HW(I,J)*PAR(1)                       640     HW(I,J)=-HCD(I,J)+HW(I,J)
               DO 630 I=1,4                                          RETURN
                                                                     END
C) Prediction of relative transition probability (OH & YbF pure rotation)
The strongest interaction between the light and a molecule is via the interaction
between the electric dipole moment op. and the electric field of the radiation:
            Helec. dipoole= -mE(t) = -mE(t) cos

               transition moment.     When |> = |’’> then “transition moment” =
                                       the permanent electric dipole moment of the
               Location of charges
                                       molecule. A typical value values are: OH ~
                                       1.5 D; CaF ~ 3.5 D; CCCCCH ~ 10 D


                                     When |> and |’’> are different
    Internuclear Sep.
                                     vibrational level of the same electronic state
                                     then “transition moment” is proportional to
                                     2nd term in expansion. This term is typically
                                     110-4 or less smaller than the permanent
                                     electric dipole moment. (i.e much smaller

When |> and |’’> are different electronic   states then the “transition
moment”


        Franck-Condon Factor                             Typical values 0.1D- 10D
C) Prediction of relative transition probability (OH & YbF pure rotation)
                         Note: not necessarily a square matrix
   “b”                                                                  Eigenvec for a

                                              An “a” “b” matrix of
                                              transition matrix
                             Eigenvec for b   elements calculated
                                              using the same basis
                                              functions used to
   “a”                                        produce the eigenvecors
                                              for “a” and “b”

Need expressions for matrix elements of dipole moment operator. First consider
no nuclear spin; Hund’s case (a):




If the spectra are recorded in the absence of a static or electric field then all of
the levels with different MJ values will be degenerate and the transition moment
will be the sum over all possible MJ values

                                         We need expression for YbF, OH and CaF; one
                                         nuclear spin
C) Prediction of relative transition probability (OH & YbF pure rotation)

  Hund’s case (abJ)




If the spectra are recorded in the absence of a static or electric field then all of
the levels with different MF values will be degenerate and the transition moment
will be the sum over all possible MF values

For the two nuclear spin case :
C) Prediction of relative transition probability (OH & YbF pure rotation)
  Need to write program: Hund’s case (abJ)
                 SUBROUTINE OHINT12(nfl,nfu,basisu,basisl,vut,vlt,enut,enlt,              Transfer eigen vectors (“vut” &“vlt”), energies
c
    1            FLT,ELo,FUT,EUP,BF,tf,fin,ic,TK)
                 calc transisiton int.
                                                                                          (“enut” & “enlt”) and basis (“basisu” & “Basisl”
c   last modified March 27 2007 For Graz class –OH
                 implicit real*8(a-h,o-z)
                 DIMENSION FLL(8),FSIL(8),FJL(8),FOML(8),TF(200)
                 DIMENSION FLU(8),FSIU(8),FJU(8),FOMU(8),fin(200)
                 DIMENSION VU(8),VL(8),vut(10,8,8),vlt(10,8,8)
                 DIMENSION TOM(8,8),basisl(10,6,8),basisu(10,6,8)
                 DIMENSION TEMP(8),Enut(10,8),Enlt(10,8),BF(200)
                 Dimension EUP(200),ELo(200),FUT(200),FLT(200),IDU(200)
                 dimension IDl(200)
                 OPEN(3,FILE='int12.PUT')
c                speed of light =sol
                 sol =29979.2458D0
                 ic=1
                 elowest=0.0d0                                        Find lowest energy for evaluation of Boltz. Factor
                 do 601 i=1,nfl
                 do 601 j=1,8
601              if(enlt(i,j).lt.elowest)elowest=enlt(i,j)
                 WRITE(3,250)
250              FORMAT(1x,'low F',3x,'En Low',5x,'up. F',5x,'En up',
                 1                    5x,'tran. freq', 5x,'Boltz. Fac.',5x,'intensity')
                 do 1000 ik=1,nfl
                 do 1 kk=1,8
                 s=basisl(ik,1,kk)
                 fll(kk)=basisl(ik,2,kk)
                 fsiL(kk)=basisL(ik,3,kk)
                 fomL(kk)=basisL(ik,4,kk)
                                                   Put basis functions into individual arrays
                 fjL(kk)=basisL(ik,5,kk)
                 fL=basisl(ik,6,kk)
1                continue
c                loop over upper number of f values                                                   Loop over upper and Lower
                 do 999 ikK2=1,nfu
                 do 2 kk=1,8
C                PAUSE
                 s=basisu(ikK2,1,kk)
                 flu(kk)=basisu(ikK2,2,kk)
                 fsiu(kk)=basisu(ikK2,3,kk)
                 fomu(kk)=basisu(ikK2,4,kk)        Put basis functions into individual arrays
                 fju(kk)=basisu(ikK2,5,kk)
                 fu=basisu(ikK2,6,kk)
2                continue
      do 225 ii=1,8
      do 225 jj=1,8
      TOM(II,JJ)=0.0D0
C     SKIP FALSE ROOTS
      IF(FJU(II).LT.0.0)GOTO 225
      IF(FJL(JJ).LT.0.0)GOTO 225
      IF(fsiu(ii).ne.fsil(jj))GOTO 225
      fphase = fju(ii) + 0.5d0 + fL + 1.01
      i3 = dint(fphase)
      phasf = (-1.00d0)**i3
      aa=sixJ(fju(ii),fu,0.5d0,fl,fjl(jj),1.0d0)
                                                               Generate transition moment Matrix
      ultj2=dsqrt((2.0*fu+1.0)*(2.0*fL+1.0))
      aaa=fju(ii)-fomu(ii)
                                                          `
      i2=dint(aaa)
      phas=(-1.00d0)**I2
      ultj=dsqrt((2.0*fju(ii)+1.0)*(2.0*fjL(jj)+1.0))
      fiq=0.0d0
      bb=threej(fju(ii),1.0d0,fjl(jj),
      -1.0*fomu(ii),fiq,foml(jj))
224   continue
      a4 = phasf*ultj2*aa*phas*ultj*bb
      tom(ii,jj)= a4
225   Continue

C   NOW LOOP OVE EIGENVECTORS
    DO 500 KK=1,8                     BEGIN        MATRIX MULTIPLICATION
C PULL OUT LOWER eIGEN VECTOR
   eLOO=EnLt(ik,KK)                                                                                Eigenvec for a
    DO 237 ID=1,8
237 Vl(ID)=vlt(ik,ID,KK)                                                      An “a” “b” matrix
    DO 499 KK2=1,8                                                            of transition
C PULL OUT UPPER IGEN VECTOR                                   Eigenvec for b matrix elements
   eupP=Enut(ikK2,KK2)
   DO 239 ID2=1,8
239 VU(ID2)=vut(ikK2,ID2,KK2)
    do 240 kkK=1,8
   temp(kkK)=0.0d0
   do 240 k=1,8
   temp(kkK)=temp(kkK)+tom(kkK,k)*vl(k)
                                                        Multiply TM by lower eigenvec,
240 continue
C) Prediction of relative transition probability (OH & YbF pure rotation)
     tint=0.0d0
     do 242 k=1,8
     tint = tint + vu(k)*temp(k)
                                              Multiply by upper    eigen vec
242 continue
     tint2=tint*tint
C     store if greater than 0        Sq. trans.moment
     eUP(ic)=EUPP
    eLO(ic)=ELOO
C    BF=BOLTZMANN FAC
    el=ELO(IC)-elowest
    el2=(ELO(IC)-elowest)/sol
                                     Calc. Boltzmann factor
    kt=TK*0.695d0
    BF(ic)=DEXP(-(el2/kt))
C    DF=DEGENERCY FACTOR
    DF=2.0*FL+1.0                      Degeneracy factor (the sum over the 3j involving MF)
    fint=(tint2)*BF(ic)
    tftt=EUP(IC)-ELO(IC)
C get only positive trans. freq.           Calc. intensity.
   if(tftt.le.0.0d0) go to 247
C only keep non-zero intensites            Calc. Transition freq.
   if(fint.lt.0.01d0) go to 247
    fin(ic)=fint
    tf(IC)=tftt
    fut(ic)=fu
    flt(ic)=fl
    idu(ic)=KK2
    idl(ic)=KK
                                                             Print out results
    write(3,243)fu,elo(ic)/sol,fu,eup(ic)/sol,tf(IC),bf(ic),fint
243 format(f6.2,F12.3,f6.2,2F12.3,2f14.4)
    ic=ic+1
247 continue
499 CONTINUE
500 CONTINUE
999 continue
1000 continue
   Lowest -doublet

-38.120


                              F”          F’             Freq.   Boltz. Int.
-38.140                2
                              1.0 -38.201 1.0 -38.145   1665.411 1.000 1.440
          +            1
                              1.0 -38.201 2.0 -38.143   1720.481 1.000 0.288
-38.160
                              2.0 -38.199 1.0 -38.145   1612.291 1.000 0.288

                              2.0 -38.199 2.0 -38.143   1667.361 1.000 2.592
-38.180
                               Output from program:

                       2
-38.200
          -            1
              J= 3/2   F
-38.220




                           1600.0       1650.0        1700.0
                                     Microwave Freq (MHz)
                      1    F”         F’            Freq.   Boltz. Int.
88.250
         -            0   0.0 88.093 1.0 88.252   4765.581 0.533 0.178

                          1.0 88.093 0.0 88.249   4660.192 0.533 0.178

88.200                    1.0 88.093 1.0 88.252   4750.676 0.533 0.356




88.150




88.100                1
         +            0
             J= 1/2   F
                          4650.0        4675.0        4750.0
                                     Microwave Freq (MHz)
C) Prediction of relative transition probability (OH & YbF pure rotation)
 3/2                       1     G=1
                           2
                           0
 1/2
       N=1                 1    G=0
                                                                        174YbF N=0- N=1
                                            2.00




                           1
 1/2                                        1.50
                           0
  J    N=0                 F




                                      Int
                                            1.00




                                            0.50




                                            0.00
                                              14450.0 14455.0 14460.0 14465.0 14470.0 14475.0 14480.0 14485.0 14490.0
                                                                              FREQ
Thank You !

				
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