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Fundamental Frequency Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 اﻟﺠﺎﻣﻌﺔ اﻟﺘﻜﻨﻮﻟﻮﺟﯿﺔ ﻗﺴﻢ اﻟﻌﻠﻮم اﻟﺘﻄﺒﯿﻘﯿﺔ اﺳﺘﺎذ اﻟﻤﺎدة :- اﻟﺪﻛﺘﻮر ﻋﺒﺪ اﻟﮭﺎدي ﻛﺎﻇﻢ ﻓﺮع اﻟﻔﯿﺰﯾﺎء اﻟﺘﻄﺒﯿﻘﯿﺔ اﻟﻤﺮﺣﻠﺔ اﻟﺮاﺑﻌﺔ Chapter three: Infrared Spectroscopy (IR) 42 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 Infrared Spectroscopy (IR) Fundamental region (2.5-15.4mm) IR Near IR (0.75-2.5mm) Far IR (15.4-microwave) Vibrational energy of diatomic molecule:- We are all familiar with the vertical oscillations of amass (m) connected to a stretched spring of a force constant k whose ether end is fixed. The simple harmonic motion with a fundamental frequency:- = ° If two masses in a diatomic molecule m1 and m2 we used the reduced mass \ = in quantum mechanically, the vibrational energy is given by = + υ =0,1,2,3 −−−− ° ° Where υ is the vibrational quantum No. The energy in Cm-1 = =( + ) ° =( + ) \ ° -1 Where ° the freq. in cm . These energy levels are equally spaced \ and the energy of lowest state = ∶ ° ° 43 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 An harmonic oscillator In addition to rotational motion , molecule has vibrational motion which is an harmonic. Hence a further correction to the centrifugal distortion to be made to account for the vibrational excitation of the molecule. The empirical equation of the potential energy of the diatomic molecule is given by ( ) V =D 1−e This equation called morse equation. V =morse potential. This potential represent the actual curve to a very good degree of approximation except when r=0 ® Vm gives a high finite value while the actual value is infinity. v At r® ∞ ® Vm®De De: dissociation energy. v At r=re ® Vm®0 V D re r 44 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 Let x=r-re x<< 1 very small =1− + − + ⋯ ……………….. 2! 3! ( ) ( ) Or =1− + − + ⋯ ……………….. ! ! We take the first two term ( ) =1− = 1 − = [1 − ( − )] = [− ( − )] = ( − ) ≅ 1 − \ ≅ ( − ) = (2 )( − ) = ( − ) \ 2 Where : 2a De=force constant for small displacement. \ = = in (Hz) : freq. for small displacement. Substitution Vm solving this eq. using pert buation theory we gets G(v) = ω v+ −ω x v+ ω x :an harmonicity correction and constant. ω >> ω x The effects of an harmonicity:- 1- The energy levels are not equally spaced. 2- Selection rules for allowed transition Dv=±1,±2,±3,…… v When Dv=±1® G(v1)= G(1)- G(0) 45 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 First fundamental freq.= ω -2ω x v When Dv=±2 ® G(v2)= G(2)- G(0) First over ton fundamental freq.=2 ω -6ω x ω D = 4ω x We can derive the equation of D 1 1 G(v) = ω v+ −ω x −−−−(1) 2 2 The maximum energy at =0 Then G(v) =De ( ) = ω -2ω x v+ =ω −2vω x − ω x 2vω x =ω − ω x ω ω \ = ω Sub in eq. (1) \ ( ) = + − + 2 − + − + = − 2 2 = − ( ) 46 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 = − = D = Dissociation energy Example(1):- The fundamental vibration spectrum of Co molecule at the wave number 2138 Cm-1 and the line which represent by the transition v=2® v=3 at wave number 2091 Cm-1 calculate 1. The harmonic constant ( ). 2. The unharmonic constant ( ). 3. The Dissociation energy. The solution:- 1 1 G(v) = ω v+ −ω x 2 2 G(v1)= G(1)- G(0) ( ) = 1+ − 1+ − 0+ − 0+ 3 9 1 1 ( )= − − + 2 4 2 4 ( )= −2 2138= −2 −−−−(1) By the same way G(v2)= G(3)- G(2) 47 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 7 49 5 25 G(v ) = ω − ω x − ω − ω x 2 4 2 2 = ω − 6ω x ∴ 2091= −6 −−−−(2) 2091 = − 6 − − − −(2) ﺑﺎﻟﻄﺮح 2138 = − 2 − − − −(1) -1 -47 cm = -4 \ω x = cm =11.75 cm Sub in eq.(1) -1 2138 cm = (2×11.75 ) ∴ =2161.5 ( ) 2161.5 4.44×10 = = = 4 4×11.75 9.94×10 1 1 =9.94×10 ×3×10 =29.82×10 1 D =29.82×10 ×6.63×10 J.S sec =147.7×10 Joul 147.7×10 ∴ D = =123.56×10 ev=12.356 ev 1.6×10 48 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 H.W.(1):- The absorption spectrum of HF molecule in the IR region , the first line at 3961 cm-1 and the second line at 7751 cm-1 calculate 1. The dissociation constant 2f. 2. The Dissociation energy D . Population of vibration energy levels According to Boltzman distribution = ∆ / = / DE For first three transitions (1) V=0 ® V=1 ( Fundamental frequency ) DV=+1 DE = ev=1 - ev=0 DE = 1+ − 1+ − − DE= (1−2 ) (2) V=0 ® V=2 DV=+2 DE=2 − 6 DE=2 (1 − 3 ) (3) V=0 ® V=3 DV=+3 DE=3 − 12 DE=3 (1 − 3 ) 49 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 Example(2):- HCl molecule is absorb the radiation at 2885.9 cm-1 by using Boltzman distribution calculate the relative Number for first vibrational level and the ground level at 25oC suppose the No. of molecule at ground level is one. The solution:- = ∆ / = / = . × . × × × / . × × =9.245 ×10 H.W.(2):- HCl molecule is absorb the IR radiation at fundamental vibrational frequency =2890 cm-1 calculate the force constant ? H.W(3):- Calculate the force constant for OH which can be the spectrum absorb at IR region . Infrared selection Rules The selection Rule of IR if the vibration quantum Number change 1. DV=±1 under harmonic approximation. 2. DV=±1, ±2, ±3 for anharmonic oscillator. 50 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 Over tones frequency:- On other transitions may be take place furthermore the fundamental frequency, at 2V1, 3V1, 4V1 1. 2V1, 3V1, 4V1 for transition began from v=0. 2. 2V2, 3V2, 4V2 for transition began from v=1. 3. 2V3, 3V3, 4V3 for transition began from v=2. st And the frequency of the 1 overtone = 2 e(1-3xe) The absorption due to these transition called (( Over tones frequencies)) The following figure represent the overtones frequencies for Hcl molecule in IR region. Zero point energy = − 1 − V =0 ®1 Fundamental frequencies nd First over ton 2 over ton 3rd over ton 4th over ton freq. freq. freq. freq. V =0 ®2 V =0 ®3 V =0 ®4 V =0 ®5 V (cm- 0 5000 10000 1) 51 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 Combination bands and differences bands:- The selection rules allowed by other transitions to produce combination bands and difference bands. The combination bands is arise by added two or more fundamentals frequencies or over tons frequencies such as:- + ,2 + , + + These frequencies become allowed but the intensity is very low. The difference bands is similar to combination bands but the transition appear in complex spectra. − ,2 − , + − Born-Oppenheimer principle ((The electrons movements is high speed compared with nuclei movements , then can be considered the nuclei is constant when studies the electronic properties of molecules and the energy of electrons is independent on nuclei energy and the total energy is equal to sum of vibration energy)) Plus Rotational energy plus electronic energy . = . + . + . Vibration – Rotation spectroscopy of diatomic molecules The vibration and rotation motion happen in any molecule at the same time, and for any vibration level consist from 52 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 many Rotational levels. The following figure explain the first two vibrational level v=0 and v=1 and the rotational levels. The vibration – rotation spectrum produce by transitions from rotational levels of vibration level like v=0 to another rotational levels of vibration level like v=1 and the selection rule for vibration- Rotation spectra DV=±1 , DJ =±1 The energy of vibration-Rotation for diatomic molecule E(J,V) = E(J) + E(V) = BJ(J+1) + + ° = ℎ ( +1)+ + + + ﻣﺜﻼ ﻟﺪﯾﻨﺎ ﺣﺎﻟﺔ اﻻﻧﺘﻘﺎل ﺑﯿﻦ ﻣﺴﺘﻮﯾﯿﻦ v=0 , v=1 D J=-1 D J=+1 /J 4 3 v=1 2 1 0 “J 4 3 v=0 2 1 0 vo P-branch R-branch 53 2B 4 2B Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 \ \ \\ \\ \ \\ = + − + + + ± + The vibrational-rotational absorption spectrum Consists of two groups of equally spaced lines(R+P)branches with a gap between them, the center of this gap is uo But experimentally it is found that the spacing is not constant , this is due to the vibration- rotation coupling. When take the distortion effect ( centrifugal distortion ) in account the = = ( +1)− ( +1) + + − + , Consider the vibrational rotational transition u=0®u=1 Assuming B and D the same for both u=0, and u=1 states and denoting upper state by single prime and lower state by double prime = \ \ +1 − \\ \\ +1 − \ \ +1 − \\ \\ +1 3 9 1 1 + − − − 2 4 2 4 \ \\ \ \\ \ \ \\ \\ = (1 − 2 ) + − + +1 − +1 − +1 - (1) Use of the selection rule DJ=+1 \ J\-J\\=1 gives \\ \\ = +2 +1 −4 +1 , =0,1,2, −−−−− DJ=-1 \ J\-J\\=-1 gives \ \ = − 2 +1 +4 +1 , =0,1,2, −−−−− Where = (1 − 2 ) which is the freq. of the u=0®u=1transition 54 Dr.Abdulhadi Kadhim. Spectroscopy Ch.3 Lines corresponding to DJ=-1 called P branch. Lines corresponding to DJ=+1 called R branch. \\ \ , = + − (m=J +1 and J +1) usually D is extremely small and in such cases , = + Example(3):- The normal modes of vibration of Co2 molecule are u1=1330 -1 -1 -1 -1 cm , u2=667 cm , u3=2349 cm and u4=2349 cm evaluate the zero point energy of Co2 molecule. Solution:- The Co2 molecule has 4 normal modes of vibration the symmetric bending mode u2 is double degenerate. The zero point energy E of Co = ℎ ∑ \ o 2 \ ∑ = (1330+667+667+2349) =5013×100 ∴ = 6.6×10 ×3×10 ×5013×100 ° =4.98×10 =0.311 55 Dr.Abdulhadi Kadhim.
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