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Unformatted text preview: How to evaluate thermodynamic properties in a diatomic? Bor nOppenh e im e r approximation nuclei remain still while e move very fast H separation of degrees of freed om H = H nuclear motion + H electronic H electronic n ( r ; R ) = n n ( r ; R )
for each nuclear separation R && the electronic Schrodinger eq. is solved R R r nuclear separation as parameter electronic coordinates U(R) A2 for each electronic state n N H nuclear n ( R ) = n n ( R ) A1 + A A+A R we want to separate the center of mass motion from the internal motions 2 particle system q
trans CM 1 particle system, with m=m1+m2
3 2 ( m1 + m2 ) kT 2 = V 2 h for internal (relative) motion as vibrations and rotations 2 particle system 1 particles system with m1 m2 reduced mass = m1+m2 we expand U ( R ) around the Req, considering the motion of a rigid rotor, and small vibrations of a Harmonic Oscillator H nuc .motion = H vibration + H rotation nuc .motion = vibration + rotation H rotation YJM ( , ) = J ( J + 1) hBe YJM ( , ) H vibrationSv ( R  Req ) = h v + 1 Sv ( R  Req ) 2 U ( R ) ; U ( Req ) dU =0 dR Req 2 dU 1 2U d + ( R  Req ) + 2 ( R  Req ) + L dR Req 2 dR Req ( ) 2U d 2 = curvature is a constant dR Req 1 Harmonic Oscillator Req by remembering that we can also separate electronic and nuclear config. we obtain the total molecular q(V,T) H = H tran slatio n + H electro nic + H nuclear + H vibra tio n + H rotation = tra n sl . ( R ) elec . ( r, , ) nucle ar vib. ( R  Req ) JMrot . ( , ) S Y = transla tion + electron ic + nuclea r + vib ration + rotatio n
and therefore ( qtransl .qelect .qnuclear qvib.qrot . ) Q(N,V ,T ) =
N! N ATTENTION!!!! since the # translation states is enormous H ALW AYS more states than particles! H we ca n use BOLZTMANN STATI STICS for each individual q, we can define a reference energy: translation we integrate (co unt ) all state s f rom 0 electronic 1 = De 1s t excited st ate energ y = 2 nuclear 1 1 = 0 vibration 1 v = 0 rot ati on 1 J = 0 = 0 1 = h 2
2 1 Do + h = De 2 0 D"o D"e
1 Req We already know q
trans CM ( m1 + m2 ) kT 2 = V 2 h De kT
2 3 2 q electronic ( V ,T ) = e 1 q nuclear ( V ,T ) = 1 + e ( 2 ) 
kT +L we only have to evaluate qvib and qrot q vib e ( V ,T ) = h
2  h n  n sum over states n 14 2 4 3 = e
n  ( n + 12 ) h q vib e ( V ,T ) = n  e =e
vib  h
2 e n (  h h ) n we recognize the series 1 xj = 1 x j q e 2 ( V ,T ) =  h 1 e  for T , h = 1 and we can use e 2 q ( V ,T ) = e e dn = h Vibrational partition 0 function q h kT vib and since 0 q ( V,T ) ; 2 h
vib 2  h  h n  h the vibrational contribution to the E is given by N Rln q = kT T
2 Evib = kNT 2 e 2 ln  h 1 e T  h  h2  h ln 1 e  2 lne Evib = NkT  T T ( ) 2 h = NkT T 2 Evib = + (1 e
+ h e  h  h ) ) T Evib h = N 2 + where we used  h = 1 1 e ( ) T kT 2 h e
 h Nh 2 ( Nh e h 1 we define the vibrational temperature as vib h = k v v + the contribution to the E becomes E = Nk v 2 e T 1 ( ) W hat is the population of each vibrational level? for a given state, the relative population is fn = e
 h n + 12 ( ) qvib = e  h n + 12 ( ) fn = e  h n + 12  12 ( ) ( ( e  h 2 =e  h n + 12 ( ) e + h 2 ( 1 e  h ) 1 e
 h  h 1 e ) ) fn = e (  h n e  h ( n +1) ) or the fraction of molecules NOT in the ground state v n v e 1 fn  f0 = 1  T 1  e T n = 0 we can also get Cv (heat capacity @ const. V) E v Cv = = Nk 1 V,N T T for T Evib =
Nh 2 Nh 2
+ 2 (e e v T v T 1 ) 2 (e Nh h 1
 h kT ) : Nh h = NkT Cv = Nk
 h for T 0 Evib = + Nh e 2 kT Cv = Nk 2 e T 0 ...
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 Spring '08
 Bowers
 Electron

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