DOC PREVIEW
MIT 5 62 - Nuclear Spin Statistics

This preview shows page 1-2-3 out of 10 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 10 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 10 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 10 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 10 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.5.62 Lecture #13: Nuclear Spin Statistics: SymmetryNumber, σ. Low Temperature Limit for RotationalPartition Function Readings: Hill, pp. 153-156, 153-159 [466-472] Maczek, pp. 54-57 ORIGIN OF σ — NUCLEAR SPIN STATISTICS Quantum mechanics requires that the total wavefunction be either symmetric (does notchange sign) or antisymmetric (changes sign) with respect to exchange of any twoidentical nuclei. ψTOT must be symmetric if nuclei have integer spins (bosons) orantisymmetric if nuclei have 1/2 integer spins (fermions). ψTOT (x) ⎯EXCHANGE→ if nuclei are BOSONS ⎯⎯⎯⎯ +ψTOT (−x) number integer spin I homonuclear a symmetric functiondiatomic molecule an EVEN function nuclear spinquantumψTOT (x) ⎯EXCHANGE→ if nuclei are FERMIONS ⎯⎯⎯⎯ −ψTOT(−x) 1/2 integer spin I an antisymmetric functionan ODD function FOR EXAMPLE: O2 The 16O nuclei have spin I = 0 ⇒ BOSONS Therefore ψTOT for O2 must be symmetric or EVEN ψTOT = ψTRANS ψVIB ψELEC ψROT ψNUCL.SPIN Need to investigate symmetry of each one of these wavefunctions for O2 upon exchangeof two 16O nuclei. [NOTE: we are permuting nuclei, not atoms. The Na atom is a Boson, but the Na nucleus in Na2 is a Fermion.] 5.62 Spring 2008 Lecture 13, Page 2 ψTRANS – depends only on CM coordinate of the molecule so this function is not affected by exchange — we will no longer consider it ψVIB – depends only on distance between the two nuclei so this coordinate is notaffected by exchange — no longer consider it.[Be careful. In a polyatomic molecule, internal rotation of a methyl groupdoes permute identical nuclei and it is a vibration.] ψELEC – ground electronic state of O2 is 3Σ− g . This term symbol tells us that the electronic wavefunction is antisymmetric or ODD with respect to exchangeof the two nuclei. [The classification operator is σv × i (both body-frame).] ∑−g and ∑+u are both odd, ∑+g and ∑−u are both even. ∏,∆ etc. consist of one manifold of even states (like ∑+ g ) and one nearly identical manifold of odd states (like ∑− g ). ψROT – rigid rotor rotational wavefunctions are spherical harmonics ψJ=0,2,4,6 … are symmetric with respect to exchangeEVEN ψJ=1,3,5,7 … are antisymmetric with respect to exchangeODD ψNUCLEAR – 16O nucleus has a nuclear spin I = 0 → a boson; there are (2I + 1)(I + 1) symmetric (EVEN) nuclear spin states: “ortho” (the stateswith larger nuclear spin degeneracy) (2I + 1) I antisymmetric (ODD) nuclear spin states: “para” For I = 0 1 EVEN nuclear spin state 0 ODD nuclear spin state ψTOT = ψELEC ×ψROT ×ψNUCL.SP EVEN = ODD ψJ=1,3 ODD EVEN EVEN ≠ ODD ψJ=0,2 EVEN EVEN revised 1/9/08 10:29 AM5.62 Spring 2008 Lecture 13, Page 3 Therefore, O2 molecules in the ground electronic state cannot exist in even rotational states!!! Because ODD * EVEN * EVEN does not result in an EVEN function as required by quantum mechanics! [When I = 1 / 2, a homonuclear molecule has Itot = 1 and 0 for Itot = 1 there are 3 ortho substates for Itot = 0 there is 1 para substate. When I = 1, Itot = 2,1,0. Itot = 2 and 0 are ortho: 6 substates Itot = 1 is para: 3 substates When I = 3 / 2, Itot = 3,2,1,0. Itot = 3,1 are ortho: 10 Itot = 2, 0 are para: 6.] Therefore, for O2 ∞ qrot = ∑ (2J + 1) e− J(J+1)θrot /T J=1,3,5, … high temp. limit integral approximation = 1 ∫∞ (2J + 1) e− J(J+1)θrot /TdJ = T = hcB 2J= 02θrot θrot k only 1/2 of the J’s in thisintegral contribute!!So σ = 2!! the odd J’s! Can use σ for high temp. limit only [sounds like a fraud. Can you prove that 12 is exactly the correct factor? It involves a simple change of variable for the integral.] Origin of σ: only 1/2 of thepossible rotational states in the integral approximation contribute to the partition functionbecause of symmetry restrictions imposed by quantum mechanics. — not all homonuclear molecules have I = 0; other molecules have other symmetryrestrictions that bring in the σ-factor (even when I ≠ 0). ORTHO-PARA HYDROGEN another consequence of nuclear spin statistics H2 — hydrogen nuclei (protons) have I = 1/2therefore, they are FERMIONS revised 1/9/08 10:29 AM 5.62 Spring 2008 Lecture 13, Page 4 so ψTOT must be ODD (antisymmetric) +ψELEC is EVEN (symmetric) because ground electronic state of H2 is 1 Σ g ψNUCL has (2I + 1) I = 1 antisymmetric spin states (para) (nuclear spin singlet) α(1)β(2) – α(2)β(1) (one state: called nuclear-spin “singlet”) (2I+1)(I+1) = 3 symmetric spin states (ortho) (nuclear spin “triplet”) α 1( ) ( )α 2⎫ ( )β 2⎪ β 1( ) ⎬(three states: called nuclear-spin “triplet” ) α 1( ) + α 2( ) ⎪⎭( )β 2( )β 1ψTOT = ψELEC * ψROT * ψNUCL.SP. ODD = EVEN ψJ=0,2,4… EVEN 1 ODD ψNUCL.SP ODD = EVEN ψJ=1,3,5… ODD 3 EVEN ψNUCL.SP For ψTOT to be ODD, ψJ=0,2,4… EVEN-J have to be paired with ODD ψN.S. and ψJ=1,3,5… ODD-J have to be paired with EVEN ψN.S. But there are 3 times more EVEN nuclear spinstates than ODD spin states. Therefore, there are 3 times more ODD rotational statesavailable because each ODD rotational state can have three possible nuclear spinfunctions. In other words, the "degeneracy" of the ODD rotational levels is effectivelythree times that of the EVEN levels. Therefore, this leads to a probability distribution asa function of J that looks like: revised 1/9/08 10:29 AM5.62 Spring 2008 Lecture 13, Page 5 H2 with J = 0,2,4 … is called PARA-hydrogen H2 with J = 1,3,5 … is called ORTHO-hydrogen We will not include nuclear spin statistics in our analysis of the rotational partitionfunctions even though we are obviously making an error in calculating the probability offinding a homonuclear diatomic molecule in a particular J rotational state and in somemacroscopic properties. We make no error in Keq because the nuclear spin orientationsare not altered (usually) in a chemical reaction and therefore nuclear spin degeneracyfactors cancel. Actually qrot for H2 may be shown to be exactly the same as the naïve prediction based onthe σ = 2 symmetry number and neglecting nuclear spin altogether, except at very low T.[How low is “very low”?] MOLECULAR ROTATIONAL PARTITION FUNCTION — LOW TEMP LIMIT CASE 2: εrot > kT or θrot > T Cannot


View Full Document

MIT 5 62 - Nuclear Spin Statistics

Download Nuclear Spin Statistics
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Nuclear Spin Statistics and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Nuclear Spin Statistics 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?