# 1 2 h0 x 1 m y x h1 x x h 2 x x2

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: − + mω 2 x 2 ⋅ ψ = Eψ 2 m dx 2 2 The wavefunc)ons penetrate to the classical forbidden zone Classical turning points Formulas for the wavefunc)ons ψ n ( x ) = Cn exp ( − y 2 2 ) H n ( y ) Cn = 1 ⎛α⎞ ⎝ ⎠ 2 n n! ⎜ π ⎟ 1/ 2 H0 ( x) = 1 mω α= y=α ⋅x H1 ( x ) = x H 2 ( x) = x2 − 1 (( dn H n ( y ) = ( −1) exp y 2 2 exp − y 2 2 dy n n ( ) )) Note that the number of nodes increases with n. The lowest energy state has no nodes Energy values E3 = 7 ω / 2 E2 = 5 ω / 2 E1 = 3ω / 2 E 0 = ω 2 En = ( ω ( n + 1 / 2 )) Data for some diatomics molecule V cm ­1 K N/m ­1 Bond length pm H2 4401 570 74.1 D2 2990 527 74.1 I2 213 170 266.7 N2 2330 2243 109.4 CO 2143 1857 112.8 HI 2230 408 141.4 The zero point energy is diﬀerent for hydrogen and deutorium. This means (for example) that the deuterium bond is stronger D = B − ω / 2 Selec)on rules for vibra)onal excita)on Δn = ±1 En+1 − En = ω : A constant which is independent of n Molecular Dissocia)on En = ω ( n + 1 / 2 ) − ω e ( n + 1 / 2 ) 2 Dissocia)on occurs when the diﬀerence between energy level is zero Can be used to determine n En + 1 − En = 0 2 2 En +1 − En = ω ( n + 3 / 2 − n − 1 / 2 ) − ω e ⎡( n + 3 / 2 ) − ( n + 1 / 2 ) ⎤ ⎣ ⎦ = ω − ω e [ 3n + 9 / 4 − n − 1 / 4 ] = ω − ω e [ 2 n − 2 ] ω n= −1 2ω e...
View Full Document

## This document was uploaded on 03/04/2014 for the course CH 354L at University of Texas at Austin.

Ask a homework question - tutors are online