University of Washington Department of Chemistry Chemistry 453 Winter Quarter 2009

Transcription

1 University of Wshington Deprtment of Chemistry Chemistry Winter Qurter 9 Homework Assignment ; Due t pm on //9 6., 6., 6., 8., The wve function in question is: ψ u cu ( ψs ψsb * cu ( ψs ψsb cu ( ψs ψsb dτ * * * * c u ψsψsdτ ψsbψsbdτ ψsψsbdτ ψsbψsdτ * c u ψsbψsdτ cu [ Sb] cu Sb 6. The probbility integrl is / ζ ζ r / ψ ( r e r ζ ζ r / ( ϕ θsinθ ψ ( P r d d drr r drr e ζ ζ r / e r r ζ ζ ζ ζ ζ e ζ ζ ζ ζ ζ ζ ζ e ζ e ζ ζ ζ e ζ 8ζ P P P ( ( ( ( ( ζ e ( ( 8.76 ( ζ 8 e ( ( ( ζ e ( (

2 6. The VB bonding orbitl is given in the text s: α( β( α( β( ψ N ψ ( ψ ( ψ ( ψ ( VB ( s sb s sb The spin prt is lredy normlized by the inverse root fctor. So just worry bout the orbitl prt: * ( ( ( ( ( ( ( ( ( ( s ( sb ( s ( sb ( ( * VB VB s sb s sb s sb s sb N N ψ ψ dτ N ψ ψ ψ ψ ψ ψ ψ ψ dτ ( s ( sb ( s ( sb ( s ( sb ( ( s ( sb ( ( s ( sb ( d s ( sb ( s ( sb ( ( ψs ( dτ ( (( ( ( ψ ψ τ ψ ψ ψ ψ dτ ψ ψ ψ ψ dτ ψ ψ dτ ψ ψ ψ ψ dτ ψ ψ dτ S S S N S ( sb s ( sb ( sb ( s ( b b b b ψ dτ ψ ψ dτ ψ ψ dτ 8.

3 8. Sphericl hrmonics Y (, θ ϕ pper in the wve functions of the rigid rotor nd the electronic wve functions of the hydrogen tom. They re lso used in engineering nd computer grphics. We hve defined these functions s Y θ, ϕ Θ θ Φ ϕ ( ( ( m

4 im where Φ ( ϕ e ϕ nd the normlized functions Θ ( θ m in Lectures 7 nd 8 for l,,, nd m, ±, ±. Sphericl hrmonics cn lso be defined s: ( θϕ, ( cosθ expression N l,m is normliztion constnt. The function R ( cn be found tbulted im Y N R e ϕ. In this cosθ is clled n ssocited Legendre polynomil. The normliztion condition for sphericl hrmonics is * dϕ Ylm, ( θ, ϕ Ylm, ( θ, ϕ sinθdθ. Show tht this integrl eqution cn be cos cos. written s: N R ( θ d( θ Hint: R ( cosθ d( cosθ R ( cosθ d( cosθ Solution: * lm lm l, m ( cosθ imϕ imϕ ( ( ( dϕ Y θϕ, Y θϕ, sinθdθ N e e dϕ R cosθ sinθdθ ( ( N R cosθ sinθdθ d dθ ( sinθ d cosθ sinθdθ cos lm l, m lm l, m cos lm Rl, m ( ( ( ( N R cosθ d cosθ N R cosθ d cosθ N ( cosθ d ( cosθ b The ssocited Legendre polynomils hve the property: ( l m! Rlm, ( cosθ d( cosθ l lm! ( where the fctoril function is: n n ( n ( n this integrl expression nd the normliztion condition ( ( N R cosθ d cosθ, show tht N!. Note:!. Using ( l ( lm ( l m!.!

5 Solution: lm l, m ( ( lm N R cosθ d cosθ N ( ( l ( l m ( lm ( ( l l m! l l m! Nlm Nlm l m! l m!!! c The tle below lists the ssocited Legendre polynomils ( nd m, ±, ± : l m R ( x x ± x ( ± x x ± ( x R x for l,,, Using the informtion in this tble, nd the definitions: ( θϕ, ( cosθ ( l ( lm! nd Nlm,, determine the ngulr wve function for d ( l m! z From this expression determine the corresponding normlized wve function ( θ Solution: d z corresponds to quntum numbers n, l, nd m. ( ( (! Y ( N R ( e ϕ! im Y N R e ϕ ( i, θϕ,,, cosθ cos θ ( cos θ Now iϕ e Θ ( θ Y, ( θϕ, Θ( θ Φ ( ϕ Θ ( θ cos θ Θ ( θ ( cos θ 8 Sme result s you find in Lecture 7. Consider the wve function ψ ( r, θϕ, R ( r Y ( θϕ,,,,, ( electron. Θ.

6 Write out the individul, normlized rdil nd ngulr wve functions R, ( Y (,, r nd θ ϕ. Confirm tht ech is normlized. For the rdil wve function consult the text nd/or notes. For the ngulr wve function you cn consult the notes or use the results of problem. r/ Solution: ψ ( r, θϕ, R ( r Y ( θϕ, where R ( r e,,,, (text pge 9 nd Y ( l, m lm ( θ m ( ϕ ( cos θ problem. / r r ( 8 ( drr R r drr e Θ Φ (see Lecture 7 or r/ ( r/ 6! drr e 8 7 ( ( 8 6! 6 * ϕ θ θ ( θ ϕ ( θ ϕ 8 dϕdθsinθ( cos θ 8dθsinθ( cos θ 9cos 6cos 8 dθ sinθ( 9cos θ 6cos θ 8 cosθ b d d sin Y, Y, ( ( b Clculte the probbility of the electron occurring between ϕ nd ϕ/. Solve this problem by performing the pproprite integrl. Confirm your nswer with simple geometric rgument. Solution:

7 * dϕdθsin θy ( θ, ϕ Y ( θ, ϕ / 8 dϕdθsinθ( cos θ 8dθsinθ( cos θ 8 8 dθsinθ( 9cos θ 6cos θ 8( 8 c Repet the clcultion in b but now let the electron be locted between ϕ nd ϕ/ nd θ nd /. Does the vlue of the integrl correspond to geometric nlysis? / / * dϕ dθsin θy ( θ, ϕ Y ( θ, ϕ / / 8 dϕdθsinθ( cos θ 8 8 dθsinθ( cos θ / / / 9 6 / 88 dθ sinθ( 9 cos θ 6 cos θ 88 cos θ cos θ cosθ ( ( The volume defined by the ngulr rnges ϕ to ϕ/ nd θ to / defines /8 of the rnge covered by ϕ (i.e. to nd ½ of the rnge covered by θ (i.e. to. So you obtin the sme nswer using simple geometric rguments tht you obtined doing ll tht integrtion. A qurtic oscilltor is defined by the Hmiltonin H h d κx. Note this 8 m dx oscilltor differs from the liner hrmonic oscilltor discussed erlier in tht its potentil energy goes like x.

8 Using tril wve function of the form ψ bxg constnt, clculte the energy E ( α x ( αx h d κ x ( αx ( / F α e H G I K J ( ˆ / / ψ ψ 8 mdx ( α x ψ dx e dx H dx e e dx κx α α ( α x x e dx m m e dx bαxg /, where α is κ α α m α m α κ α α κ α m α m α m α b Using the Vritionl Principle, determine the vlue of α tht minimizes the E energy in prt. Tht is, using your expression for E from prt, set nd α solve for α. E κ α κ α α α α m α m κm α c Using your result from prt b, clculte the ground stte energy of the qurtic oscilltor nd the ground stte wve function. ψ ( / κm / κm x κm x κm x e e h Crbon monoxide bsorbs rdition t wve number ν cm becuse of its internl vibrtion. Clculte the vibrtionl energy level spcing for CO. Assume the vibrtion of CO cn be modeled s hrmonic oscilltion. Solution: Clculte the reduced mss for C 6 O in kilogrms...6 µ kg

9 Clculte the oscilltor frequency for C 6 O..9 Nm ν kg s E hν 6.6 Js 6. s. J b Clculte k B T for TK. Bsed on this vlue nd your nswer in prt, will CO vibrtions be thermlly excited t TK? Explin. kt B.8 JK K. J. kt B E so no. c At wht temperture will therml excittion of CO vibrtions become importnt? E. J Solution: kt B E T. K k.8 JK B d Clculte the vibrtionl het cpcity of mole of CO t the temperture clculted in prt c. Assume first the expression for the vibrtionl het cpcity bsed on the equiprtition theorem. Compre this result to the hν hν / kbt N AkB e kbt quntum expression for the het cpcity CV hν / kbt c e h Solution: Bsed on the equiprtition theorem, for one mole the vibrtionl het cpcity is R8.JK - mol -. Using the quntum mechnicl eqution bove: hν. J. kt.8 JK K B hν hν / kbt Nk A B e kt B R e 8.JK.68.6JK CV 7.6JK hν / k BT ( e ( e ( The two vlues re comprble. F HG I KJ Bonus Question. You hve the option to work this problem insted of Additionl problem. As mentioned during the lecture, when electron orbitls in hydrogen re degenerte, i.e. hve the sme n nd therefore the sme energy, it is impossible to obtin number of hydrogen toms tht hve their electrons exclusively in one of the degenerte orbitls. Therefore, lthough we cn sk you, for exmple, to clculte the probbility tht n electron in s orbitl is between. nd. Bohr orbits (i.e. between. nd.,

10 the results of such clcultions cnnot be experimentlly verified for free hydrogen toms. Suppose hydrogen electron is in the n energy stte. The four wve functions ψ,,, ψ,,, ψ,,, ψ,, hve the sme energy, so experimentlly, we cnnot prepre free hydrogen toms ll with electrons in one orbitl but not in the other three orbitls. We therefore propose tht the probbility of finding n electron in unit volume rdrsinθdθdϕ is in fct n verge: * * * * ( P ψ ψ ψ ψ ψ ψ ψ ψ, Using the vlues of these wve functions given on text pge 6, prove tht the probbility P is independent of θ nd ϕ. Tht is, prove tht the verge probbility P( r is function of r only. Hint: Remember tht sin θ cos θ Solution: * * * * P ψ ψ ψ ψ ψ ψ ψ ψ ( * * ( ψ ψ ψψ ψ ψ r r r r ( ( ( ( ( ( ( ( 6 6 r r / r r / r r / ( ( ( e ( ( ( ( cos θe sin θe r r / r r / ( ( ( ( ( ( e e cos θ sin θ r r / r r / ( ( ( ( ( e e r / r r r / r r ( e ( ( 6 e r / r / r / r / ( e cos θe sin θe sin θe ( ( ( ( b Using your expression for P( r tht you obtined in prt clculte the probbility tht n electron is between. nd. of the nucleus. Do you still hve to integrte over θ nd ϕ, even though the verge probbility depends only on r? Explin.. r/ r r ( ( (.. r/ r r ( 6 ( (.. rprdr re dr e r dr 6. The stndrd integrl for this problem cn be found in the text ppendix: n x n x xe n n x x e dx x e dx C

11 However, becuse n> for two of the terms in the integrnd, you hve to repet the integrtion by prts four more times.. The resulting integrl is given in mny mth sources s: x n n n n x e n nx n( n x ( n! x e dx x n This formul cn be pplied to ech term in the integrl. But nother pproch recommended by severl Chem students is Tbulr Integrtion by Prts. If you hve n integrl of the form F( x G( x dx, you simply compose tble Column Column G x F ( x ( ( F ( x ( G ( x ( F ( x ( G ( x ( n n F ( ( x G ( n ( x where F (n (x is the nth derivtive of F(x nd G (-n (x is the nth ntiderivtive of G (i.e. obtined by integrtion. The resulting integrl vlue is: ( ( ( ( ( ( F x G x dx FG F G F G F G ( ( r r For our integrl F( r ( r Column Column r r r / e r r ( r r 6r r ( 6 r ( / nd G( r e r / e r / e r / e r / e r / e r The integrl is;. So the tble is

12 .. ( r / r r e r dr r r r/ r r r/ 6r r r/ ( r ( ( ( ( ( e r e e 6 r r/ r/ ( ( e ( ( e r r r ( (.. / r e r r ( e ( ( (.. r 6r r 6 r ( ( ( ( / ( ( ( ( ( 6 6 ( ( ( ( ( ( / ( ( ( ( ( 6 6 ( ( ( ( ( ( ( e ( ( ( ( / e ( ( ( ( ( ( ( ( ( 6( ( ( 6 ( / ( e ( ( ( ( ( ( ( ( ( 6 ( ( 6 ( ( The lst step is rithmetic..

13 .. ( ( ( e ( ( rprdr / ( ( / ( e ( 8 6 ( 6 ( ( / 99 / 7 e ( 6 8 e ( (. ( ( ( [ ].7. 6 About ½ of percent.