University of Washington Department of Chemistry Chemistry 453 Winter Quarter 2010 Homework Assignment 4; Due at 5p.m. on 2/01/10

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1 University of Wshington Deprtment of Chemistry Chemistry 45 Winter Qurter Homework Assignment 4; Due t 5p.m. on // We lerned tht the Hmiltonin for the quntized hrmonic oscilltor is ˆ d κ H. You cn obtin the generl epression for the energy, i.e. µ d En ω ( n, for n,,, by solving Schrodinger s eqution Ĥψ ( Eψ (. Alterntively, one cn guess the form of tril wve function, nd use the energy eqution to obtin refined wve function from the tril wve function nd lso determine the energy. Using tril wve function of the form ψ ( constnt, clculte the energy E Solution: z z ψ d / α e α, where α is ψ H ψ d for the hrmonic oscilltor. / / ˆ α α d κ α α ψhψd e e d µ d α α ψ d e d α d κ α κ α ( α( α µ d µ α 8α E e e d e d e d 8α κ α α α 8α κ α α e d µ µ µ 8α α µ 8α κ α α µ 4α µ b Determine the vlue of α tht minimizes the energy in prt. Tht is, using E your epression for E from prt, set nd solve for α. α

2 Solution: κ α α κ α α E µ 4α µ 8α µ µ E κ 4 µκ α α 8α µ µ α κ µ c Using your result from prt b, clculte the ground stte energy of the hrmonic oscilltor nd the ground stte wve function. Compre these epressions to the epressions given in the tet nd in the notes obtined by solving Schroedinger s eqution. Solution: ψ ( / / α µκ α e e The function is close to wht is in the tet but it is not normlized. The tet hs the normlized wve function: ψ ( / α e α µκ / where µκ α. d Repet this procedure using the sme tril wve function pplied to the qurtic 4 oscilltor which hs the Hmiltonin ˆ d κ H. Note the qurtic oscilltor µ d differs from the hrmonic oscilltor by hving the potentil energy depend upon 4 insted of. / 4 / ˆ α α d κ α α ψhψd e e d µ d α α ψ d e d 4 4 α d κ α κ α ( α( α µ d µ α 8 E e e d e d α e d 8α κ α α α 8α κ α α e d µ µ 8 ( α µ 8α α µ 8α κ α α κ α α µ µ α µ / E κ 8 κµ α α 6 α µ α κµ 8

3 e Using your ground stte wve function from prt d, clculte ground stte of the qurtic oscilltor. How does this compre to oscilltor in the ground stte? Eplin. for the for hrmonic ψ ( / α e α κµ where α 8 / α α α e d e d / / 8 8 α α α α α 8α 8α 8 κµ 4 κµ e d e d κµ For hrmonic oscilltor in the ground stte (n where α α The displcement of the qurtic oscilltor is smller becuse the mss moves in steeper potentil (i.e. 4 vs.. Sphericl hrmonics Y ( lm, 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 Ylm, ( θ, ϕ Θlm, ( θ Φ m( ϕ im where Φ m ( ϕ e ϕ nd the normlized functions Θlm, ( θ cn be found tbulted in Lectures 7 nd 8 for l,,, nd m, ±, ±. Sphericl hrmonics cn lso be defined s: lm, ( θϕ, lm, lm, ( cosθ epression N l,m is normliztion constnt. The function R ( Y N R im e ϕ. In this lm, cosθ is clled n ssocited Legendre polynomil. The tble below lists the ssocited Legendre R for l,,, nd m, ±, ± : polynomils ( lm, l m R ( lm, ± ( ± ± (

4 Using the informtion in this tble, nd the definitions: im ( l ( lm! Ylm, ( θϕ, Nlm, Rlm, ( cosθ e ϕ nd Nlm,, determine the wve 4 l m! function for l nd m, i.e. ( Y θϕ.,, ( Solution: i ( 4 (! 5 Y, ( θϕ, N,R, ( cosθ e ϕ cos θ cos θ 4 (! 6 b Prove tht this wve function is normlized. 5 dϕdθsin θy, ( θ, ϕ dθsinθ( cos θ 6 4 ( ( ( cos 6cos θ dθsinθ 9cos θ 6cos θ c Grph this function in polr coordintes. This grph should hve the ppernce of d z orbitl d Using this wve function, clculte the probbility tht the rigid rotor occurs within the surfce of the sphere defined by φ nd φ/ nd θ nd θ/. Give geometric eplntion of your result. Sme s the normliztion integrl only chnge the limits 5 6 / / / dϕ dθsin θy, ( θ, ϕ dθsinθ( cos θ / 5 / / cos 6cos θ / dθsinθ( 9cos θ 6cos θ cosθ The limits of the integrtion define n octnt. Therefore the chnce of finding n electron in n octnt is /8. Consider the hydrogen tomic wve function ψ ( r, θϕ, R ( r Y ( θϕ,,,,,

5 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 tet nd/or notes. For the ngulr wve function you cn consult the notes. Solution: We hve lredy proven tht Y (, θϕ, is normlized. The rdil is normlized s shown below r/ rr ( rdr rre dr r/ 4 re dr 6! ! 6! ( 6 5 b Clculte the probbility of the electron occurring between ϕ nd ϕ/4. Solve this problem by performing the pproprite integrl. Confirm your nswer with simple geometric rgument. See problem d. Solution: By now this is cler s long s the ngulr wve function is normlized, the ngulr limits determine the probbility. Just set up the integrl s in problem d with pproprite limit chnges /4 dϕdθsin θy, ( θ, ϕ dθsinθ( cos θ cos 6cos θ dθ sinθ( 9cos θ 6cos θ cosθ Agin, the ϕ rnge is to /4 which is /8 the totl rnge. c Repet the clcultion in b but now let the electron be locted between ϕ nd ϕ/4 nd θ nd /. Does the vlue of the integrl correspond to geometric nlysis? Solution: we lredy know the nswer the ϕ rnge is /8 the full rnge nd the θ rnge is hlf. So the probbility is /6. Esy to show using the Problem d integrl

6 / / 5 9cos 6cos θ / cosθ /4 / / dϕ dθsin θy, ( θ, ϕ dθsinθ( cos θ 4 The Viril Theorem of Mechnics sttes tht given potentil energy of the form n V ( r r where is constnt nd n is positive or negtive integer, the verge potentil energy V is relted to the verge kinetic energy K s K n V. For n electron in the ground energy stte of hydrogen, clculte the verge kinetic energy nd the verge momentum squred, i.e. p. Hint: Use the fct tht for ll n, K V E n. e Solution: For H V ( r, therefore K V nd so 4ε r RH K V K K En n p RH 8 K p mr 9..8 H kg J m kg.8 J.97 kg m s ( ( ( ( ( ( b Using the hint from prt, clculte the verge potentil energy of n electron in the ground energy stte of hydrogen nd clculte the verge of /r, i.e.. r V V K V V En 8 8 V E R.8 J 4.6 J 8 ( 4 ( 4.6 J ( ( 8 4ε 8 ( 4.6 ( 4 ( 4.6 H e 8.85 C J m V J J ε r r e C.89 9 (.6 C 9 ( 4.6 m 8.85 CJ m

7 c Using the wve function epression for the s electron in hydrogen, clculte r. How close is r to r? r / r / 4 4 4! r rr e dr r e dr ( ( m m.6 m r 7.9 m r 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 eclusively in one of the degenerte orbitls. Therefore, lthough we cn sk you, for emple, to clculte the probbility tht n electron in s orbitl is between.5 nd.5 Bohr orbits (i.e. between.5 nd.5, the results of such clcultions cnnot be eperimentlly verified for free hydrogen toms becuse the s nd p orbitls re degenerte Suppose hydrogen electron is in the n energy stte. The four wve functions ψ,,, ψ,,, ψ,,, ψ,, hve the sme energy, so eperimentlly, 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 4 ψ ψ ψ ψ ψ ψ ψ ψ, Using the vlues of these wve functions given on tet 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:

8 * * * * P 4 ( ψψ ψψ ψψ ψ ψ * * 4 ( ψ ψ ψψ ψ ψ r r r r 4 ( ( ( ( ( ( ( ( r r / r r / r r / 4( ( ( e ( ( ( ( cos θe sin θe r r / r r / 4( ( ( ( ( ( e e cos θ sin θ r r / r r / 4( ( ( ( ( e e r / r r r / r r 4 ( e ( ( 64 e r / r / r / r / ( e cos θe sin θe sin θe ( ( ( ( b Using your epression for P( r tht you obtined in prt clculte the probbility tht n electron is between.5 nd.5 of the nucleus. Do you still hve to integrte over θ nd ϕ, even though the verge probbility depends only on r? Eplin..5 r/ 4 r r ( ( (.5.5 r/ r r ( 64 ( (.5.5 rprdr re dr e r dr 64.5 The stndrd integrl for this problem cn be found in the tet ppendi: n n e n n e d e d n n n e n n n( n ( n! n The second line is obtined by repeted ppliction of the formul.. You cn lso ccomplish this tsk using Tbulr Integrtion by prts (go to Internet for more informtion. Applying this formul for n4,, nd 4 r / 4 r / r / r / 4 r / 5 r / re drre 4re re 4re 4e ( r e r r r r / 4 4 ( 6 6 ( r / r / re dr e r r r r / r / re dr e r r

9 With these results r/ r r ( 64 ( (.5.5 ( r/ ( e ( r r r/ ( e ( r r 6r 6 rprdr e r dr r/ 4 4 ( e ( r 4r r 4r / / ( (.5.5 ((.5 (.5 ((.5 (.5 6 (.5 6 ((.5 (.5 6 (.5 6 5/ 4 4 ((.5 4 (.5 (.5 4 (.5 4 / 4 4 ( ( ( ( e e 64 ( e 5/ e / e e ( ( ( ( 5/ / 5/ / 5/ / ( e (.5 e ( 7.5 ( e ( e ( 5.5 ( e ( e ( / / ( e ( e ( / / ( e (.6875 e ( (

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