Problem_Set_5_Answers

Problem_Set_5_Answers - Chemistry 341 Physical Chemistry I...

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Unformatted text preview: Chemistry 341 Physical Chemistry I Fall 201 1 Problem Set 5 Due: Wednesday October 19 Answer ANY FIVE (5) questions. 1. SAB Problem 10.15, p. 392. 10.15 The spin parts of the wavefunctions for the first excited state of helium are Illa = ozA(1)a(2), tab = a(1)fi’ (2) + a(2)B(1), and $6 = B(1)B(2). Since Sz 1/! = RM 5.19, What are the MS values for the three wavefunctions? 5,211 = swfla t 35,3 (:33 “é '1 DMD (it?) A S} .‘lt ~. 3,,a loLmazca : 102.3% ‘otm =1 aka am 1 '3 among) a A a ’3’ 'L 1. SRDVLIJC: 3%”. QCDhLCQ : in) §%,1=LCL\ : 1:: elm) gal} 3%“: -_ “helm “(a .-. ill. MS=H > Cm ll: oLLn $0.1 4— ext) “((2) ' gyrfq': gs“ %.\L(l) %C2\ +§Cl3d£z§ A " : gm £5” «m + um 33,. $0) a = saw (5.} MD + “(a (‘4?) 1"” 7. A 3%,‘1-lf= 1% { clam {>Cz} - foEOLCL31§ ‘ 315,1. 4 : 5%,z%’ cLClfl $sz 4- $(‘30L(Z\} A. can 53,1 so.) + Eco Ssfidca [I A 4 = id} (-123) am + sang} ace} 375,113C : ~37.- {can $02.] .. $613 oLLaW} g Ll! : O M5: 0 (l L: fiaMA‘ulwuéa‘) (:3 fl“ 7. 9.03 eac'l) A A A SSI‘PAC: $215“ $60 Q>Ca ; figa 5%,, %LQ -; ~E (pf-z} Ft») :5EanGQ‘ 3$a1$<¢ €55” $Lfi kLa : bc0§512%€z\= ~3FC0 $02.) In 33w = JR yawn: 42¢ M :4 S N 'f‘iéttebiem 10.35. p. 394. 10.35 The first ionization potentials of Na, K, and Rb are 5.138, 4.341, and 4.166 eV, respectively. Assume that the en» ergy level of the outer electron can be represented by a hydro- genlike formula with an effective nuclear charge Z ’ and that the relevant orbitals are 35, 4s, and 55, respectively. Calculate Z ' for these atoms. “3 En 165112— (27.2Hev) 2.“- TE‘ : City-L 1v." 1 v s "’- 2: hi ZItI k (2'7.7_l| aV} 17.741 ¢V 3. SAB Prohiem M36, D) 394, 10.36 What is the total electronic energy in hartrees of He, Li, and Be with respect to the nuclei and free electrons? Ionization potentials are given in Table 10.4. Nag gami— Ea : 21,7.H av E SABJ. 3:52] Ca.\ He (up A He‘btg +4.- “ICE - E (Rafi -EL\\-GJ ‘.— urges? .» Helm»; I E7. = o — ELM“ :tE‘4JJE1 = - awe) E UM : — {TEfi—IE 1'} ELHeX=~ {mass-)4. 514341573 av .(E—h W 2.7.2ugv Cm L; (.9 _> Ld’LcQ +e' ‘IE = Hm)- em) | L'L+ (.69 -> LL1+(%>+ Q: IE1 3 E LLJ‘A'E - ECLJ‘) LL1+ \ a LL3+L°3\+C- = 6 ' ECLLz-r) 1EI+I€2+ 2933 = - E (L0 Etta: — £13+r§1+teg E (Li) = - {93°‘2+ 79ers?) + I‘L'LMS‘} G_\(. q ECLD = —’7.Q'78 ea (:3 59033 —> 939133 +6 TEL: E(Pae+\ -Ja(se\ 93¢:ch -> Pn‘fip +e' “IE1: : we”) -ecse*3 95¢“ Lop -> $29.03} +‘e7 "SIZE:D = E {Fwy} ~E (Be?) 1%] +IE1+IE3+WH = - gage“) E = - RTEV“ 1?; 4-— 1:; {5133*} E (Be = —‘ {Wall-I- xaxzn + $3.qu + 2:7.7x3}e.\;.(5e. X V . 27.2116 ‘ E 5e): ~lLI3.(uC98 Ego 8A3 Froblem @377 p. 394. 10.37 Why is the ionization energy of boron less than beryllium and the ionization energy of oxygen less than nitrogen? (at) (b) B: [He] 232 2p Be: [He] 252 70: [He] 252 2p4 N: [He] 2s2 2p3 During ionization, we are removing an e' from an incomplete Zp-subshell. During ionization, we are removing an e" fiom a filled Zs—subshell (has an extra degree of stability). During ionization, we are removing an e‘ from an incomplete Zp-subshell. During ionization, we are removing an e" from an half-filled 2p-subshell (has an extra degree of stability). ix?! 8A3 Problem 1039. p. 394. 10.39 For a carbon atom in the configuration [He]2522p2 the following term symbols are all possible: 180, 3P0, 3P1, 3P2, and 1D2. According to Hund’s rules, which is the most stable state? Hund’s empirical rules [SAB, p.386] are given as follows. L The term arising from the ground configuration with the maximum multiplic- ity (2S + 1) lies lowest in energy. 2. For levels with the same multiplicity, the one with the maximum value of L lies lowest in energy. 3. For levels with the same S and L, the one with the lowest energy depends on the extent to which the subshell is filled. ‘ a. If the subshell is less than half-filled, the state with the smallest value of J is the most stable. b. If the subshell is more than halt-filled, the state with the largest value of J is the most stable. We apply Hund’s empirical rules in the order given: 1. Of the two possible spin states, singlet or tIiplet, the tu‘plet has the larger spin multiplicity. 2. 3P is the only possible triplet state. 3. For C the 2p—subshell ([He] 2s2 2p?) is less than half-filled so 3130 is the most stable state r, B. SAB Problem 10.45, p. 394. 10.45 If we use ‘1’ as a trial function for Hamiltonian H , where _, (¢0 + 11051) \P _ (1 + a2)1/2 and $0 and (#1 are the two lowest eigenfunctions with eigenval- ues E0 and E1 (with E1 > E0), show that the variational method yieldsa=0. we, know fin ~. em 1:34.: 2,89‘ “m wwasg — R “we ws Wm M: (ma-3'”?— mm“ 3:y‘k-‘éuess " N h g i=3 “a 4M} u 7. N 31,8532‘34- o‘EAal} qémsg‘i}ms§ M i¢°+q<k3 iEaéhi. quhl} “ ‘- “View”: mam. “Eamotsmfl 9‘ 1 Swain.» T" +91%,» L7: = “2%, “Eg Skid?! +* (LE, “(aka)”?- +- 0:908 blight-1» QIE‘LSQZA-f 8g“ maths?“ CLTe normosxl‘gec) QKAL Q%%Muv& ‘ “2. SE12 =1 S‘b‘mA'tzl Shmhxé‘t = g§fi¥oAt 2C3> we Have g: “ziEaZ—t—o-l-m a}ng 9: Cl+am\"§_'fio Ayala} Now "fie \DLS‘\' Vaduz a? “ $845—$525 LESE ._ 0 1 C 1+¢L3-‘SL ZKEJE — (1+ «Wzag {30+ a.2 E} O : (\+o.7-3.‘512q*€fi - Cid-offi-‘(zofli ‘ o .. (Mn-man {be} We 0"“? S°\uv\;wv~ +0 4a.“ ‘1‘“:ka 4946" “Alng S-evLSe f5 on 7. SAB Problem 10.58, p. 394. 10.58 Calculate the splitting in kJ/mol and eV for an H atom in the 2p state in a 10-T magnetic field, neglecting the spin as in Fig. 10.8 and equation 10.43. Compare with kT at room temperature. I? $+che (Eglw E]“+: Paw?) WtRm=+‘-,D -l 4- )LS B (m: +13 EMA- ll 0 CM=OW _. Ps5 (nu—n - v. AE : 3E9) = 03.27% )t to Z C1013 = (1.2114): \Q'2$3 T A2: (@27th LEan ((9.011xm7'5 M‘v :. 8.385 x “3.21:: vac-P.“ \033 LE: (WV-'7‘! x $7333 ( tV »: Bagel x to“ tv \e'r .L... 1-"> My: (rsz In‘ It“) (magma g tum xto‘z‘J' kb-r = (q.l\‘l x \o‘“ I) (¢.ozzx\§3mu—E’§(_§§ = 234.2; kTwu-F‘ 1033 RT: (‘LH'I Xlé2)33( EV 3 2. 2.97ox thZEVV 1'bo2x1o"9.3' 8. SAB Problem 10.61. p. 394. 10.61 In a hydrogen atom, the 25 and 2p orbitals have the same energy. However, in a boron atom, the 2s orbital has a lower energy than the 2p. Explain this in terms of the shape of the orbitals. Tacit-0.1 (\afi’rr‘ibu‘H’c-u hh¢¥hflm WEarISovxS 1%. NA $~€£th %=Y/°~e P10: 38— (24’332 “*1 é C2A. VGJWJ A1$+*tbu+CaM-Qunc¥1o€ .Fz‘ __ in L3"*‘ 9“? C2? ‘CGJXIOJ A‘SHILW'L'DO‘H fiqfic'i'icnw _\Ue gee \n a. Piefi' 0—p- ‘i‘ae‘radimi $t3i‘Y‘lbufi‘I'cm LHCA‘LaMS “4°ng och'exe meg—\Tfi—e—KQfi- an e‘ in ‘0» 2A,. CI..U- gene-Lakes Chagar 4-D “We nucleus and 9.5 a; {‘ESI-LH' +Re “Mimi-we anges an fie aver 0.3.9. wouii he. ' enerfl e) 5 Pro" q fought)». oer? 1-6ka mad-iv“ w°w\&¥e HP. of‘afi'qi eheY‘Cgie'S E h d £2132. 1h?- .— .— (1.1.74er3 Gene}. obs qvgsu\+ 04‘. +9.; Alt—(puma; 'm er-HVQ {LAT%QS Esz3 4 Ehfii a g . £¢m+mue on: ">613... mod 3.0 2.0 owd mud (Md 9. SAB Problem 10.63, p. 394. 10.63 The first three ionization energies of scandium (Sc, atomic number 21) are as follows: Sc = Sc+ + e‘ E1 = 6.54 eV 36“ = Sc2+ + e— E1 = 12.8 eV Sc2+ = 803+ + e- E = 24.75 eV What are the electron configurations of the three ions? q “‘53.; . EAv'l RAB—3X 2r 10. A variational calculation for the He atom involves ‘1’ = 15' (l)- 13' (2) with (Z’)3 1/2 Z, r 15' = 3 exp — . 7mg an The resulting energy found is r 2 r s: {(2) ~ Y2 } (27.2 eV). Plot 8( Z' ) between 2' =1 —> Z’ =2. Determine the best value of Z’ . Determine the energy 8 at the best value of Z ’ . Determine the ionization of He and compare to the experimental value Of 24.580 eV [SAB Table 10.3, p.378]. P“? 9‘?” a. We Axel—ermine 7-7.1 tv on; O. .PunchLcwx 5:; %l . 1-4. 1-5 1-6 1-7 -2-8350 -2.8025 —2-7500 A p\u'\' a; €(213/Z'70. 2V VS. :2, is shown on "Rye. hex-l- P¢¢a¢. b. ‘Tk-e. M‘IMIMLLM in 2(2-‘3 occw‘s token (“3. €am+tmue§»3\) .N O N 2.. .8 .N .m> ANV w 06. NN. C10.Con+twuepe3\ng) a, ugcop 3, H.399 + e 12‘ TE‘:§U’re+\- Ewe) :te.‘ : __ '1," (rakew _ (InseV) Cf)" IE ': ~94le eV “" 71% ev l T5 = 23.\ 2V «)9 er“).- ': \Z’B-l - 2%.g%\ 06062)) (145%) 11. Write the term symbols arising frem the ground state (31603911 mufigurations Of a. Na, b. F and 3. {he waited configurafia'filsz 2s2 2p 3 p of C. on “a. e." Qb‘€:% urn—‘ifl;u ; INeZUSA) We C¢n913~ev oak} +ae 3,5. Valence P: L 1 O S: ‘12 J t ‘l-L ‘15 +9“? emu? “Perm fiwmbd Possible b. F e' cou£lv~¢uva-\-Luv\: 114m] (ZA-‘Czfls We coaguleur and.) ‘3 {gaw‘vajngn'i‘ 2-? €f’fl . From +Ra +n_ble, in Lee-Hare 1* 45:42;ch "Te-"Wt sgwdocls we .(JncA J-Re WM gxkubul Pus$thL¢ is 7’? ’ug. cure “Levin 9xaudao\$ Pug»;le c. C exid-ai e‘ m§13ura$w EHQJCZASLéflc3P3 Me‘erA} '. Sa'hce— He 2.? AAA- 3? 5‘s arem'écbutvada-cl- ' we WSuH-fl-L-Fmbh. For new- a unvolt'd" e'ho I'm. Levi-gate 2| ~A+m1c Term L wh-exg we {—wul +3.4; pussi'vbe hm shd-es lama $'PJ b stuckfioi' mni+~rnplg+ th-zs ’ S ‘. $=¢> ,Lzo :. 32¢ ‘50 ":5 ‘. 3: 3 L:o 3'1\ ES” 5P1 $20,L_=l ‘37:! 1?. 3P1 $=l,L‘:.| :. fizz’l’o a a ‘5 P21 P‘J P0 ‘D 5:9, L.:.2. 32.2 3 “D7. 33 S31 I“1.22 3:5,2.\ 3}, jb gb (“-co h+wxue¢9t l3 mgHagég \AH—e “\‘Rd' +94 1? the!» 7p? E‘s awe \hE’bu\\(ol-€I~.+. So “Q1'Z‘ (It. gcvfl'mqe $3 Electmn configuration for .A.(2p) (3p) a 3D (ML=+2, +1, 0, -1, -2), (Ms=+1, o, -1) 15 substates (J =2+:1=3, 2, 24:1) “03. 302, 301 b 1D (ML=+2. +1, 0, -1, -2). (Ms: 0) 5 substates (J=2+o=2) 102 (3 3P1 (ML=+1, o, 4). (MS=+1, o, -1) 9 substates (J=1+1=2. 1, 14:0) 3P2. 3P1, 3P0 d‘P(ML=+1,D,—1),(Ms= 0) 3 substates (J=1—0=1)1P1 e 35 (ML: 0), (MS=+1, o, -1) s substates (J=O+1=1) 351 f‘sn (ML: 0), (M50) 1 substate (J=O+O=O) ‘s, 12. The origin of the D lines in the spectrum of atomic sodium is shown in the figure given below. Calculate the spin-orbit coupling constant A for the excited electron configuration for Na of [Ne] 3p. V /cm‘1 16 973 The energy-level for the formation ofthe sodium D lines. The splitting of the spectral lines (by 17 urn—1) reflects the splitting of the levels of the 2P term. 589.76 nm 589.16 nm .3, A moxitvuz eruxxeaa 3en¢VbLt$ o. nonmetal-Cc grili‘?) —» re. a; _Q_ tied? f as - We Sf‘m macaud ax Womew+wm 3 A , of, +69, 2 (generoelhas Gk - -> Mauvueu-tc MOMQM. p-‘L " —>; P24. : - 32.9.. A 2% e flare. \; Qn‘tnir as, llw‘l‘efficfif‘uo‘w 0+ "Egg MankvaA-l: mowed- ?LA muck fie .IKA'UJL-Oiol WLo.D&vL€A—t¢— Lem E _>,—» 9 E aw?» cc Nu m'\‘ ..> Earl-t. Ala: 2'9. Ugh-ere A {5 '3'er SY'Iw-ov‘lm'l' 'hz Ceuvlima uws+cut+ (.12. Cam-H nueA‘X A A: 4> New 4&9. +o+‘&.\ 1n<3_le.V'-Mowen+um 3': L+S , 5:. -'> a 4» -> =Ct+?\o(\?+—?3 = t“‘- 4-5-5 +'2.L‘§, a, L ‘ a...” $’ ‘é‘ ? ‘- -{ B'C‘Sa-n‘h'K L(\-+—l3'k1- SCS‘I—O-kz} = .L .St 5 CS“) — LCM-13 ~$Cs+n} 1x" _. ,1; LCL-t-n ~ '5 CSi— OE « WWWa-IWA ¢ggmw MWhér‘. Wee. JMMWWM WMMW .. Exid-eé sh-Las mp MO. Hui-K e- Cm{.u-bwv-o.-\-Luv\ L‘- £°Y We Voltwtcfi— S: LL-LS\) .-. )L—$\ = \l-I—li\ ’\\--§_\ : 1;.) L ' 1 Em+,312= JiQCA 5" 1(1) : €43ch Emhuz: .17: email g: (1;). le)-J1.’(_:_D} .. -4ch “kc A”): : E“+’511- E‘wg’qL : tack: - (‘91ch z: flare-Eu we, Mal-e Hoff 5rtu~°fg\‘\‘ uupi-HM} gfi'es rise 4'0 He -Pn'ne $+PuC*'\-tfe. A°u51¢+ {files In He aA'hMIC SFec‘x'Pwu e—‘L NO.- ...
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Problem_Set_5_Answers - Chemistry 341 Physical Chemistry I...

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