Problem_Set_4_Answers

Problem_Set_4_Answers - Chemistry 341 Physical Chemistry I...

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Unformatted text preview: Chemistry 341 Physical Chemistry I Fall 201 0 Problem Set 4 Due: Wednesday October 5 Answer ANY FIVE (5) questions. l. SAB Problem 10.4, p. 392. 10.4 Since H and D have different reduced masses, they also have slightly different electronic energy levels. Calculate (a) the ionization potentials and (b) the wavelengths of the first line in the Balmer series for these atoms. (,3 . H '> i++ *e“ :E = o — E|~ :5 = - (-43.ch ev\ =13.cos eV 2:: = — t9 Ewan PH N¢+me “Pi-me. Now 1:» m Mqu me) P» mwe “‘P h = ‘ :kilvk “4} Hr!- "“ E M? “ML h = ,1”. q.\n‘\3%q7x153‘ E ihk q-lo‘l 58%?)(153‘ } H», 3.3%, 9860:0527 5-2943 5850111?” lib. -. 1.000 Z7Zl («3.4- Cofi‘xmLeAW IE 0)} : (1.000 212 1\C\%.bob av» IE Cb} = |3.Q\o eV _ _ —n Clfl : CR»- \OQjQ77.bggtpcw. [SA-B,p.353‘1 Baku/war Semis Ina-y 7- w x-\~& no =3aqagl.“ >l Lag {lira-\— line 5;. {banner say-Hes is 3—)2 -\-vn.ns\’t-§m ..L- 'R .L _ A. = C -\ g 9‘ _ a. 11 $2 \0Q3517.5956£w~ J; = 19,12,1A‘H8 w" y ‘ ( ' N “‘ WM 3 - _ -q 15,232.4Q3cw WELW‘ ‘5 M 3: LD‘SE.L\'7° MW» ‘ 'Rb: 10%,70'Lq2575. c—m" [$Prf5,y.%§‘$] L32 1. -3. = E. R 3 D 2"" 257’} '53 D .L = (Laqj‘2c7.qz75c.u2‘3 '\ 3L >x: (089.2CHMWL 3152B Problem 18.63 p. 392, 10.6 A muon is an elementary particle with a negative charge equal to the charge of the electron and a mass approx- imately 200 times the electron mass. The muonium atom is formed from a proton and a muon. Calculate the reduced mass, the Rydberg constant, and the formula for the energy levels for this atom. What is the most probable radius of the ls orbital for this atom? pvt-$4:- we, oksevva Mia-l“ “lete damage tx\9reSSlows .poy- R ,E omJ v-Mr a.“ 'mvolve ‘Hate reaLuCexl mass}... M Widow-FM? BM : (Zeus/me) Mp _ 200 m (Loewe;- Mp3 i100fl§+|} : m 03.103 333 '7 x15“ PM 3 2,252,614“ ass-2x16“ 3 m} “5-72. (.27, 1x16” y»: “3.0:Q'2sdzs 0 xx 151%}: M WWW WWWWQVM/ my“: - m -23 ’ ~ - 12M: (1.6Ll1qz50 x10 3(LOQ7 373 Igsqubcwt“) q.1oq 38‘37ute”3‘ ? m = cm" co, Enz-i.(?-7.Z\13qb\9\() 1M3 hz Em :— <1-(o‘i2. 9250 Xmas) (2.7.2.113‘16 rev) 2(q-quegq—7Xt63W re" erwuam ' E w —3\ «em? : 0mm 339 1 xw )(szfltj 71% PM} man. @29 O X “'28 m? = am 3. SAB Problem 39.7. p. 392. 10.7 A hydrogenlike atom has a series of spectral lines at A = 26.2445, 19.4404, 17 .3578, and 16.4028 nm. What is the nuclear charge on the atom? What is the formula for this spectral series (1.6., n1 and 112 in equation 10.16)? w e '5““Cu--\~ midi—Q g=LerQ1 'L’”.L’ k 90 {h‘1 “1 \Me CL$vame We. 3\’5 %_\\/~e,m are We .Q_\\"S“- U; memkerg (3L Btr§€$ Che» wL‘VLx a SlucL-CA-c “g VaJueA . On $O\4\/|th3 .Lor Q}, we hang {1: l 12 >.{..L -_l..15 60 h?- 7. u “z We have assume. dL—C—Q—{revd- values 4- h. Wm n2: n\+\,ml+1é yup—'5, Vila—H awuL \uuk (4—4!“ Lung—Eeosmmé 43;. Q} - 41 26.2445 nm 7L2 19.4404 nm 43 17.3573 nm 44 16.4028 nm Rm 1.0973731534E-02 nm’1 ' “Z E\nc.@. ‘E 22$ 1%, £QW$§5‘~&\,€\T when mm: 1 value. “(1 "Z? mu,5\—‘oe, scav- Hits Seems W1fimW§WNWWW$§M¢V¢WmWN£¢NW - E fHL “1:3,H)S,(¢)... g 4. SAB Problem 10.9, p. 392“ 10.9 Calculate the wavelength of light emitted When an elec- tron falls from the n = 100 orbit to the n : 99 orbit of the hy— drogen atom. Such species are known as high Rydberg atoms. They are detected in astronomy and are more and more studied in the laboratory. ’Rtgékev"? erw‘la Pu”? H J.— 2/R J... -J... ’3 H 33'qu \uc" \Me use RH: \bq)t.17.g?s§6 QwC‘ {EMS} PIES‘S] .L __ toq,e77.€93eqm" {4. - 4.. 1 cm W? L. 2 2.1.29 gqq‘zqg xréflam“ :\ A - Cm 1474:: 8461 2%% '3‘ “3‘ M90 '5 ‘3 . ‘ \ 5. SAB Pmblem 18.10, p” 392. 10.10 Calculate the Rydberg constant RH for a hydrogen atom ‘ and the Rydberg constant RD for a deuterium atom given the value for R00 given in Appendix B. l “on = wane-2. M's-aqua: PR» R = #100 Mo. “our )1»: VMEM WP+me ,& = M2 ’ _ l m“ MP'F‘Me 14. n3 M F 1,12 ~{ , “V‘- (Loq j -7 3 g .43 ‘+ 22M ’ g ’ l 3‘4“ 1.512 Q23\xL°-21kg a? = W D We NW. —- Im r- .___£a"" mid-Me i : MA 1 Me m4.» me ' 1+ me :1 v ’R _ § ‘ . n 3)- C104 737,3153x4c " 1+ q."°§38°|7xlo-Mk: > M 3 Jan's 5‘66 0 >4 La'“ k3 QB = \°°« '70-;- ‘4261 MM“ Idol-e: Tu gar? (RH. mac) (Rn givem 6n {35% of. $A9> me = 62.; 0°; x \b‘g‘ Eta vs-zeAS +1: la LL$eJ~ Ln Drag find-mus 0b,\’v.en «have. 6. SAB Problem 10.20, p. 393» 10.20 Calculate the expectation value (r) of the radius of 21 2s orbital and a 2p orbital for a hydro genlike atom. Is this the result that you expected? 02A. G-O. 311 v- l 2 7. - 32C — (Rio 2 “(no “(PJ‘KU‘ { at».k e 110 Yoo J— “7. Cum- 52 -fl. R719: .__‘_:l (i— z{z~t}eu% 1(23 1 ° q" a °° 3 Z ‘1'- <r>_ S 'Y‘ {2- e 4o dr %~ 0‘. QB -0 3 ‘° 3 ‘* 7- 9 "El (T5 : .L S {Ulr- — + u fie “ear 3 4° °‘° “: 8 . 7t 4‘?) : Q3010 2 ‘ 7. P a” l 23" —z ‘1! s '- ‘l 20.0 R74 = “ram - Man" (A) (no 3 “56 Y,» (a ’2 me tnr _%v s! 5] —. “<6 t‘ “21%” a 2“ (33(8) a, co 1 4T»: fe‘rt 021‘ 3w- Y:D < g 8" g “a r - J. _ Y E A’“ v I“ a) at C ‘° -20 °°w+\'mu.-ed‘\ 5 41"» : _L ‘3‘. 1% ab 3 G (at) 4*“ 5.3» 4E ‘ V In+ui¥iVel~y one. {new}; axiaec“ He 2A. 0.0 La have :1 shame? (Y) , however <f5a&><*>z . 6n tws \aaU‘fi-‘m .(Lécae ‘NLA‘LGA dL$+rtiawchm bytng .(_¢~r fig 2A «RA ZIP 93‘1hs I$Prp> 3:» are. $0.5 (2.539] we $ee +3; mkx.mkwt L.“ a2 2’? YQ-A:L dL$¥P|bw¥Cw oogm 337 CL skurx‘ur Y._ v-nu“ menus More, SMcJLzr Y‘v-uJues we \pus$[\aLe (Fe-p +8“; 1,? VILLA fight],un mole-X *3 2L “QALOA iL$+WILK¥LW 44‘ng a 75%; a man H- <r7zv < (v72 A ' (10.20 cavA- \ nqea 23 W’Xcfi-s 0?— Pm Anal. 19,“ (c3) are skawvx 3n 43¢ (33*ch am +Re nafl yo.%e. H-Cs (JecLy 4-he Maximum 1» fig 1? “mild Als¥wku+RnA eCch‘S «2+ «- SkovLev ruktwsavwl as a rain“— (“>943 < (out. ’ 9. m m N. o m v m N v o o...o mfio owd mNd on: n>.m>tca C\°,20 un+\Y\u.e.A3 3 (£)d 7. SAB Problem 10.34313” 393. 10.34 Since the outer electron for Li is quite a bit farther out than the two 1s electrons, this atom is something like a hydro- gen atom in the 25 state. The first ionization potential of Li is 5.39 eV. What ionization potential would be expected from this simple model of an Li atom? What effective nuclear charge Z ’ seen by the outer electron would give the correct first ionization potential? (ab IE : €n=°° -- E‘uz .1-|’- a. 2A 3" Cs Lu “(33:51 It: 0.. (- 3.60430’ 11. itE 7. 3.\-!02. -e.V O an :21 = o - i-QllzCeeoegfl} A 22. Q1 : 22113.) 034.0603) sz '1 (Lg—0:0 (a eVU For Her there is a difference between the average distance <r> between the electron and the nucleus and the most probable distance rmp between the electron and the nucleus. Determine: a. <r>a11d b. rmp. ‘ :‘f‘5‘l’ we Nae-e4 *3 #‘Lhflx‘l-uvx q|o 4-51- Fmeklb.| P-357 all 5MB. wt kww W - 5‘ "‘ J7: Ht % to \H h“ no 2‘ .—- 8 wt =2 l C (m9 Y t 0 15¢ - <2 Y° -_ (z A. "1 $09 — to b tn (brig L13 pVOM fins CV) and. ('23 we hue . 3 V '1‘" &Lo '2 611-3!" 'E’ 31— e “0 W130. I] '2:- ng = Z (23 )1 ‘9. as a: " -Z‘C &\ro 2 Ll (7'! L e. T. a3]; a -1: . _ ‘1 (21° 1 A e “a mum A: 1 Q31; 0 We JttSanu¥Lw .guwc‘l‘imx Lur +£4.14; S-l-nJc-Q 0L1+e+ \S Z .. Fla T- rl «lg : A V1 3 «o a (’0 Q 41'» ‘: § ‘r ‘oolfl' = A1 EY- Y?‘ e. gig-6 dew Tzo fzo w - _ r (V) : 25(7- nge ‘15 L‘- :. “EU-d 2" g 3 A r: 3 Ll “ — D no“ 0‘0 Ll?) . \‘ Y‘MP occh‘S when; YMCA 15 «maximum .“1 a‘Plecr31 1. 9" §K2Yze mo} \Y‘gr W: rzvmv OM’ R “P -_Y A¢§zr -r1(%3}em =0 zsz-il-z-‘C'ge'ufi, :0 ' 3‘ Tut: E no: ’ a. The Schrodinger equation for a H—like atomic species can be expressed as 7111 + I? zez _E 2m ré’rzrl’l/ 2mr2l/l (47rgo)rW_ W. Prove directly that the ls-atomic orbital W100 : R10(r)160(62¢) satisfies this Schrodinger equation and thereby obtain directly the energy corresponding to this wavefimction. (cl/Log:- A e‘ZE—f “Wm A:(%3Yl 0 rvrsjc w.‘ °b$evv¢ A L:L 1*!“ 2 0 suite L2 ‘ALPQY‘J‘S “4‘3 ‘3“ a (“‘3‘ ‘i’ Pay—Lied er\vo.+iveS l‘See. 3A?) ~01)“ Cq.‘¢,o‘) J P.3331 - (Sour eXL-(L-Fuveu+ia,\ efibmuti—Cm; C41,“ be. wY‘I'H‘eJA a”: _Lé_7: wt 3 +32}; 1L - ~1mE— V 3x" m X 1" mnst M - 7&3: film Z Jimw‘w ~2- legsu ': Ar‘C'e .5- -2; 2L Mew-A SLx—«éxyeao r qe 3r 7' : A -=t —:L'~_TZ ~‘2 ‘7 Iii-fleetime .2- le(\ “on 33%;? o “ :A{‘§+EV—é1{€-a€ so fif an -i-r I! = -21- %7- \- "‘ “it its} 9 ° 1 is LA... ((43.1;sz ~33“. «vfiike “a (CL. Cow-\flVMA-erfi a Menisc- we QMgQVV-e, *HLGI‘P 2 «me :— .___.~ .L 1.. ‘R CUE Rec} ab sure”. \a{.—\~ mud sick 4. m cumumud fimugiru. Can be, udTV‘A'th 1L5 {‘?§fl+ g: Elkoa +£ 'q’wo 3 “2"”; I‘m.» r 1. flor qo 0': i731 c.01- -k2 E = Jul _3 1m ca” —_ 2.7.2.“ 'sqb \ 2V wk. Maser-Va +9.15 Enexevfl 15 In advaamaurk- @144 $1545 29‘“ “DJ—7 when n:\. 10. Determine the probability of finding the electron in the ls atomic state of He+ between the nucleus and 2610 5 Does the value obtained make sense based on the ls—radial distribution function? 1% Cash PCB-9 20.; = g m Ch Yzo ZAD .21- " = “'3 *‘e “mu Y‘:0 a q“ a E - 7- “7’ P Loezmg- ‘4 :23 3314 L3 an ‘37-0 Q: - PfQ-bzafi- J;- ‘ch‘fia‘a :D Now we. use; fietemmexc wbaxémmkcs +° aL-bmim (2)”1I3yze‘ydy = — 418‘8 +1 = .98625 Uh ¥ov 'l‘Re LA. afi'bmlc 941—452. Tut: EL: wk.me reuf Haj“ C3193 % °bur-2.5 vamp: (be YCLXLL‘ Vlwc&e Carol‘ch have 35 T20 Q0 '3. "‘9 “(Mp $5 a Value 54’. He'm’l‘agvoul alched‘o lud— Shun/gr flak \ Make} Sense. 11. Find the radius r* such that there is a 90% probability of finding an electron in the Is atomic state at" He+ between the nucleus and r*. 'FCO-EYX\‘ T. 8 Pic AV v:o V" -ql- u = A18 *Ie 1° &r Y':° I "er- ‘Lr , Ms}. M 0.0 0‘9 v (o * 2( ‘3 L5 _ P 78:? W Raga % S “age “t3”; W‘Rw‘:fl* 0 5:0 me n “V - : "I; S “fie "34"} : b-QG ‘3“ View we use $C¢€fi$\$lc Nevkpia‘gfi QM; Vow“? ‘1: m&;\ 'ifié “\“gv‘wm £185 \Emfi'vl% Ls {chad 4—0 0&0: 2% (2)_1j:'3223y26‘ydy = .9 [ :‘e—Y‘eprfi :1: “is a; ' II L ‘ qe I t" W .0 N v’ 12. For a H—like atomic species (H , He+ , ...) the radial function R(r) satisfies the differential equation 32 2 a £(£+1) 2mze2 2mE 2 + —- — 2 + R = ~ 2 R . (3r 7' 37" l" (471790)th h (a) Start with this radial Schrodinger equation and a guessed solution of the form R(r) = (l + br)e_‘" and find a and b such that this solution works. (b) Normalize the solution in part (a). (c) Find < r > for this solution. ((1) Find rmp , the values of r where the radial distribution function D(r) = r2 R20") is a maximum. C333 '93 vemava “Wide 2L2 Mawéemcg claw le/QAM S‘tée °£I+€xe wards-Luvs 4kg ckx‘ve‘r an ai%ngL‘Lu\e um 4&{ vigk’r srclt 1%»2 efiudc-«W (he. a? $an11ue has no Maandamcaan v3 , w-z Se-l- i=0. Merl“, R': egg Q": M‘R CL? ) AT?— {Mi +b_qtr}e'“* a R": {—1 (www.mrtyggh Sic} ~Q.\o +a—hr oak Sim- {Q2— 2119 +Q2'lor} I] N — 33R QO‘. €"+£K'+ :{az-ZQL-i-qzlavlséox v +};{~q+b—q\ov }€-QT +. 1%: { l+bv3 ere“- Clo: Chlrre‘Q’k1 me." Q"+ 7:; (’4. 2:: R: {oz—24b +4210Y' #2; +12 -2alp - \- v a Y + 33‘. +- 2.? b 2”” Q°T Qo Cuzcow‘nnueA‘W FM" 4&2 wald- kLnAsicle. all Hie echdr-ln +5 be a—C— HQ Lear-M Cankth CH-b‘r‘\ e’“ i we (BTDLLP’FE'Y’WS R“+ .24: +23 3 ={Cq1+alba+C—uqb+13n " “a Y “to 3; -- “3+3; +r( “ a} we see Rd- “qmb-l-Zib zu (I) Qo —Q+~L+%_ :0 C21 Q0 4-Way“ 81% CH (103 (19.. ccn+xnmecX23 C63 (r) t S VDCT‘\A.\' (:9 W 2 ‘2. —‘Lo.‘¢' U : Ade g f' "’ (haw-3 9— °\-T wH-ex 6L: i no 2A., ’° -10» " - A: S‘V‘g—Zmrq+azv~5.}€ 0K1- Yzo ~ ={ua‘}{.§£_ -943; H2 3‘. E cum (ms taco" " '— _‘~_\_ {g __ 2.02M} +129} 0L \B 7,2, 6% u -. i §zq- cm 4420} -. 15918} = E. mm; but“ CL ” 3 3) .% .e ‘ As Q (Lmflfle check ’ W 3% “kn 003$ , yESQ , we. have, <f>ha: “git—9.0 %\+ +2; [\__ QLQHBJ} 2 h 4%: ziécfiwb-M 5%. <T>10 - caqu E C \2. Luwanqeg‘g 3 WM? 3 a» 953‘ xvi;th AIDwa 1° Ax- ”:Dmc- firm é van—N" 2:2” “ma 0. e 3; 3C?) :59. ii ‘f‘2- 20:634- qz' Vq3e~2°w d’Dgfl 2 _{ (2.?— (p curl + Lice—‘63) ~20; [T1~ZAf_$-\-o.zv‘q)} 9.2.0“.- CU‘ m9 = V’ i 2— Coax +- Um?" TL '2. mv+qazv1—za3r3}€—mr Ar -2qr QDQH : 1v {\—- Klax + L‘ALY‘L ~q7’x—3}e Cir “mg 3D\u:\7l;rm$ Y=D m5, (=00 correspeml 4'0 W‘:W‘"M¢L M INN; “WM/9°“ 3 3. l.— uox +qq‘r’L-a Y‘ = 0 La? 310w“ m 1— M my “>5” We $c\v-e—\-C\CS numerxgglhzt WW: 9. (2.91%63\q° .2. ...
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Problem_Set_4_Answers - Chemistry 341 Physical Chemistry I...

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