exam_2_key - -1- 05 March, 2007 , Name M/ 650%? EKG WV!...

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Unformatted text preview: -1- 05 March, 2007 , Name M/ 650%? EKG WV! Social Security Number [42; CHEMISTRY 481 (Biophysical Chemistry) Midterm Exam II Instructions: (i) Take it easy and relax!!! If there are any difficulties or questions just ask the instructor. (it) This exam consists of 08 printed pages. Please make sure that your copy consists of this number of pages. (iii) Reference may be made only toa single formula sheet (two sides) (iv)Work out each of the following 05 problems. (v)OnIy one copy of this exam is to be taken per student and it is due immediately at the end of class. (v1) Be sure to define all symbols used and indicate your answers clearly for maximum credit. All answers should be expressed in SI units. (viii) Please see last page for physical constants and conversion factors. Good Luck!!! "i think you should be more explicit here in step two." 03-05-2007 course\chem481\07\exam\001\mw100 -2- Problem 1: (20 points) Consider an electron in a hydrogen 25 orbital having the following radial wave function: 1/2 R(r) = [2 — i] e"’2"° 8610 do where a0 is the Bohr radius (52.9 pm). Calculate the mean distance (r) of an electron in a 25 orbital from the nucleus in terms of the Bohr radius: /y 5" /6¢ (46¢. af7/AE 2S Ofé/ya’qf‘dvéyd/Lye,’ #0,.” fddéu (norMM'zaJ) Wat/(750040531 I'S'. ( / _ _/_ .._C ‘r/qu 77L”)9'Q(r) 576,02 (‘2 40) e ( hi/VC/UAJCfi-Ch) (HI’Zre 40-: 5.9.7 pm] (Eaér rad/Ls). 7/76. madam ao‘sfw—m <r> ,g (7/0“? 5; 74,75 6%, JIM}, um: 5/?”- <r> : ffitj r 3L3 é/y : fffzg: pyfi s/‘nm/aa/garldr . ’0 44 - , :: f q/-—’— (9-5 e‘P/aqo P[~/-’—— <3— L>e-r/Qdo Padr‘ 9M3 - 0 3403 do) / 3%? 40 0» fl? fag/Jae QM Of-fi}? quafm'aq/is 5mg;a(~€r¢6/)éew 7%? W M $ 3&5an hem/howls f; Who/Mafia?“ :><r>: L 00 ___,: :2 —f/.>4 1 :z (34% <2 402 << ° N 0* Sfls. _ .1. 9a ‘ g sf 4 —:r + r“ “M6 I“ng (40) 0 ( do 40:26 — an —r a0 ‘ 4f r3e /4OJ/" -1] r4{fir/0‘24",Ljafoch‘e—ré'oélr1 % 40 & 4° 0 J We (43¢ 14¢ ;0/¢ZrM Air/um: oz) / f x”e—q"%x = [4— 0 17+! 4 9M5. ' WHO}, fives; <r>:/:q—3>{4-35 - it 41 + .4. 31 ° (Margy 40 We )5’ 402 0/40) 4 — .1. - 4-é4‘7'_ 4—24 4‘ 4 3px;. . (3403)< 0 a“ J” W ‘9 > <r> = 640 -3- Problem 2: (20 points) Consider a hydrogen atom in its ground state, for which the wave function (normalized) is: 1 —ra W(r)=1}—3€ /° “0 a) Find the root mean square distance W)” of the electron from the nucleus, in which ( ) denotes the expectation value. where a0 = 52.9 pm is the Bohr radius. b) Find the most probable radius of the electron from the nucleus. flq. <1) The exp? afgq‘r’orr Ara-w of (r 2) [Sift/40 é7 0W») 04> = firm ” WM c wmfmma norm-2,4) =-<7’_'/—gda r; Z‘r/flo [CL/f slhade ___— - " 77' ' 277 3% = era/m / f a 0 0 a _ 0 H /\ 4\ a“ b W vi “a! \_/ I /I\ \ I C K; /\ p fl l 0 \_/ II A in» M \l” (\ \k 3, <r3> = 3402 :> <ra)"7- = 7"?40 Ma‘s : 7r? (57'7Fm) ‘ = 7/‘éem 1—— chsr) 5) 7772 W “Mama/Ms *6 mwimm ,» PM” d'rfifFI-blflh'm ,‘FUAJ 0,1,0»: 2 PCP) =- 47fr}/7I’(r)/l 369+; : +7”; / >6—2r/ao z 4. r}Q-3F/ao 7740? go? olPC _ , , QM5_ 4,. r) FZHL :0 $9 MAXIMUM 2—; A4057" prdédb/e rad/us i’ ; "9" 0/ 23:. ‘3‘ 40 E :o (403){r (€/>< do +C6/F/)(9~r‘3 r:r* Efls' "Rrael + ;r*‘ :O —4— Problem 3: (20 points) An electron occupies a 2s orbital of a hydrogen atom with the following radial wave function: L3 1/2 2 _ i e—r/ZaO 8a0 a0 a) Calculate the positions of the radial nodes where the wave function equals zero in terms of the Bohr radius (10: R(r) = b) Sketch the appearance of the radial part of the wave function as a function of distance: -44, g_ 7,642 rag/M wavefufia/f/M faf’at 5:71; aégmzh a; aria/7MB éaa “Mire? pw/i: " / : r, ~ _/‘___ - F/¢+ 40 K (F) gqoz ( do > g ‘ . WWW” (é) ' MM“ _ a _ 00”") A) 777‘)- N’ fl ’47 ’mjdgs 4’4 Where 9%? Ndue 7£L4N 7:; 222-:er . (MW/y 1% Cr) :0 as. r ~> «9 , W/vm'fi. one 7%; “6% i . [CUQSb/ZL (Sail/‘3‘ fig, 41‘. krflsr/ZZ, _; a ‘-‘ may Lug?!” I»; 00 3 2'» yxgbnfl‘x'c’w “57"” WaverNC/g‘m A; 73am) whgfi flaw/7" 2 /5 2am) => I r :2 qu (mad.ng b0/€> (ma My... WM @ x‘s J'qu #0 Ivar/240%} aims/W“. we can 6/42/06 2/5 farm ’_ W 7%? web/é zéwda'm why/wigs 7%er far/S a bade“ 47/ r: 240 ( 679a»; 3%};Oj #0") q’g {Md/‘MMZM/m mgw owéj {If/e4]; am/g‘fwb _; "5,15? ., 97¢ W4 thw (/7794 M7 %_,,u} and 1'3 CyNo/ed) -5- Problem 4: (20 points) Let’s summarize some of the key results obtained from angular momentum theory. Be sure to define all symbols used! a) Write the Schrodinger equation for 3-dimensional rotations of a molecule (in spherical polar coordinates) in terms of the corresponding Hamiltonian operator and its eigenfunctions: b) What are the eigenvalues of the energy corresponding to the Hamiltonian in part (a)? 0) Write the eigenvalue equation for the square of the total angular momentum (in spherical polar coordinates) in terms of the corresponding operator and its eigenfunctions: (1) Write the eigenvalue equation for the component of the angular momentum along a given (z) axis (in spherical polar coordinates) in terms of the corresponding operator and eigenfunctions: (5—4 h. 42¢) TAZ 'SCA/w'ldhvigf‘ figmra’l fdf‘ ?'d’"’“°h5/'0’1¢( raflZ‘mw (gp,}5_) /5 . A H “7‘ = E 1r A; (I) z E (V) 31, v if (9,42) 1XM[(6,¢) fi K M Man/— v _ r 4/ gfififfq édfl‘gmwr‘ \ .Sfl/fief/éo-g AW¢5 (rm;- ‘5) 75¢ 2174th 0% 1% Away; awry MAM/owmg M 0717,01 M I. E1 = X/IHZ :2 1:0/ /, j--- °<> (anju/m—Momm/‘um 2]: 7WuM' number) & Moménj «f /'I7¢rfiq ($6.) 4) 7})! XML/due €7M’é‘0h 741"“? *dM Maid/4f“ MOMMéM S7uotN-d1'1: fl) EWIQQ c»): 14”“)‘32/ (694’) me 2 “‘6 R édrmom'cs f : a) 0;,— -_ 00 (mama/aw makime 7:4:52Mr) (Fly/s-) FF,» 6’? mean” of '6’? d/hzu/d/f Mdmcjhéufl mfld‘n at, swufl/e C 2-) 424;) fie éyznVa/«e e7uavf70n is I. /‘ M (I) a 4 low M! M2 Mg = tf/ -,?,L/)_ [-1) in? (angu/ar MarvehA/M ~/n>‘/leoyé'm 70MHM nuuéen) -6- Problem 5: (20 points) Let us consider the familiar 2px, 2p)” and 2p: atomic orbitals, as in the case of the biologically important element nitrogen (N). a) Sketch the boundary surfaces for the above p—orbitals, being sure to indicate the sign (positive or negative) of the orbital lobes. b) Derive the mathematical expressions for the angular functions correSponding to the individual 2px, 2py, and 2pz orbitals starting from the corresponding spherical harmonics. Hint: Yé”(§2) = V3/4Jr cost) and Y::)(S2) = $V3/8n: sin 66*”. . film-of 4"“ vo 7/42 bLM‘Q<$‘)«"Y* ~ " .4 Pkg/MB r3“ (J fa 4kg: A; 2) 'V (a $17 P I; X; <?“Py} M Q-pxp; flré/ylofl/g. W. 0 (c gigguj .2__,_-_!§ {5,4 fm any 07/ 5,65 onJ-rrj rt" g/&N&MJ_$ swaA C) N) gunfl‘ - 60») ,1: 7lhe bout j; Jig/face; b) NOW (aw/>12; 7’50 *1qu !,>."\2L: Lax/P 47L "n76, p~orb¢¥als £445 £55) W4} W W'{;¢f I‘“ 754— prb‘feoHdhs of ‘f/kJZ/f «nag/0w Mammfum m3 1 'l} FOI‘ (71, (:0 Mill mi: 07 fhws anc§ulicjtf [KT-MEI)"; : C, - (5%) F2 oz yj’zcaf-p) of C0569 0< ?/f‘ (Shae, Z: (2.05 <9) For [9x 4nd Pg Wz— wn s/cULr flog», ffiafi My! XNQIEW‘NWL; ,0ij 07C (7%? camp/d JEUN air/2M 6 .r Cl) C!) Fx 0( m{yfl (3,45) + 7g, (6%)} o< sine (6“pfe“‘p) 04 siné COS 45 (WA/NZ; gulf): cogcfifi Sz'rz<b) M X/p ($I'Ncew x: fsinacoscp) “V” W A I“ W NW) + 7.3” (6%)} A 5/37 &5;q 45 (Moi/«4 {iflé :1 cogcfit : sir) (I X fi/V‘ (§/'bcz :1: FS/PheSl-nfl _8_ PHYSICAL CONSTANTS" (ROUNDED TO FOUR SIGNIFICANT FIGURES) Quantity Symbol Value Speed of light in vacuum (' 2.998 X 10“ m s" (exact) Permeability of vacuum 110 471" X 1077 N A‘2 (exact) =- 12.57 ><1()*7 N A": Permittivity of vacuum 60 1010c: (exact) e: 8.854 x10“2 C2 N" or? Planck constant 11 6.626 x 10'341 s . 17/277 n 1.055 x 10*34 J 5 Elementary charge 6 1.602 X 10‘19 C Bohr magneton 11.13 9.274 X 10‘24 J T‘] Nuclear magneton [2N 5.051 X 10“27 J T71 Rydberg constant R7, 1.097 x 107 m7] Bohr radius (10 5.292 x 10~11 m Hartree energy Eh 4.360 X 10'18 J Electron mass me 9.109 X 1073] kg Proton mass mp 1.673 X 10~27 kg Neutron mass mn 1.675 X 10‘27 kg Deuteron mass md 3.344 X 10‘27 kg Avogadro constant NA 6.022 X 1023 mol_1 Atomic mass constant mu 1.661 X 10‘27 kg Faraday constant F 9.649 X 104 C mol_1 Gas constant R 8.315 J K‘1 mol—1 8.315 x10‘2 Lbar K‘1 mor] 1.987 cal K‘] moi“ 8.206 ><10‘2 L atm K“ mol_l Boltzmann constant k 1.38] X 10'23 J K~1 Acceleration clue to gravity g 9.806 65 m s—2 "The best values and their uncertainties are given in Appendix B. SOME NUMERICAL CONSTANTS AND CONVERSION FACTORS 71' = 3.141 592 65 2.54 cm inch—1 e = 2.718 281828 4536 g 1b" In x = logx/Ioge = 2.302585 09logx 4.184.] cal—l (exactly) 101325Nm'2 atm—1 1.602 X10‘19JeV'1 105 N m—2 bar‘1 1073 m3 I:1 1.013 25 bar atm’1 133.32 Pa torr“ ...
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exam_2_key - -1- 05 March, 2007 , Name M/ 650%? EKG WV!...

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