exam1 - 18 February, 2008 Name Mlaoéflflg findwm (KM!)...

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Unformatted text preview: 18 February, 2008 Name Mlaoéflflg findwm (KM!) CHEMISTRY 481 (Biophysical Chemistry) Midterm Exam I Instructions: (0 Take it easy and relax!!! If there are any difficulties or questions your instructor as (ii) This exam consists of 09 printed pages. Please make sure that your copy consists of this number of pages. (if!) Reference may be made only to a handwritten single formula sheet (two sides). Please see exam proctor if there is any question of what is appropriate. (iv)Work out each of the following 05 probEems. um - - w “H'm‘xxmmm’aflm axwfl<wflww¢wama a, (v)0nly one copy of this exam is to be taken per student and it is due immediately e at the end of class. ( (Vi) Be sure to define ail symbols used and indicate your answers cleariy for maximum credit. All answers should be expressed in SE units. (viii) Please see last page for physical constants and conversion factors. Lots of Good Luck!!! "But you can‘t go through life applying Heisenberg's Uncertainty Principle to everything." 02-1 7~2008 courselchem481l08lexam1001lmwaO Problem (20 points) The quantum mechanical operator for the linear momentum is given by 13 = —ihd 1 dx. For each of the wavefunctions indicated below, determine whether it is an eigenfunction of the linear momentum Operator, and explain briefly why. a) The function y(x) = eikx 13) The function y(x) : 3%t c) The function y(x) = 6””r + e‘“ mW,M«maimwmmmmmm (Q p715-) S'pi‘s, Spa/“S. A, film affirm-giggly agaimfifieg an $43 egg” fay.) $52293 7‘6) yiexis’ #12: prpdfwaf“ mt flea; izéfsawyz’w; 53mg! :93 ariagvemr'm ofvfifi: 4 ~ {5 gr]; _: 5}"; a j r_ 5"“;gwifcmpné10fl ape/‘cxj‘or \ e - V6? 35%: 66) For 7%: £237 €63!“ Mame» 23W; fifflffifigc’r #7234, (am) ,__ *4; (wk) 6"“ x . 1“ 'é‘gfi}? 55:3 r: + 1% a“ " a." Q‘Iiiléx W7 afiehfa”o§§ofi 0?!- S/F/‘er NZ fl“; £53K) gaij-d7LQ W333wa (.efi'emva we) dmd/ 7’7’7e «930(7) enffwrx) cigm «5) “513 £1; (6%") = #52 («4%) 6‘“ 4% e -¢'é>< = (07%;? 6’, ’ = ‘ #3:? ewékfi ' (“‘éx 75 am ggégfigm'f/qn} (517,071 57:70: we déffiifl! fhfifl/ddjyfil— Do of %a éiyfith/ua m1 fie eienyé’wfdfw 6) “mi (afig’éx 7L awgkle : we“? (fifefiéx - céeflféx) dx I (Jflaaé) (fig/(x _ Q—JkX) ; 1Lé$<€ 145%: _ a-Jéx) é/gdrgfg%akx 2L-€_--6k5< is H Z- a done‘ 0&711’1'9; a? can SW(W gtzfzehva/c/e) mwfifigyfly'7 W76 54! Ne fwualfon , 44-» Zé/ww7fuwwé'on 524cc; we. ,5 E i g E is é .z E § § E: E i :1 g 3 g 3 g E i Z i 2 E E “3” Problem 2: (20 points) Let us consider a deuterium atom in its ground state for which the wavefunction is a Ina—mo :riza0 w(r)=[ 3 where the Bohr radius is an 2 52.9 pm. The wavefunction is not necessarily normalized. a) What is the average distance of an electron from the nucleus of a deuterium atom? b) Calculate the value of (l/r)“: (2) Calculate the value of (13)“3 for a deuterium atom: L I 4; fig? fag as gfi‘ifiv. Q, A y «m! WK»; v z (A. at [1 (x J ' ,w “1.x. .' w- ».n . . “‘3ij g" 3"” 5 ‘ :7 T» ff m5 I ‘5?" 5 r7 (fiAéK-ja/j { \I f r C £6244; _) " “9:: If: 2;: l I' 2; ":32" M sfw‘g‘ av“ ” 055 b _q 6;“; f5< “ We? c) 0 oz) \. / “3’79’53 Off 27%3/ C) )(47-7‘) 2: ’ / Z7403 (2/613)? «a? ” z 5‘ 1: 6.x" / ~23" / wwwrifg ,1 _1 ix}! 90 “) <r> 2 __153¢(r~};5‘ r“ VIZ/cm flak/W .- Vz) —- .---’1. ii / Wm; 5/ :- ~29 ‘3 H; 2-: (I )(fiflégf {fix} frag? )3 3m a. _. 90 9(2) : / 427" x" W's 2 .25., r / r _ -« (76/6?) {1‘65 ) r Jr 9%.?) WC” f 36 V40 a? :(“L 4:?" E! #70”) (3470)?" c? :x”! 404-- L772:ng zé c9570 - 3 - 3,22%}. — g: (53'7/72913) <f‘) : 7?"? 52m L... 9% W.) 13> <M~> =- [MN u;— WWW oz) ‘ - / ‘27 _ 2M. 3%”? 4—7- 530) r 6ft" 2%gj4—lyfruz 2%aJ/V, o * - .1... 37an (rm/4&9)?” 40 __ / E E E g // _\/4”,¢7":)5-{ a 3.?3 E L’Hjfi“ (.l/Q'a) é :33 d 242‘s ' 2 [5‘ (5‘9, 5? [20227) 3 (p3) : /, flax {a 5/9,?)3 I cleats. <r~3>/3= 3? 2 be £3 £2 g E 2 Z 5 a A i g 2' :5 s Zg g «I’m‘1'“7"XQa‘W-fimwfiwvwrr'-x"Aw-x'A'A'A'A'A'xxxxxxxxmxxmmwnw‘xuyasnx m!mvm:m¢m¢wxww Problem 3: (20 points) Consider a particle in a 1—dimensional box for which the wave function is: 111,100 : %sin(£§x) a) Show that the above wave function is an eigenfunction of the Hamiltonian for the region between the walls where V = 0: b) Using the result of part (a), what is the energy of the nth level? mmm";xmmmmmmwm‘mm wpflvwwmwwm “ms; mmmwnw Wm“ H i Wigwam» «wpa‘x‘k‘x «ism-u-m-a-rauzzxamam-aw Warm-aw umqq-m m (MWS‘) 551*} 72’ 55"“ W 7‘32 (0 ’1' 5”“ aye/174M» stflmME—a g/igraaiom w? ram 716 556,64” may; #7714} 2 5/? 7?», H’”fl,93‘”fvfi¢4‘w I’L— r~—~ k- ' '. \—» 547%.;«3 :; _ _, ZZW-Ziéfl/W ( 3/4931/0‘59‘2 $656?!” 77’6 #flm‘g/fi’mfir‘ 5,081?” 2': (71/3/4377 : 43:52 3 z-z-zafi %/ amVTLv 9mg); 7/5Wér/0re’ we need #27 flbXQ 7‘66 27v! Myavfipfl df @406): #7? _ .3. ’3 I77 [0 X é , d2?» .. —(m& *2— M {ZZ 5' “nix [#5 9K3 ‘- L) (g— (z— HY L. IE I 73 P Q N /\ fix) \4. w\ \n S: F\ is P \i L #— zany :9 W = iiw/vaez" (€er * 7/»: 6,02%- : giii) 7%” Z? 34? (>0 53% féw/M’m Km § *_ (fig/*5.) 3“ «51) 725.6 @W t . F, ‘* r \ 39 “y: fig 77% ’Zgfi‘g/ ’4 ’5’”?%/ 7456 @/ grit/ajwrfl ééqéfifW'fiw ,764): / 92947;,7 ,174704 = 5,771,709 #5 550100 zed/6V? (at) Z mmmwyflxwmmhm‘firf‘fi'asvé‘3mif;$.23:3&1!.££$1t.11a'1iwrKlzrzt‘t'wit'acmy.923sza'mzra'dm'mwmm wwwmwmawrmmwm'm'A-m'n'xw‘ms -5- Problem 4: (20 points) An important pigment molecule found in plants is fi—carotene: \\\\\\\ Assume that the electronic properties of [3— carotene can be considered in terms of a particle-in-a-box, in which L = 2.0 mm. Given the free electron model, what is the wavelength of light absorbed by (by V” carotene? ,. .. Mm“. mwmnwww...w th'mmzwmw Wmmwwwm M521,” 4% “fire? “539$ gi‘i‘a’rm Mafia? {Emma W fiwffi “5" “g”- “‘3”ng are!“ firfiu" (7.4: '55 m mic. frfiw$3yfi '67! WWW“; Ara/6%), €354, fadéflré’b’pfll gawfiwflmgg/‘W‘zw. Wé-fi @efiz’ny /€V€/s mfg/Ryézfi 6;“ E = L 22 = / .2 e N”, w a “gmé‘a‘ ; ) l / mgrgérgm geFW/ga éer’weea $4? fw— afi’gww (Que/{g xiiiwg: Ag: Eni/ FE»; _ .2 9 A; -— fl "~29 m [(n ) : Jaw“; : 2/774]! 6 ( 8.924% $254446 7’62“? M é/cwn (y) Jami/e 5.44%; j-m me a Vfifflaf 52:2 7/” é/Pa‘rm; wé/ofi aw dwrwéfirfi Mg! acmsz :5? WW aw /m,% 07% /M%b(, {Ida 54,5, fl¢(av?/.2Ly! /a W flame; flmflwiaja/ej {:41 gfasirms arm a“ W /QW. 15/ we 1659“? W; “7? 570,45. _ ’i jgf 7L 5 m2 4 71/ 7L fl =7 "—— T/v v.5, Czar/26K AE:£.E.: A grad—9‘ FMS- $ A : gngc (20140}; ‘ R ‘2 _ — .___ (s)(¢.//owo“?£és (ammo in) (3-mx ’0 MK) 6”)" [Qflf/JQJRQX /0"3‘7‘\7‘s) ———~ “m -3 "v? gas/Es : 3-. 723 7 ><’ /O ’97 é‘pfig - “:5: 532+ am (flew?) ("U470 Mm - fyprw'mc/yf) WWM‘WMMHW.WW..WXA1 Hymnme man,» «mmmsxonvwww -6- Problem 5: (20 points) Infrared spectroscopy can be used to determine the types and amounts of secondary structure that is present in proteins, due to the influences of hydrogen bonding on the vibrational frequencies of the peptide bond. Let us assume a quantum mechanical harmonic oscillator model for the NmH stretching vibrations of the peptide bonds of a protein. a) Calculate the wavelength of the infrared radiation absorbed by the N—H bonds in a—helical regions of a protein. Assume that the force constant is k = 640 N m“. b) Calculate the wavelength of the infrared radiation absorbed by the N—H bonds in non—hydrogen bonded regions of a protein. Assume that the force constant is k m 680 N m"‘. c) Explain in simple physical terms the influence of hydrogen bonding on the N—H stretching frequencies of proteins: (Mp. A"; ’54)...“ v" 9 o e» .5.- ifc'ilffl; fimfiwwa oscsfifiiflf‘ W XEWS. Wfixfi’fi é” :1 a, = (Mam 3 a» = 455m w w : {Egng m 6—— mW.;... 9’3’ éydwfiin rJfom (9% 746g .seme éer‘weeq &? /€v€/£s;7mfim %: A? = Eyf/ u gy' = [(y%/+é.)-(V+é)] fi‘od Wisteria 45' = 30.) it; 3 #:7/2‘ A 7;; m {9/ .J K /\ z: arc {geiyflrce tdnzi‘ae'fis‘y‘L fl) .2}? 7%? anyway": W“#;rauégg fr; Muée/o’éea’) 2% == 6%? N»??? g A. I (%/7)(fia??zwai%$_i % 4&5 MAW“; g‘ __*_ \éa m; 6‘s 2 3,5145% /o'"ém a‘ '2 ,5) ,2}, $541? {saga-m“ WK ,6 1-“ égo xii-532‘?! “a; 54; ~ 1 .. L A :f“ m; : W7 2‘ $3 5,3 x [/4é773A/0 g)? A K g X/ %S) «.636? A/m" xvi-9,2? m Ayggtffli/«géa $5?) 695%; /‘/“‘ flj/zfiégvfif :: mmmammwamma \ 4f 7506 ova 506% fig??? J 18$ KKW for éygfr’fijfign bigmg‘wg NT/f 5/“0 was % y/‘éflflgzflfimég fleyuafi {g Q; fig 3 5 Ag/ .41 fig 59”"? 613 #707465 57574;“ as Lb EXW 133g ygiw/ggw f4, A55; (gar?) fisfiwi/EL‘ 736a (1975/ “7.673%; 23: a: as fiéaf 26/ mm eejfwwaq 7i w c? fidm 5 /fn 574g”.ny as!" (be/Lug: ad: 5/5,,f-i3/wfxfli :52 6 Mi? :1 7% a agnggfa w‘zfim , cf E E i E i E 3 ...
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This note was uploaded on 04/02/2008 for the course CHEM 481 taught by Professor Brown during the Spring '06 term at University of Arizona- Tucson.

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exam1 - 18 February, 2008 Name Mlaoéflflg findwm (KM!)...

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