exam2 - 10 March, 2008 -1- Name Mia/74d firm/on (Lee; )...

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Unformatted text preview: 10 March, 2008 -1- Name Mia/74d firm/on (Lee; ) CHEMISTRY 481 (Biophysical Chemistry) Midterm Exam II Instructions: (0 Take it easy and reiaxill If there are any difficulties or questions your instructor as always ready to helpil (ii) This exam consists of 09 printed pages. Please make sure that your copy consists of this number of pages. (iii) 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 probtems. (V)Only one copy of this exam is to be taken per student and it is due immediately at the end of class. (V!) Be sure to define ail symbols used and indicate your answers clearly for maximum credit. Al! answers should be expressed En SE units. (viii) Please see last page for physical constants and conversion factors. Lots of Good Lucktll 03—1 0—2008 courselchem481l0819xaml002lmw100 "I Dom HAVE THE Test 1 saneouteo FOR TODA‘C MY Doe An": lT. “ m « wx-ixfixwvmrmn.m; . he g g t Problem 7: (20 points) You are a quantum chemist interested in molecular dynamics simulations of proteins. Consider a harmonic oscillator whose mass is equal to that of a proton, with a force constant is 855 N m'l. By substituting deuterium for hydrogen it is possible to shift the vibrational frequencies and thereby investigate protein secondary structure. What is the change in wavelength that would result from doubling the mass of the particle? vmwmm /« .6564!“ .ég J; (/QMS‘ 4 1341 M’ 72% A! 5,92%- a» 5/242- :: 9772 *7/97’ . ggaé. 2*: AE ‘ r: aSc///Mr' ; ,él‘ 4:"! fidfce 9/7 h“ (“E— ’79455‘ . - I g _: arc. “1 h? :. ‘/lé?3>< M”? 4 (2,?) (2-?7é2f/oj:2.434§w 'fi? ._z$r,. . . M W.Mm_vu 3 2m (Xduéé M5194.) 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Consider the spherical harmonic for l = 2 and m, = 0: Yéz)(9,¢) = 1(«ms—Injkosz 6 — 1) Show that the above is an eigenfunction of the Hamiltonian operator I? for a free particle on a spherical surface and indicate the corresponding eigenvalue of the energy. .....m..m".MMMMW..WM.mwm-mmWmmmm A" We. Wm M AxxA-IMVmb‘kwvm'lwewazmi’rfi’flx"< 474‘ 7/241; f5 x/Vyf/f fang/301?»? : we né‘éd 7‘6 :‘M‘zaw- W .fifimlvgfeb 7L (free pdrfi-C/e) m Wfiéfe #(éfip) .2. (9})(9lqb) 3 >/(>)<—f2..) 41509) 33) 17> z 1.. 93> .1. 9— 5,972 r 2r} r + r} g ' 7’" : I R g: / )9 « i i "—4 "‘ Mgr/v9) + (3;? 19' 51/7626 ——m p a . . f’ I‘Ja/e W your fisfrmjsua bat/.2. been Iveran m fizrfimfg yj”(.x;25 ; £71 (Tn/z 1.23%! 2n 5 M W4 054/? ‘52? "a/pdrf ” :97 17‘; flog”! 5e 1 can siI/Qrc’c/z 3,07%, #32 \{/ (5/),6‘ % 4 i (3694.261—1} : E7 (fry 5/05 I 9i fizzy / , 6/ . , _. < 1/5.. —-— {509:9 (6644, 9 (w- 5/», _. E x > /é?.7‘ 46 ) yd _/ / _ f 5f, " :1 é) _—_- (m)( )L ) /E77'%[5/a 6644. } E70 (’6)( )»/_:.§:. 1 Sl/«b;6(_5,',? a) 7L m€(25;0(9¢a;6)} : 3,97% Ki?“ gfig- :> (—9 JUL/#97:; wfiw‘e-O = E? W 7b2reforQ) ( 5‘ (a) 9-) 3,97%, 4 a.) _ {/5 6/ (p z E (a? 4) Varyifi‘canIan’A’L/VO (45¢) “(kw/“3‘ yo ( ) ,L >/d Q ) air-Er? . M 9m refilwfi’m («J/74:13 6 3 1 . (C ‘ VW) (5*);Wj’bh gfis - E‘ :: ___._...__ .__._n -5- Problem 5: (20 points) A biochemistry student is interested in the quantum mechanical harmonic oscillator as a model for the secondary structure of HIV protease, an enzyme involved in AIDS. Calculate the following observables in terms of the vibrational quantum number v: a) The mean potential energy (V): 7/ \/ b) The mean kinetic energy W): (Hint: V(x) = 5-ka and E = (T) + W» (/fiD/‘Sob Q1 pfs . JPJL‘E, 3/7716. wig/h q, 5:? a) '70 ___g,¢:/ c2 giigi’giifl iii/£3:fikflfiéfiflfijfii‘éi‘gfiéiqug .131 gmwggm __w___ elf/3: 7" flags? 2‘0 __ en: r Fm M 42%? ° __ r 7_.- A - .. __ file/9,2 xvi-"p.131!!! 345- ;. WM -_.- _ mm. mm.“ M Gawain g ?’ a? g ; :(;:L»m*"“:mmmm::mmur A M444vV‘A.WmeMMWmmmmnmm..mmmmm"m‘Mmmmmfim..mm.wmmmmmmmazmza‘wqmwmam. (chi, 397‘s- -__..MM w WWW mflg K72_E_._§ymfwlf _.,__ _ ..__ ._ _.__ ____‘_mwwefg_é_dmézéwh_z§fifiWLM -_'—~--~ Quantity Symboi Vaiue Power of ten Units Speed ofiight c 2.997 925 58$ 103 m 3"] .G 8.314 47 10”: L bar K"l moi”1 8.205 74 10—2 L atm K4 111101“1 6.236 37 10 L Torr K~1 11101-1 ............................................................................................................................................................................................................................................... .. l mc 9.109 38 10-33 kg m 1.672 62 10*” kg 1.674- 93 ' ............................................................................................................................................................................................................................................... .- :xmmar ‘ ‘ Magneton Bohr MB = @231/2»:c 9.274 DE 10'“ J 1"”1 nuclear ,uN = {ah/23w}, 5.050 ’?8 10”” J T4 gvaiue 33 2.002 32 Bohr raiding v K 1 I a0 = MrsuiiE/mee2 5.291 77 10‘“ m ................................................................................................................................................................................................................................................ u Fine—structure constant a: = Mela/2h 7.297 35 10“3 a“; 1.370 36 102 SB 11 1‘21 1211011 8035 ant vai‘tational censtant www.muW-wwwmmwwwawfimmw *Exacr value Wm amak aflgfi‘zamfi A, (1 alpha H, r; em N, 1' nu Y, D uJP'SHOn B, fl beta 9, 8 theta E, 6 xi (13,915 Phi I", y gamma I, 1 iata 1—1., 7: pi X, x chi A43 deita K, 1c kappa P, p the T, I,” P3i E. 8 epsilon A, EL lambda E, a sigma 9s W omega 23,4: zeta M, ,u mu T, r tau Wathemaficai geiaiiens 3.141592 653 59 2.718 281 828 46 TE: (-3: Logaritth and exponentials lnx + my + = inxy... 111x —~ lny : In (x/y) alnx : 311x“ lnx m (1n10)10gx : (2.302585...)10gx e.re)’e:___ : ex+y+z+m eNJeJ’... = ex"? (ex)a = aux em = cosx :i; isinx Taylor expansions fcx) : Z i, m ":8 Lac—a)" ex:1+x+%x2+... In)“ =(x—1)—§(x—1)2 + %(x—~1)3 —%(x—1)4+ In(1+x) = x —%x3+~§-x3 3 1+3: Derivatives d(f+g) = df+ d3 d(fg) = .fdg + gdf m=1—x+x2... ... 1 f g m iféfc: d: dg d: [92] {a-x] [Q] ~ “i fix: 321,632.” (By/ax): = 1/(ax/ay): dx" _ n—l a; .. me «i ax__ 0:: as _ ac dlnx ._ __I__ (ix x Prefixes z a f g; zepto atm femio pica 104‘ 10’“18 16“[5 i042 I: name 10‘” El micro milli 1041 i'fi I 0‘3 {I semi I 0‘3 _ integrals )1—1 Ix’fix = x “+1 + constant f§dx = Inx + constant 0° _ 1:! f0 x"€'“xdx -- Efim Isinlax 113: m %x — (715a) sin 2m; + censtant sin (a — b); _ sin (a + b)x 2(a —b) 2(a + b) Jsin ax sin bx dx = + constant if 33 at b2 2 0 _.:z erfz =Egjzeydy useful reiations At T = 298.15 K RT 2.4790 kJ mol‘] RT/F 25.693 mV RTID lO/F 59.160 mV kT/hc 207.226 cm—1 kT/e 25.693 mBV Vern 2.4790 x 10‘2 n13 H101“ = 24.790 L moi—1 Conversion factors 1 6V 1.602 18 x 1049} 96.485 Id 11101“1 8065.5 cm”1 i ca] 4184* J 1 atm 101.325’k kPa 760* Torr lcm" 1.9864 X1043} 113 3.335 64x 10‘30Cm 1 A 10“lflm* 1 T 304 G* i L atm 5 101.325 ]* 31°C = T/K “$73.15* (*Exactvaiues) Unit reiations Energy 13 : 1kg m2 s"2 “—" 1 AV 5 Force 1 N w 1 kg m 5‘2 Pressure 1 Pa = 1N rrr2 = 1 kg m" 3‘"? = 1 J m—3 Charge 1 C = 1A 8 i’otentiai difierence IV = 1 J C“ ‘»= 1 kg m2 s‘"3 A“ 1:3 as; k M G T deci delta kilo mega giga tam. 107‘ 10’ 103 10" 109 1953 E. 5 i E E, g g E 'E' mam-mmwwwmn-vsmmumm "wave mm. WNW“. wwwsw awmmmw:wmam-u-mvuu waxmwm-wwm ...
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exam2 - 10 March, 2008 -1- Name Mia/74d firm/on (Lee; )...

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