exam2_solutions - Chemistry 354L Physical Chemistry II Fall...

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Unformatted text preview: Chemistry 354L: Physical Chemistry II, Fall 2006 Name UT EID Nov. 3, 2006 Exam #2 Note that the problems are of different length and levels of difficulty. Also each is worth a different number of points (Total=100). Scores: #1 #2 #3 Miscellaneous information: 1 ev = 8065.5 cm‘1 = 1.602x10'19 J h = 6.625 x1044 mzkg/s (or 15) h = hr’21t, )4; = h, l= dsinQ Ephom = hv= hex“). = hcfi = ha) Ax Ap 2 71/2 p = I‘I’IZ = w’ 19 (my) = I 111‘ q! = 1 H = -(fi2:’2m) 32:23::2 + V H: P2me + V, P = 4a axax H lPM) = En ‘Pn(x) H ‘P(x,t) = if: B‘I’(x,t)/6t 1Pn(x, I) = ‘I’n(x) exp(-iE,,t/h) (A) = (WWW) = I will I‘I’) = Z I‘PnX‘I‘nl‘I’) = Z 0» {‘15:} c" = (warp) = l [pulp dx (A) = Zion/1n where p» = cf Plane waves: In =A ei’h, El = k2h2f2m sin kx = (em — e"‘“)/2i, coskx = (e’b’ + e"’“)f2 Particle in a box: 111,, = (2a)“2 summit), n=1,2,3, En = n2 if 7:2!2mL2 = n"- h2/8mL2 Emmy = (Elf/2m) (ansz + nfofi) Harmonic oscillator: viz/(215:2) _ 4J2 11‘"=[2"n!:mx2] Hn(Jc/Ax)el ,n—0,1,... Ax: =(JE/hyi E” = ha) (n+I/2), co = Jaw—p Planar rotational motion: H = L2f21 = #le azraaz, L = on ma mm = A: em": Em = mzhzflf, m = 0;] ,:2, I= 2” r02 p= mima/(mamz) L = m Spherical rotational motion: L2 = 10+ 1) n E. = L20; c) Find the energy levels of these stationary states. OULY we 000 n 5mm writ/WE En = ’KQ(V\*'Q\ “2 (33:5: 92" En: huCZ-«dgj nzolltifn d) When light is used to excite the particle, it gives off radiation as it jumps from higher energy levels to lower levels. Sketch this spectrum of emitted light? Could one tell that the potential is half a harmonic oscillator from this spectrum? The” LEUELS Me" spawn 135/ 455: 2.1 kw S’ECTRMM : JNTENSrW tn.) 2 H- b #30: W15 Look; 31457“ LIKE HIE" spam-1 oF H Face OSthen'me um 59 oamxm Fié'QuékJC'r OF 2d Problem 1. (40 points) V A particle of mass m is confined to a potential given by V(x)={k_x2/2 for x>0‘ no for x<=0 21) Write down the Schrodinger equation for this potential, and state the boundary conditions. 50426 0mm gamma-N L 1 2:: )6 Yea + if;ka Hz) : E 100‘) (Savanna: convinces: WOKSO} : O V’bcfi‘w} = o b) What are the normalized stationary states for the particle in this potential? Space” ‘f’stoX =0, use" Cm.) one? Hnue’ THE 000 n S'mfP'S, cam-c A N006 It)?” X=O .; _ z t It’lz‘ntl’rx'cfe'l i H.060 a "‘0" F01?- x20 0 Fee x<o Ya (K) " Problem 2. (20 points) A diatomic molecule has two particles with masses m; and m; separated by a distance R0. The molecule is confined to a plane, but is free to rotate and vibrate. a) When the masses are equal, (m 1 = m; = M), what are the rotational and vibrational energy levels in terms of R0 and M? u-F-I wwan Enzmcmm “455* W :13 ESL K '(VH’GA W733»): EM: magi/LI (a) “1.411- E-m " /l‘[email protected] b) If m; >> m; and 1112 = M, what are the energy levels of rotation and Vibration (in terms of R0 and M)? M: mt = 1' MM- Pl my“ .__..M .H. FOR m, >> m1 Ull'san'noad'- E5“ = {1% (“WEB r MK." ammo = E...“ = ’zma.‘ 4 c) When the molecules from (a) and (b) are in the same vibrational state, which has more energy? Ulgmwz ( Eh“): 'L “2'1 (WU/a) > tEVM (VHsz ' En“) {2013mm 1. Bf“: “11*. > "1‘12" 2 "516) (2.” JUN“! “(name (a) emu “me: SHQLCL’YL fieldermo H855 st Home“ EMMY d) Classically, a diatomic molecule rotates about its center of mass. Sketch this classical motion for the molecules described in (a) and (b). (a) “MT-ml. / F01 “"OLE‘WM (ml, 00%..) :5 Man“ the" Loom of: THE‘ MOLFCMLE'.‘ gums noLEcaLEf (to) actions 1%va me: new mass m, . 5 Problem 3. (40 points) A rigid rotor with moment of inertia, I, is confined to rotate in a plane. The ground state for rotation is Wow) = 1/4231". Each excited energy level has two degenerate states, which can be written in exponential form, or in terms of the sin and cos functions: sin(m¢)/JE cos(m¢)/\/; exp(im¢)/ 1/27: We” = f =123 m or m a a a or wmsinioos(¢) = { a) Which of the stationary states, 1110, ‘I’mup, and Wmswm are also eigenfunctions of: (i) angular momentum L and (ii) L2? (i) L“; = 4+. 6/3“th L‘s, = -t‘-57£w(5m) = 0 1: o a 0' *0 2 0' We L Ken : "‘1; 34¢ exam/m L111:le awéw ethi/m = 1 Mk stem/J3? ‘ Alma ezhfl/fl z 1’ hi; PHI?” = L1“: tf/KEH' t, L “15%” = “it a4» ’fi’fl “’44:.le " 'l-‘ht VII? G's/"Sad (W) L‘ a‘”"°‘= -t" 57w“ ’3'? 5%; (w) t'LLm‘ VIE? Sid/£45 (we) : 13 m 1 q/hw/us It a '- ‘1’; mm 443' ME EIGWFu-lcnm-JJ OF L M L‘ ‘f’ns‘m‘ Hanna mawwowms oF L} our do?” L. w 6 b) What are the expectation values of (i) L and (ii) L2 for we, the Irma“, and the 111;” states? tutor. rm: Bé'Sf-(LTS OF («a3 L. N.) = 0W.) <‘f’oid‘f’o> =0 tilts) = o W.) Gentle.) =o LWF>= mm“) Oman“) = int L‘lt...*'>= m‘tn‘”) -: <+..""’It=l~t..°"> = new} I“. < Vain/H5 {L- I Vnshvas> = of V11. 5‘21] ' fits-L sac-Sfaéhax =0 L"! KW“) = L'm‘ mew") .:. (ea’ywlul flew“) = Hi" c) When a magnetic field is applied, the degeneracy between the excited states is removed, and the energy of the 111mm stationary states are shifted by (a small amount) (“E = BL, where [3 is proportional to the magnetic field strength. What are the new energy levels of the rotor? Sketch the energy level spectrum. THE me ME Elam Eastman OF L amt 616.:de} int on». man mm “to Frau , E...- “‘72-: I term Wt: Erato , SE... = zfimi M 1m ,. mt d) The rotor is prepared in an initial state, 111.(¢,t = 0) = sin(¢)/1f:_r . In the absence of a magnetic field, this state would be a stationary state. With the applied field, the time dependence can be written in a simple form, using the basis functions, Wm”. Write down this time dependent wavefunction. ‘f’. (Gum) = ’31? 5mg) = Yum [e£¢_e—ca3 " [If/IW_‘+,-IEM'] Ex! 1 THESE ‘f'n ME mnwmr SDHES urn ENEBMES Est ' 1‘41: 1/31: l l I"! I 'f 01 I“! if (0-1.) = 1 [35¢ 8";(EJSEJM: . ‘ __ '— 2 {R _ 6-1.5 e'LCt.‘&.€g)t/‘ 1 c) Find the time dependence of the density. At what frequency does the density oscillate? ,0ch = Hana) We) : km.I:C—ifieKEwEEJt/t_eaaeim-SEAf/filelie—{(EnS'EHIi _e'3”e~i(€a-55-lt’*] —. Sad 1 .- 7_ cos ( 2g JfEHSEB-(EuSEBt/kll xme- magazine/t3] —=> 9: 23m: = Ym ‘ SML(G * Eff/:3 ‘Thé' owsmascmwm com 1H6 W meant,er a: 2’5 ...
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