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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" spam1 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 gammaN 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, camc 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? uFI
wwan Enzmcmm “455* W :13 ESL K '(VH’GA W733»): EM: magi/LI (a) “1.411 Em " /l‘1@.1 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) “MTml. / 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 EIGWFulcnmJJ 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] ' ﬁtsL sacSfaéhax =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 (01.) = 1 [35¢ 8";(EJSEJM:
. ‘ __ '—
2 {R _ 61.5 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_eaaeimSEAf/ﬁlelie—{(EnS'EHIi _e'3”e~i(€a55lt’*]
—. 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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This note was uploaded on 08/30/2008 for the course CH 354L taught by Professor Henkelman during the Fall '06 term at University of Texas at Austin.
 Fall '06
 henkelman
 Physical chemistry, pH

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