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Unformatted text preview: Chemistry 354L: Physical Chemistry 11, Fall 2006 Oct. 6, 2006 Exam #1 Name UT EID Note that the problems are of different length, and different levels of difficulty. Also each is
worth a different number of points (Total=100). Scores: #1
#2 #3 Miscellaneous and possibly useful information: h = 6.625 xio'34 mzkg/s (or J s) Ephom = hv= hc/A = hcf/ = ha)
h=h/27t, xp=h, x=dsina 1A=10"°m p =11112 = @111 1 eV = 8065.5 cm'1=’il.602x10'19 J
(11w) =I1p‘ 1;! = 1 (normalization) Ax Ap 2 h/2 H = (h2/2m) 62/6x2 + V Plane waves: H‘I’n(x) = Enwnoc) 111k = A em, Ek = kzhz/Zm P = ih a/ax , H = P2/2m + V sin/(x = (e’b‘ — e""")/2i HW(x,t) = in 81P(x,t)/6t cos kx = (eik‘ + e'i’“)/2 (A) = (\PA\P) = [901111 Particle in a box: pp) = z Macaw = Z en pp”) t1!” = (2/L)”2 sin(mrx/L), n=1,2,3 . ..
c" = (111411041112, w dx E, = n2 762 18/2qu2 = n2 h2/8mL2 W = anAn where pn = cf Em,y = (7127:2/2m) (nf/Lf + nyZ/Lyz) Problem 1. (50 points) Consider a particle in a box where the potential is zero between x = 0 and x = a, but inﬁnite
elsewhere. The system is prepared in such a way that the wavefunction is 1.006) = Nx2(a  x)2
where N is a constant. a) Find N so that the wavefunction is normalized. .1 = I‘WY’ Fan 9 uoﬂnnuze—o “HdEFu’Jcnod
if?»x*caxm<oxtcaxyvax
25:54: x“8 foiq +6xsaz__q_x{a3 + X46141 Ax
= ’0‘ in x“ We m + %><°a=% me + V; x‘d‘i:
= Nth/q“ V2. * ‘/¥'9’3+V5] 4‘ L
.__M;[703l5+$"+0‘+20H26Ia‘\ = Lia“ A). 639.
630 b) What is the expectation value of the energy in the state 1p(x)?
Q
r .— 3 ‘ "kz 6‘
<c> <~rmw> of tr ( 4,“ as?» L A
5”“ low2m x‘a‘wlxﬁlzm «new 2M0 A
" 'ﬂ‘tf 11x"~ 36x50” 38x“a‘ l6X30\3 + lx‘ﬁ” (ix 1M. A)": 63O/a“ 5 <E>: ﬁla7[63o] = £11522 Ma“ (0—? c) What is the second lowest value, E2, that a single measurement of the energy of the system
can give? For). I.) perm“: m) A {on
En : K1312 1.. Zma" . m
2. n1 1. 1.
E2, : k l _ Ll_ : II t Zma Ma." 1
2mm d) What is the probability of obtaining the value E2 when the energy is measured given that the
system is described by the wavefunction 1p(x) given above? (Write an expression for the required integral, and then use symmetry considerations to ﬁnd the value). P2 = «W? = CE = A) {SNPM/ob ° XIQXY
. ° “T T 1001766 ’mnT Wk) IS S‘fﬂhlpr—IC mm?
X: 4/2 SO THAT QIL at
of LidaWAX ‘~ “all; ‘(L‘kdx . . P120 e) Lets assume that a measurement of the energy is made and the value found is the energy of the
ground state, E1. What is the wavefunction for the system after the measurement? (F 174E Eumv «s nmqmmo 65 E. , THE QWEFuthoQ (5 no THE 622.0qu 5719/”? km . Fr: mom Problem 2. (20 points) A certain linear, conjugated hydrocarbon molecule with 6 7t electrons is found to have a lowest
frequency absorption peak corresponding to light with wavenumber x71 = 2.8 x 104 cm]. (Note
that the wavenumber is proportional to the photon energy, Ephoton = hcfl) a) Estimate the length of the molecule. Make use of the fact that butadiene, which can be
approximated as a box of length 5.8 A, has a lowest frequency absorption at x73 = 4.6 x 104 cm'l. Note that butadiene has 4 7t electrons, so its lowest frequency is due to a different‘transition. I» z \ 47. 7
AE: = k“): = EqEs = 0W2
2" n HOLE’ng
"' *1 L 12. 2‘
AE‘VACVB: Es— Ez. ‘ '17,: (QVL—a‘
Tblcmt. 1M6 anno‘. 1315. :2; _ ; L
6E3 DB .5an v d BamomJE ———— 3
Sussnmmc 1).. a ,nun L3 To Fido Ln 3A5}
d ——1—L——— L
La” 5 t 79’
a = —— ' _. 8 ‘
Ln" #uo ' " L” 5?. L8 ' b) Estimate V; for the absorption of next lowest frequency. alum "Mr, D’ = 2.8::10“ cm“ mo Acu, =EqL—_J. 4 1m: new Louar Trimsmou IS FIZom EVE,f ~ 11 ‘eL 1. 1.7. P
W E—iilb‘z l“) — “M L91. 2“ . 9140 U; I l.7 2.8 xloi'cnv' = $8 1”) *cm" Problem 3. (30 points) Consider a particle that is restricted to the region 0 < x < 00 by a hard wall at x = 0. That is, the
potential energy is U = 00 when x < 0 and U = 0 otherwise. a) State (mathematically) the boundary condition(s) that the wavefunction must satisfy and show
that q)(x) = B sin kx is a solution of both the time independent Schrodinger equation and the boundary condition(s).
Show VGA90a Wm :3 O Gowomr cowome 1 2
H‘hEY: “ExtISS/dlcx = Lia13$!«BKK LM bx 1'“ _ tsp ‘f (5 n SoLuﬂoA 72> 7m: 5.5. um ENERGY En ' /2u« . ‘f sn’nshes THE (Souﬁm‘lai'
' mummy b) Is q)(x) = B sin kx also an eigenfunction of the momentum (support your answer with
calculations)? 7;) L$0 ﬁe) : 351..) ko _ 0 CHECK u: (3%”: 9‘? NHEYLE p ISKJCO‘JSWTI p? : "d195— I’Sswhx " "Zk‘K 'W
3" gees ch ’\
’9’? p ‘f aﬁp‘f 60 V :3 NOT A) mam Paw/mo
OF rlOHEJJ’RAM. c) What is the probability distribution for this particle (sketch as a function of x)? ...
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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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