exam1_crib 2009 - CHEM 374 EXAM I Spring 2009 Name: (DO/D...

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Unformatted text preview: CHEM 374 EXAM I Spring 2009 Name: (DO/D This exam should have a total of 6 numbered pages including the back page which contains reference material 1) Consider a green photon which has a wavelength of 500 nm. a. (5 points) What is the momentum of the photon (in kg mfs un'ts)? {[9 >—% (-3: [(3) find“ Ci"qu «we [WW/QJ .55: we 13% b. (5 points) What is the energy of the photon (in J units)? g>lrxv 7“ “CA 2 PC 3: ‘twélfi Kat/sf-QJ‘ c. (5 points) What is the frequency of the photon (in 3'1 units)? V 1 9/“ “1—: mam“1 327,5 “a s“ d. (5' points) What is the wavenumber of the photon (in cm']l units)? 2) (20 pomts) A given particle has the following expectation values: (x) = 0, (x2) = a2, (p) = 0. What is the smallest possible momentum squared expectation value which this particle could have (show your work)? La. Extra Credit '(5 points): A given particle—wave has a (normalized) Gaussian probability density 1 —x=r(2a1) H e (Ix/2.717 of this particle? , Where a z 1 A. What are the standard deviations of the position and momentum 4x5 LO “"7 WWW 3) Consider a wavefunction which is proportional to sin[(21:/ a)x] and extends over the range — % < x < g (and is equal to zero everywhere else). a. (15 points) Normalize the above wavefunction (show your work). To» “Wis a. We. w~ gWMM SD 7‘ L - ,l M 1' 1 SW 2-. —a/2 0 /\ an e. (5 points) What is the probability of finding the abo e particle at some positive x Value (anywhere in the range 0 s x s a/ 2) ? bi? Km? @th? ' 4) The ground state of a particle that is confined to a box which extends Over the range 0 < x < a has a normalized wavefunction of ‘I’l(x) = \EsinKn/ak]. a. (20 points) Evaluate the expectation value of the kinetic energy of the above particle (Show your work). /\ L A _—RL 7.. ,1 t he :— 5‘7‘ 9—3"— de a zit-La [218% _ 2% We" b. (5 points) What is the expectation value of the momentum squared of the above particle? A < b ’ ‘ K; 1 J:\ 30) <f§§e <h>1lm :2. 1 c. (5 points) What is the expectation value of the position of the above particle? l “ In L) \ q 4w ~=— 2% e7 Sgwwey EQUATIONS, CONSTANTS AND CONVERSION FACTORS: Photon energy = hv - Photon momentum = h/Pt Frequency = v = c l )t = c 17 = w / 2n: “2 2 2 Operators: fc=x° 16:256—— Kx=2px = 2k 367 Vx=V(x)- H=K+V l x m m x Expectation Value: (A) = 1111*}?de Probability Density: ‘11“1' (for normalized 1I") rth2 Energies of a particle-in—a—box: En = 8 L2 where n = 1,2, 3,4... m Uncertainty principle: O‘AO’B a Q; = 1KA2) - (A)?- Us = 1.} (B 2) — (3)2 The commutator of 21 and 5" is ELF]: 3E} "- 321 , for example, [5E, 16 = ih FUNDAMENTAL CONSTANTS: 1 eV = 1.6 x 10"9_J R m 8.314 J/(K moi) k3: RINA = 1.38 x 10'23 J/K h=6.6x10'341_s h=hlzaczl.1x10'34ls me 2 9.1x 10'3' kg (electron mass) mp = 1.7 x 10'” kg (proton or neutron mass) NA = 6.0 x 1023 molecules/moi c = 3.0 x 103 mfs :rt= 3.14 CONVERSION FACTORS: 1 A =100 pm = 0.1 nm =10"°m 0 °C = 273 K E a T00 111/ 3 17(cm'l) EULER’S FORMULA: - em = cos(6) + i sin(0) M = ccos(cx) dx‘ DERIVATIVES: W = —csin(cx) x de“ a, —— = ce dx fsin(bx)dx = Sgl-coswx) fsinUJx) cos(bx)dx = $st (bx) fsin2(bx)dx = lx -— isin(2bx) INTEGRALS: 2 4” b fxsin(bx)dx = Egsinwx) — 35m 2 u r . 2 _ x _ xsm(2bx) _ cos(2bx) fxsm (bx)dx— —4 my) ———-8b2 _ .[m xze'x2 “khaki = 0'3 «ll—n: ...
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exam1_crib 2009 - CHEM 374 EXAM I Spring 2009 Name: (DO/D...

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