This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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/sfQJ‘ 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 zitLa [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 particlein—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 = Sglcoswx)
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: ...
View
Full
Document
This note was uploaded on 03/02/2012 for the course CHEMISTRY 374 taught by Professor Donben during the Spring '09 term at Purdue.
 Spring '09
 donben
 pH

Click to edit the document details