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Unformatted text preview: Exam 1, Chem 4A, Fall 2005 September 23, 2005
Professor Cohen Name
'SID Lab GSI Name Lab day and time Problems
1. (20 pts)
2. (10 pts)
3. (30 pts)
4. (20 pts)
5. (20 pts)
Total (100 pts) m
a Work all problems to 3 signiﬁcant ﬁgures
0 No lecture notes or books permitted
 N0 calculators with wireless access
0 Time limit is 50 min
a Show all work to get partial credit 0 Box your answers Equations and Constants Coulomb’s Law E 2 It??? 60: Permittivity of vacuum; qn: particle charges; r: distance between particles 1 and 2
Beer’s Law A = ecl A: absorbance; 5: molar absorptivity; c: concentration; I: pathlength Transmission '°/oT = i X 100% T: Transmission; I : output intensity; IO: input intensity Absorption A = —'loglo (T) A: Absorbance; T: Transmission Energy of Light E = hl/ = % c: speed of light; h: Planck’s constant; 1/: frequency of light; «\2 wavelength of light Kinetic Energy of a Particle E = %TnV2 v: velocity; m: mass
Photoelectric Effect
E = hl/ — (I) h: Planck’s constant; 1/: frequency of light; (P: material workfunction Mean and Standard Deviation — 1' 331'
x: n Saw—5:)“
i n—l
ray{measurement i; it: mean; n: number of measurements 8: Gaussian y = 1 e"($"/‘)2/252 a 27r p: Gaussian mean; a: gaussian standard deviation Confidence Interval [,t = (T: :l: % p: Gaussian mean; a: gaussian standard deviation
Ftest
2
s . .
F calc = 32 if Fcalc > Ftable s1 and 52 are statishcally different ttest where standard deviations are statistically equal _ [51*52l nlnz _ 321(n1—1)+322(n2—1)
tmzc — spouted 774+qu Spooled — ﬂ1+TLz—2
DOF = mg + 713 — 2 DOF: degrees of freedom ttest where standard deviations are statistically different  2%???” _lg.iﬁgﬂiéﬂﬁ_
tcalc S§/n1+sg/n2 DOF “(al/nl) (82W)? 2 n1+1 ”2+1
N0 = 6.022 x 1023 molec/mol Avogadro’s number (10 = 5.292 x 10—11 m Bohr radius
k3 = 1.381 x 10‘23 K" Boltzmanns constant e = 1.602 x 10“19 C Electron Charge
me = 9.109 x 10’31 kg Electron mass m1) = 1.673 x 10‘27 kg Proton mass
mn = 1.675 x 10'27 kg Neutron mass 60 = 8.854 x 10‘12 CZI‘lm‘1 Permittivity of vacuum
h = 6.626 x 10"3415 Planck’s constant c = 3 X 108 rns‘1 Speed of light
R = 8.314] Inol‘lK‘1 Gas constant R = 8.206 x 10‘2 atrn mol‘lK‘1 Gas constant FTable T—Table mmmmmmm lmimma
mmmmmmmmmmmmmm mum
mammalian mum mmmm— 
maxim”m
.13
.5
.E!
‘B , mammmmmlm mm
mum1mm
368 _l.mlmmm mm 2.54 248 2421231 2.23 2.16 2.07 1.88
.11
m 2.01 194 1.88 1.83 1.75 1.67 157m ———
Ionization Ene ; (kl/mol)
Electron Affinity(k]/mol) 45.504 1. (10 + 10 points) (a) Draw a diagram of the potential energy for CsCl and the ions Cs+ and Cl‘ versus internuclear
distance. Label the major features. (b) Estimate the dissociation energy of CsCl (in kI/mol) relative to the separated atoms, use a bond
dissociation of Re=3.222 A (1 A = 1010m). State any approximations you make. AEA = AEQOQbe‘ AEpo AEL " (Tries ‘ 1%) = 313L235. .NA. “4““ — (IEQS— EAL) h Luv gar I000?
23 Mb"
(Laozxm‘ﬁX\.cozx/o"")z m (5.022% {a}? [0003)
= QTY (8.854rlo"‘c‘3“m“)teia55 Xlo m) Mot — (37533920); gmrgx.
m i  Hons cm. RDQM' CNN/BM .— : \HCN ‘J/mm
‘ :_ i 4 kT 2. (10 points) The element Pt has a relatively high workfunction of 5.65 eV (9.0513 x 10’19 I). 200 nm
light is incident on the surface. What is the kinetic energy (in I) of a photoemitted electron?
,_ = \n 'U — <13 :hE<z§
l. _ gs. (ﬂex/O “Islﬂ’éwog M/s) (WW gx/o X) (200 x) 04 m) = 01.339 x/o"°':5' — 7.066 “CWT 3. (10 + 10 + 10 points) A large amount of research has gone into developing organic molecules for use
in thin film solar cells. Below is the structure of an organic dye molecule first synthesized in 2001, as
well as its absorbance spectrum and the emission spectrum of the sun. W (a) For a hypothetical solar cell coated with a monolayer of the dye (a layer one molecule thick), the
absorbance at 500 nm is 0.0340. What percentage of transmitted light does this correspond to? A = ‘ﬂojlo (T) T (0‘0")qu Solar Emission Spectrum
Dye Absorbance Spectrum .W. wwmmx WV..~.W E
5 * 400 500 600 700 . ”Mam Wavelength (nm) 100 \ggile‘xgth (m Absorbance “I‘l‘QOgmL—r) : 0,6114? —A
‘0 =T— i%T=°I2.S°7o E (b) Assume that each monolayer is 1 nm thick and the molecules have an effective concentration
of 6.40 mol/ L in the film (1 L = 1 dm3 = 1024 nma). What is the molar absorptivity at 500 nm?
(report your answer in units of M’lcm'l) A=££c ——> 5—, :L  €9.03qu _ 0.0053 (loan/M >=53I25 ME: c, ’— UnnBCéHOM)— nm‘M [Olen/M (c) Why is this compound useful for a solar cell? What property would one optimize to make it a
better material for a solar cell? Mam We “mm o. shrug absewphbq band W 400 100”».
(«time h. Sun has ashug missy‘m Tim Method would be locl'l‘er \? il mm More ﬂashi at ﬂung/zy—
wmkxxgm law it“, Sun has} a 51mg €Mess£m [may own;
“llOlOnm‘ 6 4. (10 + 10 points) The gem stone ruby a is crystalline material composed of A1203 that contains a small
amount of chromium which results in the intense red color of the gem. The FBI has intercepted two
separate illegal shipments of rubies. They believe that the rubies came from the same source and
analyzed one gem from each shipment. The mass of chromium in each gem are shown below as well
as the number of times each gem was tested. Gem Amount of Cr ( ) Mass of Sam le ( ) Number of Measurements
A 0.056 i 0.002 3.27 20
B 0.06 :i: 0.01 1 3.27 15 (a) Are the two gems from the same source based on a 95% Confidence Interval. Justify your rea . 7 ‘
F V = .32.. 3 (053:? _. 25 T: 9 EU“ JAM Standard cﬂnuw'aﬁans owe S‘l‘a’rtshtalvb Jig/ml
 R54 (~23) 470%: (0.0560.00! : .._Q_O_L__: owq g 0.004 _ H 5265 f
a. I . 2 _ ‘  '
“:22 + 0‘“ Sash auwwb‘ W 0‘0”“ [5 ull D0 , 202 .5 11 _2= ‘Hlﬁlx/o 23 4.:Isrwo"_z= lame: 14.9;
(2%?) (cl?) 4M5” +q'W/W0'" Z'Tﬁ?x’O’2
2‘ ”(T 2: It. DOF ”‘5
L5,”;
VYJ “I
mmnzo‘" ”75%: team c {a  W51}: H» G Cam in it... Me wide; is no+ StankPrcqntIj
(was) (2.3) d:l‘€om\+ and More 5% but Sam 50urce. (b) Sketch 2 gaussian curves: one for gem A and one for gem B. Label the mean, and label and shade the region between + 1 and  1 standard deviation. Be sure your drawings clearly show how
J A
f close ’\ precise the measurements are for each gem.
0.057 ‘ f 5
Gem A 63'" All)... a. o.
0958 7 cs 0!. . co? 5. (15 + 5 Points) (a) Shown below is a plot of the electric ﬁelds of two blue, 400—nm light waves (with a small vertical
offset on the solid curve in order to visualize both waves). A single photon of blue, 400—nm light
was measured with a photodiode detector placed in a spectrometer and has an intensity of 1. On
the figure below draw the electric field that would result from constructive interference of two identical 400nm waves traveling the same path in space and '
o What is the wavelength for the new wave you’ve drawn?\ qwﬁM \
0 Label the wavelength distance on the drawing. 2
§ Ilezl TQM=1 =1 0 What is the intensity of the new wave?
a What is the amplitude of the new wave? 2 (b) Shown below is a plot of the electric fields of two blue, 400 nm light waves that are 180° (7r) out
of phase with each other. On this ﬁgure draw the wave that would result from the interference
of these two waves. Flatt 2:44. ...
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