CHM 424
EXAM 2 CRIB - COVER PAGE
FALL 2007
There are six numbered pages with five questions. Answer the questions on the exam. Exams
done in ink are eligible for regrade, those done in pencil will not be regraded.
1 coulomb = 6.24×1018 charges
1 volt = 1 joule per 1 coulomb
h = 6.6×10-34 J s
c = 3×108 m s-1
1 in = 2.54 cm
n=
c
v
1 amp = 1 coulomb per 1 second
ε0 = 8.85×10-12 C2 J-1 m-1
k = 1.38×10-23 J K-1
1 amu = 1.66×10-27 kg
NA = 6.022×1023 mol-1
E = hν
⎛ 1
1
1 ⎞
= ( nl − 1) ⎜
−
⎟
f
⎝ Rin Rout ⎠
2
n1 − n2 )
(
R=
( n1 + n2 )2
1
1
1
=
+
f dob dim
( n1 − n2 ) ( n1 − n2* ) ( n1 − η2 )2 + η22σ22
R=
=
*
( n1 + n2 ) ( n1 + n2 ) ( n1 + η2 )2 + η22σ22
mλ = 2nt cos θ
sin θm =
hc 1 2
= mv + ω
λ 2
RTE =
RTM =
mλ
a
A = − log (T )
sin (θ i − θ t )
2
sin (θ i + θ t )
2
tan (θ i − θ t )
2
tan (θ i + θ t )
2
n
tan θ B = 2
n1
sin θc =
n1
n2
M=
dim
d ob
n1 sin θ1 = n2 sin θ2
dθ
m
=
d λ a cos θm
ρ=
Δθ =
dl
dθ
= f
dλ
dλ
0.61λf
r
2λ
Na cosθ m
R=
λ
∝ mN
Δλ
CHM 424
EXAM 2 CRIB
Name_________________________________
FALL 2007
Score_________/150
30 pts. 1. Provide the following definitions or facts at 6 pts each.
a. Label six of the regions of the electromagnetic spectrum
b. List the types of transitions that occur in each of the regions you labeled in (a)
(i) gamma: nuclear, used to identify isotopes
(ii) x-rays: inner shell electronic, used to identify atoms in solids and provide information
about adjacent bonded atoms
(iii) ultraviolet: valence shell electronic, used to identify atoms in a sample and their
concentration, molecular electronic transitions involving single bonds, used to determine
structure and concentration
(iv) visible: valence shell electronic, used to identify atoms in a sample and
concentration, molecular electronic transitions involving conjugated double bonds, used to
determine structure and concentration
(v) infrared: vibrations, used to determine functional groups existing within a molecule
(vi) microwave: rotations, used to determine bond angles and lengths of small molecules
in the gas phase
radio: magnetic resonances, used to determine bonding in molecules, electron spin
resonance, used to determine location of unpaired electrons
Exam 2 Crib, 2007, page 1
c. A grating monochromator is adjusted to pass 800 nm light in the first order. What other
wavelengths in the UV/visible region will be passed?
The monochromator will pass 400 nm in the second order, 267 nm in the third order, 200 nm in
the fourth order, etc.
d. What is the most common spectral source used to obtain ultraviolet (<350 nm) absorption
spectra?
deuterium lamp
e. What logic operation is performed by the following gates? (AND, OR, etc)
___NOT_________
___AND_________
___OR__________
___XOR_________
___NAND________
___NOR_________
Exam 2 Crib, 2007, page 2
30 pts. 2. Perform the following calculations at 10 pts each.
a. Consider 300 nm light.
3 × 108
= 1015 Hz
(4 pts) What is its frequency in Hz? ν =
−7
3 × 10
(4 pts) What is the energy of one photon in J?
E = hν = 6.6 × 10−34 × 1015 = 6.6 × 10−19 J
1
= 10−15 s
(2 pts) What is its period in s?
15
10
b. An 8-bit DAC has an output range from 0 to +10 V.
(i) Including zero, how many unique voltages can be produced by the DAC?
28 = 256
(ii) For the input binary number 00000101, what voltage is output by the DAC?
10 V
× 5 = 0.195 V
256
c. A fluorometer is constructed with a xenon arc lamp having a source size of 0.1 mm, a
monochromator entrance slit width of 1 mm, and a source to slit distance of 11 cm. What values
of dob, dim and f (in cm) will magnify the source to just fill the slit? (b) What optical property of
the monochromator will determine the lens diameter that just fills the grating?
M =
a)
dim
= 10
dob
dim = 10dob
11 = dim + d ob = 10dob + dob = 11d ob
dob = 1
dim = 11 − 1 = 10
b) the monochromator f/# will determine the lens diameter
Exam 2 Crib, 2007, page 3
30 pts. 3. Answer the following two homework questions for 15 pts each.
a. Write ln(10) as a binary number. Use six bits to the right of the binary point.
ln(10) = 2.3025851
-2.0000000
0.3025851
-0.2500000
0.0525851
-0.0312500
0.0213351
-0.0156250
0.0057101
21
2-2
2-5
2-6
ln(10) = 10.010011b
b. The silver iodide bond energy is 255 kJ/mol. What is the longest wavelength of light that is
capable of breaking the bond?
255 kJ mol-1
6.022 × 10
23
mol
-1
= 4.23 ×10−19 J
E 4.23 × 10−19
ν= =
= 6.38 ×1014 Hz
34
−
h 6.63 ×10
λ=
3 ×108
6.38 × 1014
= 470 nm
Exam 2 Crib, 2007, page 4
30 pts. 4. A signal is sampled at 2 kHz by an analog-to-digital converter. The digitized data has two
frequencies, 0 Hz (dc) and 1 kHz.
(a) An RC low pass filter with f3dB = 1 kHz is placed between the signal source and the
analog-to-digital converter. The amplitude of the 0 Hz component drops to 0.1 its original
value, while that for the 1 kHz component drops to 0.707 its original value. What are the actual
(unaliased) frequencies?
(b) Based on your answer to part (a), if the sampling frequency is changed to 6 kHz what
frequencies will be observed in the digitized data?
(a) Possible frequencies can be determined by an
"accordion." Possible frequencies yielding 0 Hz
are 0, 2, 4, 6, etc. kHz. Possible frequencies
yielding 1 kHz are 1, 3, 5, 7, etc. kHz.
An RC filter will attenuate f3dB by 3 dB, or
0.707. Thus, the 1 kHz component is most likely
un-aliased. 3 kHz and higher frequencies would
be attenuated by larger values.
An RC filter will attenuate 10×f3dB by 20 dB, or 0.1. Thus, the 0 Hz component is most likely
10 kHz.
(b) Changing the sampling frequency to 6 kHz
requires construction of a new frequency
accordion. Since the 1 kHz component in (a)
was un-aliased, it will appear at 1 kHz. To
predict the alias of 10 kHz, locate it on the
graph and run a vertical line. 10 kHz will be
aliased to 2 kHz.
Exam 2 Crib, 2007, page 5
30 pts. 5. Suppose an optical component such as a lens with refractive
index, n3, is surrounded by a material (such as air) having a
lower refractive index, n1. In many instruments the
reflection due to the refractive index differences cannot be
tolerated. It is possible to dramatically reduce the amount
of reflection by coating the lens with an intermediate
refractive index, n1 < n2 < n3. This is accomplished by
making the layer of n2 material 1/4 λ thick. That is, if the
light is at 600 nm, the layer is made 150 nm thick.
The second requirement for an anti-reflection (AR) coating is to choose the value of n2 so
that R1,2 = R2,3. What value of n2 will accomplish this? (The answer should give the value of n2
as a function of n1 and n3.) Hint: at one point in the derivation it is prudent to take the square
root of both sides of the equation!
⎛ n −n ⎞
R1,2 = ⎜ 1 2 ⎟
⎝ n1 + n2 ⎠
2
2
⎛ n −n ⎞
R2,3 = ⎜ 2 3 ⎟
⎝ n2 + n3 ⎠
n1 − n2 n2 − n3
=
n1 + n2 n2 + n3
( n1 − n2 )( n2 + n3 ) = ( n2 − n3 )( n1 + n2 )
2n1n3 = 2n22
n2 = n1n3
Exam 2 Crib, 2007, page 6