Chemistry 660 Exams--Fall 2003
Prof. James Ingle
posted 12/19/03
average on final was 75 with a range of 59 to 89
Key for 2003 final
(10) 1. best - matrix modification, dilution, separation
okay but often do not work well - saturation because interference effects do not always level off, standard addition because calibration curves or interference effects are often not linear
no good - internal std because difficult to monitor both signals simultaneously, internal blank or background correction because not an additive interference
methods such as adjusting the ash temperature or the L’vov platform are not
general techniques and not really techniques for compensation of interferences
(8) 2. 2.00 (218/324.7)½ = 1.34 pm
(8) 3.The blank noise is higher by a factor of 10 because the background count
rate is 100 times greater for element B compared to A. This behavior is common
because the background count rate even for clean blank varies with m/e because
of masses that are there from the plasma species such as Ae and entrained
species
(12) 4.a. fF = 0.5 = kFtF kF = 5 x 108 s-1 or 0.5 ns-1
b. EFB/EF = 2 = 1 + Kq [Q], Kq = 1000 M-1 = 1 mM-1
kq = Kq/tFB = 1000 M-1 /1 x 10-9 s = 1 x 1012 M-1 s-1 = 1 nM-1 s-1
(12) 5. a. For A = 0.2, T = 0.63
for A = 1, T = 0.1
With both wavelengths, T = (0.1 + 0.63) /2 = 0.36, A = -log T = 0.44
b. both A increase by a factor of 15 to 3 and 15,
for 410 nm, T = 10-15 is negligible compared to T at 400 nm
for 400 nm, T is 0.0010
now considering stray radiation, T’ = (0.0010 + 0.0002)/2 = 0.0006, A’ = -log T’ = 3.22
an alternate approach is T = (0.0010)/2 = 0.0005, T’ = T + r = 0.0005 + 0.0001 = 0.0006
(8) 6. K = [A2]/[A]2
as the total concentration (monomer plus dimer) increases, the fraction of the total in the dimer form increases. Hence, the calibration curve (A vs total concentration) is not linear and will curve upward yielding a positive deviation. At low concentrations, the monomer is dominant so the first part of the curve is flatter because the monomer does not absorb.
(12) 7. W, H, W, R(l) or K(l), Top (optical efficiency for all components in the light path), pathlength (axial versus radial), position in plasma viewed
(8) 8. Find the same wavelength on both axes and draw a line
between these points which show where the excitation and emission wavelength are
the same.
(10) 9. To answer the problem, one must show how absolute detection limits are converted into concentration detection limits. 1 pM is 1 pmol per liter or 6 x 1011 molecules/L
for 1 pL, the number of molecules in this volume element averages 0.0, and for 1 fM, the number of molecules averages 0.0006. Hence, it is difficult to detect much less than 1 pM because on the average much less than one molecule is observed. With SMD the detection limit is likely to be similar to worse than with conventional fluorometry depending on the exact situation, background signals, and the volume viewed with SMD.
(12) 10. With a new slit width that is twice as big, nE would double and nBE would increase a factor of 4
hence, S/N = nE/[nBE + (nBEC)2]1/2
initially, nBE = (nBEC)2 so C2 = 1/nB & S/N = nE/[2nBE ]1/2 = 1, also nE = 20, nBE = 200
after doubling slit,
S/N = 2 nE/[4nBE + 16 (nBEC)2]1/2 = 2 nE/[20nBE]1/2
The new S/N equals (2/5)1/2
12/9/03
Question 7
One element and the same standard or concentration for both instruments.
Exam 3 - Final
I had prepared a key for exam II but did not link it properly and it is now below.
also see latest information page for correction to lecture notes - http://www.chem.orst.edu/ch560-1/ch560/latenews02.htm
Three pickup times
| pick-up time (my office) | turn-in time (my office or mailbox in envelop) |
| 9 am Monday 12/8 | 5 pm Tuesday 12/9 |
| 9 am Tuesday 12/9 | 5 pm Wednesday 12/10 |
| 9 am Wednesday 12/10 | 5 pm Thursday 12/11 |
Emphasis on written lecture notes, in-class lecture material, and class notes, and supplements
Material Covered
all reading assignments in chap.
7, 8, 10, 12, 13, & 15 and links to articles about ICP/MS
Problem sets 10 - 13
Supplements 4, 11-15
11/23/03
Key for exam 2 2003
1. a. (sS)s = nS½ = (sS)f = > nS
nS = ξ-2
b. 1.0 x 108
2. a. The amplification will also increase the blank standard deviation by a factor of 10 resulting in no improvement in detection limit. The amplifier could even add additional noise and make the detection limit worse but this is not very likely..
b. The S/N will theoretically improve a factor of 100 in proportion to the square root of the integration time. However, when one noise is decreased another noise may become dominant. Because more photons are being observed with the longer integration time, it is likely that signal noise remains limiting but it is signal flicker noise. Signal flicker factors are rarely any less than 0.0001 so that the highest S/N would not be much above 10000.
c. The standard addition is meant to deal with multiplicative interference in the sample solution. Signal flicker has nothing to do with the sample solution and can only be minimized with instrumental solutions.
3. a. 1.03 ± 0.21mM
b. t alpha greater than 3.737
4.
a. 70.7
b. 40.8
c. types of noise not observed with photon counting including amplifier noise and excess dark current noise and quantum noise in the PMT gain (multiplication noise)
6. a. signal flicker noise
b. shot noise - not just signal shot noise
c. signal flicker noise
7. The reaction coil promotes mixing of the sample and reagents to allow reaction of form a species to monitor with the detector. There will always be some mixing do to dispersion as the solutions move through the tubing, but without the coil an mixing might be much poorer and the amount of the monitored species formed could be much less and difficult to detect.
8. a. this is an issue dealing with control of an experimental variable that affects the calibration slope - internal standard, double beam (instrumental correction is too general), a matrix modifier is possible
b. this is an additive interference - internal blank, separation, other selection parameters than wavelength
Additional information about exam 2
11/13/03
questions 6c
RSD improves means that the RSD gets smaller
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11/12/03 -
Question from student - "On problem 2 part a, the units for calibration sensitivity are not the same for both the times that it is mentioned. The first time it is 1.0 V/(mM/mL) and the second time it is 10 V/(mg/mL). Are the units, V/(ug/mL) correct since the calibration sensitivity is the change in signal over the change in concentration?"
My answer - The calibration sensitivity the second time is supposed to have the same units as the first time or 10 V/(mM/mL)
11/10/03
a partial key to exam 1 is shown below
11/9/03
Exam 2
There will be a lecture on Wednesday but not Friday.
There will be two pickup times.
1. Wednesday 11/12 at 10 am at the end of the lecture - the exam is due in my office at 10 am on Friday 11/14
2. Wednesday 11/12 at 4 pm at my office - the exam is due in my office at 4 pm on Friday 11/14
Emphasis on written lecture notes, in-class lecture material, and class notes, and supplements
Material Covered
all reading assignments in chap. 5 & 6 and in appendices A & E
Problem sets 7 - 9
Supplements 9 & 10
11/10/03
key for exam 1
1.
a. 0.049 sr, the area of the collimating mirror (or grating) controls the solid angle - it does not matter if the source is extended or a point source
b. 0.29, the reflection losses are for two glass-air interfaces of the lens, two glass-air interfaces and two glass-water interfaces for the sample cell, and the transmission of the filter or (0.92)2(0.92)2(0.997)2 0.4
c. 1.39 x 107 photons/s
d.
e.
f.
2.
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1. solid angle |
2. Rd |
3. Rth |
4. spectral bandpass |
5. throughput or Mo |
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a. double the area of the grating and focusing and collimating mirrors |
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NC |
21/2 |
NC |
2 |
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b. double the focal lengths of the collimating and focusing mirrors |
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c. triple the exit slit width (not the entrance slit) |
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3.
a. the fraction requested is the thermal emission signal with the excitation source off over the signal with the excitation source on which includes thermal emission and luminescence (these signals should be corrected for the appropriate blank signal
b. The ratio will generally increase with wavelength. The thermal emission signal depends on the number density of the excited state which in turn depends on the fraction excited as determined with the Boltzmann distribution. The fraction excited and emission signal will increases in general as the wavelength is increased. The photoluminescence will be fairly independent of wavelength if the source radiant power is independent of wavelength.
4.
5.
a. the primary advantage is less power or current suitable for battery operation, low voltage, small tungsten lamps are available but require much more power.
b. LEDs provide a limited wavelength so several might be needed for analysis at many wavelengths, actually white LEDs based on a blue LED and a phosphor are available
10/25/03
question 3
This is not a tricky question. When an atom is in a flame and one shines light of the appropriate wavelength on the flame and atoms, one can see both emission due to thermal excitation and photoluminescence (emission due to photon excitation or absorption). Hence the total emission signal is due to two components and the question is about figuring out what part of the total signal is due each of the components. For example, it may that 3/4 of the signal is due to thermal emission and 1/4 is photoluminescence or due to photon excitation.
10/24/03
correction for problem 1
in the specifications,
1. it should read the source radiant intensity is 1.00 x 109 photons per second per sr
or Source radiant intensity: 1.00 x 109 photons/s-sr for the 254-nm line
2. the sample cell pathlength is 1.00 cm
3. in problem 1g - NEP is the noise equivalent power as defined in the book or the notes or supplements
Click here to see the table in midterm I
Check here for information about the midterm over the weekend
Question session on Friday at 9 am in lecture room - attendance not required
midterm 1 - handed out at 4 pm Friday October 24 at my office - Gilbert 251
Emphasis on written lecture notes, in-class lecture material, and class notes, and supplements
Material Covered
all reading assignments in chap. 1, 2, 3 and 4 and in appendices A, B, C, D, & F
Problem sets 1 - 6
Supplements 1-8
Concept Organizers through chapter 4
Not revised below this point for F03
Final F02
key posted later, for now see late news page
Key to Midterm 2 F02
1.
a. 103, 104, 1.1 x 104 counts
b. square root of above numbers, 31.6, 100, 105
c. 0.00010 x signal or background counts, 0.1 and 1
d. total noise is 105 and basically due to shot noise as flicker noise is insignificant
S/N is 9.53
e. DL is limited by blank noise which here is background noise and more specifically background shot noise from b)
DL = ksbk/m = (3 x 100)/1 x 105 = 0.003 micromolar
f. (0.003 micromolar/5) = 3 x (1 x 104t)1/2/(1 x 105t)
0.2 = 1/t1/2 , t = 25 s
g. S = B =104, total noise is limited by signal and background shot noise and equal to 141 counts, S/N = 71
h. contribution of a given noise based on the square of the rms noise because add the squares to get the total noise
(sS)f2 = 9 (sS)s2
nSx = 3nS1/2
nS = 9 x 108 counts = 1 x 105c , c = 9 mM
2.
a. photon counting is for very low level light measurement because it discriminates against amplifier-readout noise and dark current noise that is not shot noise from thermal electrons emitted from the photocathode. It does nothing for the S/N compared to more normal current or voltage measurement systems when signal or background noise is limiting.
b. signal (source) flicker noise where the noise or drift will be correlated in the sample and reference channels and cancelled out with a ratio
c. signal shot noise, for a given A, the noise in A is proportional to one over the square root of the reference signal for signal shot noise
3.
a. multiplicative because the signal produced by the analyte is affected by the sample matrix
b. additive because a direct spectral overlap
c. an internal blank is used to provide a good blank signal and the incorrect blank signal is due to additive interferences
4.
a. 0.95 = 1 - 2a so a = 0.025, ta = 3.182 for n = 4 , 0.98 = 1 - 2a so a = 0.01, ta = 4.54 for n = 4
m = x bar ± ts/n½ , 0.032 = ts/n½, = (3.182 s)/2; s = 0.0201; m = x bar ± ts/n½ = 1.00 ± (4.54 x 0.0201)/4½ = 1.00 ± 0.046 nM
b. texp = (x bar - m)/s/n½ = (1.05 - 1.00)/ 0.050/4½ = 5.0, for n = 4, 5.0 > t0.01 (4.54) so less than a 1% probability that the difference is due to random error and a systematic error is highly suspected
5. a. false, b. false, c. true At any analyte concentration there is always the possibility that the measured signal will be less the the minimum signal that defines the detection limit. If c = DL, there is a 50% chance of detecting the analyte or a 50% chance of not detecting analyte. In this case, where c = 2DL, analyte will be detected more than 50% of the time but there is still a significant probability (on the order of 5-10%) that signal will be low enough to conclude not detected.
6. Important variables or processes include injected volume (sample loop volume), flow rates, diameter and length of tubing, volume of the detector cell, dispersion, and kinetics of reactions used to produce the monitored species.
I expected there to be a general statement about what defines the sample throughput (samples per hour). It depends on the ability to resolve one sample peak from another. The maximum throughput is proportional to the flow rate (mL/min)/(volume of the dispersed sample zone at the detector). For example if the flow rate is 1 mL/min and the sample zone at the detector is 0.5 mL, the sample throughput is about 120 samples per hour. Smaller sample volumes do help but the zone volume at the detector is much larger due to dispersion into the reagent solution.
http://www.chem.orst.edu/ch560-1/ch560/ch560exams.htm
Last updated on December 19, 2003
