Chemistry 421/821 – First Exam Spring 2010 page 1Name Answer Key
1)(15 points)
a. (5 points)Describe the primary difference between the operation of a photomultiplier tube and a silicon diode (photodiode) detector.
A PMT generates (and amplifies) a current using the photoelectric effect. A photon hits a photoemission material (Cs2O, etc) coated cathode where the ejected electron is attracted to the anode (90V potential difference) and the current is measured. Conversely, a silicon diode (photodiode) prevents current flow because of charge separation between the p-region and n-region. The photon provides the energy need for the electron to flow into the P-region.
b. (5 points)Describe the primarydifference between the operation of deuterium lamp and tungsten filament lamp.
A deuterium lamp generates light by using an electric spark to generate an excited deuterium molecule that relaxes by decomposing and emitting a photon. Conversely, a tungsten filament lamp generates light by black-body radiation. The tungsten filament is heated to a high temperature (> 1000K) at which point it begins to glow and emit photons.
c. (5 points) Provide one example on how a specific property of light is used to make a monochromator (be specific).
One of the following, with a description on how the property separates white light into individual wavelengths:
Refraction prism
Diffraction grating monochromator
Reflection reflection grating
2)(16 points)
- (6 points)Illustrated below are two Gaussian curves describing data collected on two different instruments. What do the curves indicate about the relative accuracy and precision of the two instruments?
(a) has better precision (smaller standard deviation) but is less accurate because of a systematic error of ~25. (b) has a lower precision (higher standard deviation) but is more accurate since no systematic error is apparent.
- (5 points) What is an easyway to detect random error?
Make repeat measurements and calculate a standard deviation.
- (5 points) What is the best way to detect systematic error?
Use standard samples and/or calibration curve
3)(15points) Please explain the following issues in UV/vis spectroscopy:
- (5 points) How stray light affects accuracy.
Measure tramsmittance (T) = P/Po. Stray light can make P or Po appear incorrectly larger at the detector because the light is not passing through the sample and being proportionally absorbed. The result is an incorrect high or low tramsmittance with a corresponding decrease in the accuracy of the analytes concentration.
- (5points) How dark current or shot noise affects precision.
Dark current or shot noise is the random emission of an electron (current) that is not caused by a photon. The result is a random change in the current, tramsmittance and correspondingly concentration. Therefore, it decreases the precision.
- (5 points) How the proper use of a filter can improve accuracy.
A filter can be used to narrow the bandpass and/or remove stray light. Again, we are measuring T=P/Po at a specific wavelength so removing any extraneous wavelengths with differing molar absorptivity will improve the accuracy. The amount of light absorbed at different wavelengths will vary (generally lower) resulting in the observed P being lower than expected.
4)(15 points)Please answer the following questions about calibration curves.
- (5 points)What is the difference between limits of detection and sensitivity?
Sensitivity is the slope of the calibration curve, while the limits of detection are the defined by the ends of the linear regions of the calibration curve or the minimum/maximum concentration of an analyte that can be detected at a known confidence level:
minimum analyte signal(Sm) - mean blank signal(Sbl))/slope(m)
- (5 points)You are designing a method to measure K+ in urine samples that requires a Na+selectivity of k 0.025 and obtain the following best-fit lines to the calibration curves:
K+: y = 23.7x – 0.03
Na+: y = 1.6x + 0.007
Were you successful? Why or Why not?
No, it was not successful because k = 1.6/23.7 = 0.068 > 0.025
- (5 points)Given the calibration equation in (b)for K+, what can you tell me about the dynamic range for K+ detection in urine?
The equations tell us nothing about the dynamic range since the dynamic range corresponds to the linear region of the calibration curve which requires visual inspection of the full calibration curve.
5)(10 points) You are designing an echellete grating to be used over a wavelength range of 190 nm to 800 nm. You want a minimum resolution across this wavelength range of 0.01 nm. Assuming 1.0 cm of the grating is illuminated, how many blazes per millimeter are required?
R = / = 190 nm/0.01 nm = 1.9x104
= 800 nm/0.01 nm = 8.0x104 use largest R
R = n/N= 8.0x104= (1) x N lines/mm x 1.0 cm x 10 mm/cm
Lines/mm = 8x104/ (1.0 cm x 10 mm/cm) = 8x103 lines/mm
6)(10 points) Calculate the equilibrium constant for
AB
Given
Molar absorptivity ()Species / 260 nm / 510 nm
A / 23 / 4.7
B / 7530 / 55,000
and an absorbance of 0.238 at 260 nm and 0.102 at 510 nm in a 1.0 cm cell.
K = [B]/[A]
A260 = (1.0 cm x 23 x [A]) + (1.0 cm x 7530 x [B]) = 0.238
A510 = (1.0 cm x 4.7 x [A]) + (1.0 cm x 55,000 x [B]) = 0.102
0.238=23A + 7530B A = (0.238-7530B)/23
0.102=4.7A+55,000B substitute for A
0.102=4.7((0.238-7,530B)/23) + 55,000B -> 0.102 = (1.1186 – 35,391B)/23 + 55,000B
0.102x23 = 1.1186 -35,391B + 23 x 55,000B -> 2.346 = 1.1186 + 1,229,609B
B = (2.346-1.1186)/1,229,609 = 9.982x10-7
A = (0.238-7530B)/23 = (0.238-(7530 x 9.982x10-7))/23 = 0.0100
K = [B]/[A] = 9.982x10-7/0.0100 = 9.98 x 10-5
7)(6 points) Please circle the correct answer.
a. (2 points)Which compound is not likely to have an observable UV/vis spectra?
b. (2 points)Which compound is expected to have a higher absorbance (larger )?
c. (2 points) Which compound is likely to have a higher fluorescence?
8)(13 points) Please answer the following general questions about fluorescence and phosphorescence:
- (2 points) What is a common ground state to excited state transition?
to* or n to *
- (2 points) UV/vis measures P/Po, what is measured in fluorescence?
P or PL
- (4 points) What is the role of vibrational relaxation in internal conversion and intersystem crossing?
Vibrational relaxationprovides a mechanism to transfer an electron to a lower energy electronic state (internal conversion) or reverse the spin of an electron (intersystem crossing) by overlapping vibrational states.
- (3 points) Why is the discussion of fluorescence and phosphorescence focused on deactivation processes?
Fluorescence and phosphorescence occurs because the excited state relaxes by the emission of a photon. This is a relatively slow process compared to the release of heat. Therefore, any process that enables the rapid release of energy via heat inhibits the fluorescence and phosphorescence process. So, the discussion of the deactivation processes highlights what prevents fluorescence and phosphorescence or what needs to be avoided to allow fluorescence and phosphorescence.
- (2 points) Please identify one approach to increase fluorescence.
Decrease temperature, increase viscosity, decrease concentration of quenching agent (O2), increase rigidity of structure, form metal complex, increase number of rings, increase number of resonance forms, or decrease paramagnetic properties or heavy atoms (I, Br)