Chem 155 Midterm Exam Page 1 of 10 ame Signature 1. Mercury (Hg) is is believed to be hazardous to human neurological health at extremely low concentrations. Fortunately EPA Method 45.7 cold vapor atomic fluorescence spectroscopy (CVAAS) is up to the task. a. Your parameters are the following: i. CVAAS has a linear dynamic range of about 0 to 000 ppt (ng/l) Hg. ii. CVAAS requires about 100 ml of sample iii. You have 1000 ppm Hg as a primary standard solution. iv. Assume that you can pipet no less than 100 L and use no volume larger than 100 ml to minimize waste in the following: If necessary If necessary b. You expect the concentration of your environmental sample to be approximately 100 ng/l. Convert this number in to ppb and compare it to the EPA limit of ppb. Is it above or below the limit?. 1 ppb = 1 g/l = 1000 ng/l so 100 ng/l = 0.1 ppb 100 ng = 0.1 ppb c. Give the concentrations of calibration standards that would be appropriate for the analysis of such a sample. (If you don t know the answer to the above question assume 1000 ppb) e.g.: 0.0, 0.1, 0.5 ppb e.g.: 0, 100, 500 ppt d. Describe how you will prepare a diluted stock solution from the 1000 ppm [Hg] standard stock. This will be diluted below to make the calibration standards. 1 st dilution nd dilution rd dilution C1 V1 (pipet) V (vol flask) C unit Final 1000 ppm 0.1 ml 100 ml 1000 ppb 1000 ppb 0.1 100 1000 ppt diluted stock solution conc. 1 e. Describe how you will prepare the calibration standards from the 1000 ppm [Hg] standard stock. Target (final) calibration standard conc. / ppb Conc. Of soln. pipetted / ppb Volume pipetted / ml 0.00 1 5 100 0.100 1 10 100 0.500 1 50 100 Final volume / ml
Chem 155 Midterm Exam Page of 10. Which is a larger interval, the 80% CI or the 95% CI and why? The 95% CI is larger because a larger interval is needed to generate higher confidence in the hypothesis that the true mean lies within the given interval. 4. The following set of measurements were made for determining the iron content of a vitamin tablet (mg per tablet): 19. 18.8 18.7 The expected value is exactly 18.0 mg per tablet. Is there evidence at the 95% confidence level that the iront content of the tablet tested differs from the labeled amount? In your answer, assume that the method is valid, i.e. it is free from significant sources of bias. 19. 18.8 18.7 18.9 4 ( 19. 18.9) ( 18.8 18.9) ( 18.7 18.9) 1 0.1 ( 18.9 18.0) 0.1 5.04 5.04 4.0 Yes, there is evidence of difference at 95% CI % confidence interval freedom 50 80 95 99 1 1.00.08 1.71 6.66 0.8 1.89 4.0 9.9 0.76 1.64.18 5.84 4 0.74 1.5.78 4.60 5 0.7 1.48.57 4.0 6 0.7 1.44.45.71 7 0.71 1.41.6.50 8 0.71 1.40.1.6 9 0.70 1.8.6.5 10 0.70 1.7..17 0 0.69 1..09.85 50 0.68 1.0.01.68 100 0.68 1.9 1.98.6
Chem 155 Midterm Exam Page of 10 4. When is the method of standard additions helpful? Standard additions is applied when there are matrix effects matrix effects are a difference between the sensitivity of the instrument to the analyte in the sample matrix relative to the standard matrix. 5. What is an internal standard and when is it helpful? An internal standard is a chemical added in equal concentration to all samples and standards that is completely resolved by the instrument from the analyte and all interferants. The internal standard signal is used to monitor and correct for fluctuations in the instrumental sensitivity. 6. What is the function of a reagent blank? A reagent blank is prepared to see if any of the chemicals used in the sample preparation will give a significant signal for the analyte e.g. if there is contamination or interference from the sample preparation step. 7. What is the purpose of a spike recovery analysis? A spike recovery analysis tells you if the sample preparation step is responsible for the loss of analyte. For example, if during sample preparation some analyte is decomposed or lost through filtration or vaporization, a spike recovery analysis will show less than 100% recovery.
Chem 155 Midterm Exam Page 4 of 10 8. Fill in the blanks below: LS = lower state US = upper state GS = ground state hν + US LS + hv stimulated emission gain hν + LS US spontaneous emission loss US LS + hν absorption loss 9. Draw an energy state diagram for the Kr, F and the KrF* excimer as pertains to it s function in the excimer laser. a. Identify the atoms or molecules in all four states. (1) b. Identify the two states involved in the stimulated emission process. (1) c. Draw an arrow indicating the transition that produces laser radiation. (1) Kr* + F Kr F* Upper State for Laser Radiation Produces Laser Radiation Kr + F Kr F Lower State for Laser Radiation 10. What is a population inversion and why is it necessary for gain in the laser?. A population inversion refers to the condition of there existing in the gain medium a larger concentration of upper states than lower states. This conditions ensures that the rate of stimulated emission exceeds that of absorption.
Chem 155 Midterm Exam Page 5 of 10 11. Fill in the blanks: ame of EM regime: Wavelength Predominant Excitation xray 0.1 to 10 nm Core electron Ultraviolet 00400 Valence electronic Visible 400800 Valence electronic ame of Spectroscopy UV or UVVis Vis or UVVis Infrared.540 m vibrational IR or FTIR 4 Radiowave 0.5 10 meter uclear spin MR 1. Draw a Jablonski diagram for an typical fluorescent / phosphorescent organic molecule and identify: a. the ground electronic state, S0 b. the first singlet excited state S1 c. the vibrational levels within S1 d. the triplet excited state T1 e. transitions for absorption of IR radiation f. transitions for absorption of UV radiation g. Fluorescent emission h. Phosphorescent emission 4
Chem 155 Midterm Exam Page 6 of 10 If a CzernyTurner monochromator has the following specifications: Holographicallyruled diffraction grating with 049 grooves per mm. 1 m focal length Grating position such that the diffracted angle is 45 degrees Operation in first order 10 6 nm mm 1 049 mm d. What is the reciprocal linear dispersion effective bandwidth of this monochromator (use appropriate units)? cos( 45 deg) 1 1 m 1000 mm m 0.45 nm mm 0.05 nm 0.45 nm mm e. What slit width would be required to achieve a spectral bandwidth of 0.05 nm? 0.145mm 1. Give an example of a light source for the following: a. broadband UV D emission lamp or Xe arc lamp or Hg emission lamp b. Visible and near infrared Quartzhalogen or quartz tungsten halogen c. Monochromatic visible Laser or hollow cathode lamp
Chem 155 Midterm Exam Page 7 of 10 14. Describe the operation of the PMT using words and a sketch. When a photon strikes the photocathode a photoelectron is emitted. This photoelectron is accelerated by an applied voltage to the first dynode. When the accelerated electron strikes the dynode it emits a cascade of secondary electrons. These secondary electrons in turn are accelerated into the next dynode and so on until a large and easily measurable pulse is generated. 15. Elaborate the acronym for PMT and CCD detector: P hoto M_ultiplier T_ube C_harge C_coupled D_device 16. Draw on the diagram below to indicate the sequence of events that leads to light detection in a CCD pixel. Fill in the boxes with explanatory text. 10V 10V 10V Al contact SiO nsi 1. e are repelled from Al contact, photon (hv) is absorbed in depletion region.. electron hole pair is generated. h+ accumulate at () biased Al contact and e are expelled
Chem 155 Midterm Exam Page 8 of 10 Spring 007 Terrill Statistics: lim x Mean : x Average: x avg Population Standard Deviation: Sample Standard Deviation: s x x x lim x x avg 1 Bias or absolute systematic error = x avg s Relative standard deviation = x avg Confidence limits: x±ts n x±z n
Chem 155 Midterm Exam Page 9 of 10 Spring 007 Terrill Propagation of Error: Addition/subtraction rule: If: a b± b c± c d± d Multiplication/division rule: If: a b± b c± c d± d Then: Signals: a b c d Then: a a b b c c d d S S b m C General transducer response S min S b Blank Minimum detectable signal Standard Additions: C x C min Blank m bc s mv x Detection Limit C x V s_0 C s V x V x = sample vol C x = sample conc. V s = spike vol C s = spike conc. V T = total vol V S_0 = xintercept Beer's Law: A b C log P P o C k P P o Beer's Law Absorption Emission Boltzmann Partition Function: i o g i g o e k T E k 1.8 10 j K 1 Boltzmann Constant
Chem 155 Midterm Exam Page 10 of 10 Spring 007 Terrill Electromagnetic (EM) Radiation: Constants: h 6.66 10 4 j s Plank Constant c.00 10 8 m s 1 Speed of light in vacuum k 1.8 10 j K 1 Boltzmann Constant 1 a 6.0 10 mol Avogadro's umber To convert ev to Joules multiply by: 1.60 10 19 J ev Equations: E h Photon Energy c Velocity of light in vacuum Interactions of EM Radiation with Matter: c medum n 1 sin 1 n sin I R I o n n 1 E e Diffraction and Monochromators: n n d ( sin( r) sin( i) ) D 1 h n n 1 d sin y c n medium max T 10 6 d cos( r) n F Effective bandwidth = w D 1 nm K Refractive index Snell's Law of Refraction Fresnel Formula for ormal Incidence Reflection Photoelectric effect Wien Displacement Law for Wien Blackbodies Displacement Law for Blackbodies Grating Equation normal incidence Grating Equation Reciprocal Linear Dispersion R f F d n Resolution Speed