Chem 454 / 554
March 7, 2001
Exam 2
Answer any 10 of the 11 questions below:
Each question is worth 10 points
Hint: that's 5 minutes per question on average. Answers that take 10 minutes to write probably
have more detail than I'm seeking-- keep it focused. If your answer only takes one minute to write you are either very brilliant, you are short on details or you are missing the point of the question.
USe back of page or extra paper if you need more space.
1. Beer's Law--
A 0.0125 millimolar solution of a dye in a 5.00 cm cell transmits 44.2% of the incident light at 546 nm.
- a. What is the molar extinction coefficient of the dye?
- epsilon = -log10(0.442)/ (5.00 *0.0125)
-
b. What concentration is needed to absorb 99.9% of the 546 nm light in a 1.0 mm layer?
conc = Abs/(path*eps) = -log10(0.001)/(0.1* epsilon, above)
2.
An Acetylene-Air flame has a maximum temperature of 2400o C.
- Lithium can be detected using the 670.8 nm emission line. This is a 2p --> 2s transition.
- The ionization potential of Li is 5.390 eV (1 eV = 1.6 x 10-19 J)
-
- a. What fraction of the Li atoms in the flame are excited and thus able to emit?
- b. What fraction of the Li atoms are actually ionized,
- and what would be the expected signal loss due to ionization?
N(j) / N(0) = (Pj/Po) exp (- Ej /k T)
k= 1.38 x 10-23 J/K h= 6.62 x 10-34 J sec c= 3.00 x 108 m/sec
part a --atoms
- E*= (hc/lambda) = 6.62 x 10-34 J sec * 3.00 x 108 m/sec /670.8 x 10-9 m
- kT = 1.38 x 10-23 J/K *(2400+273)
- P3p/P3s = 3
- N(3p)/N(0) = ______
partb ions
- E(ion) = 5.390 eV * 1.6 x 10-19 J/eV =
- Presumably N*/N = 1/2 (two spin states for 2s electon, unique closed shell for Li+)
- N(ion)/N(0) = ________
- signal loss is basically fraction ionized (e.g., 10% ionized= 10% loss in signal)
3. What is the expected line width in the flame emission spectrum of the element Q if
- a) The excited state, Q*, has a lifetime of 0.72 nsec. and emits at 487 nm.
- b) Q* atoms are traveling with a mean velocity of 3.2 x 104 m/sec
- c) Q* atoms collide with other species in the flame every 3.75 microseconds?
- Dl/ l = v / c c= 3.00 X108 m/sec
part a-- delta-t x delta-nu = 1
- delta nu = 1/0.72x 10-9 = 1.4 x 109 Hz
- delta lambda = ______
- lambda = c / freq
- so d(lambda) = -(c/freq2 ) d(freq)
- d(lambda) = (lambda2/c) d(freq)= 487X10-9)2 * 1.4 x 109/3.00 X108
- = 1.1 X10-12 m or 0.0011 nm
part b-- doppler shift
- delta-lambda/lambda = 3.2 x104/ 3x 108 = 0.0001
- delta-lambda = 487 nm x 0.0001 = 0.05 nm
part c (as in a but delta-t is 3.75 x 10-6 sec)
- could be 0.5-0.1 times that value if energy change averages every 2-10 collisions
- will obviously be much less than in part a (i.e., negligible)
- delta-nu= 0.3 x 106 Hz
- d(lambda) = (lambda2/c) d(freq)= (487X10-9)2 * 0.27 x 106/3.00 X108
4. We are likely to use a Cold Cathode Lamp for AA Spectrophotometry and a Tungsten halide lamp for visible molecular spectroscopy.
- a. Briefly describe each type of lamp and the characteristic emission.
- b. Why is each lamp appropriate to its use, yet totally inappropriate for use in the other type of instrument.
AA lamp involves an electrical discharge (plasma) in an inert gas. This sputters some cathode material so we get excited atoms, emitting. The result is a line spectrum (typically Neon and the selected metal atom.) Key point is that we get extremely narrow spectral liens that match the sample's absorption spectrum perfectly.
The incandescent bulb gives us a continuum spectrum. We need that to find an absorption spectrum (we need to look at all wavelengths.) For spectrophotometry (molecular) we need to set wavelength to desired value so the lamp must supply light at all wavelengths. We use a monochromator to actually select the wavelength.
If we swapped bulbs, the incandescent lamp would not have enough intensity for good AA measurements and we'd need a carefully tuned narrow band monochromator.
5. Show and describe briefly the overall layout of a double beam scanning spectrophotometer for use in the visible and ultraviolet region. Label key components.
refer to figure in text for the answer to this one
certainly need to show lamp, monochromator, order-sorting filters, chopper and two paths, PMT.
.
Define or explain the use of the term below-- in particular, distinguish between related devices.
- colorimeter
- photometer
- spectrometer
- spectrophotometer
- fluorimeter
colorimeter -- can be visual comparison, white light or a filter
photometer-- use a sensor and meter to measure light intensity, filter typically
spectrometer-- measure light intensity at a specific wavelength (dispersive element required); would include array isntruments. Still has the idea of a meter that measures intensity
spectrophotometer-- specifically measures absorbance at a specific wavelength
fluorimeter-- measures light emitted by a sample when sample has absorbed light.
7. Flame vs. ICP
- a. Briefly describe ICP (including the full name)
- b. What is the major advantage of ICP over flames?
- c. What is the major disadvantage of ICP over flames?
- d. What is the role of a nebulizer in both cases?
- ICP is Inductively Coupled Plasma-- plasma is a gas with significat ionization, rf power is supplied (coupled) from an external coil (iductive coupling.) Basically this is an extremely hot (active) gas; much hotter than flames.
- Hotter-- so much larger degree of excited states and more emission. Also, many elements are excited by ICP that fail to emit in typical flames. Also valid-- fewer interferences, wider conc range, and ability to do many elements simultaneously.
- Disadvantage-- realistically cost and complexity ($150,000 vs. $20,000) May also have higher degree of ionization than a flame.
- Nebulizer-- forms an aerosol from a solution, so small and uniform droplets of solution are supplied to the active region, evaporated and atomized.
8. X-rays as a method of elemental analysis
- a) what is the origin of an L- X ray (an atomic level description)
- b) Describe one important method for excitation of the sample
- c) How does a solid state device detect the X-ray and determine its energy?
- d) Both XFS and AES give the elemental composition of a sample. Describe one important difference in the treatment of the sample or in the nature of the results.
An L- xray occur when there is an electron vacancy in the L (n=2) shell and an electron from the n=3, 5, ... orbital jumps into the n=2 vacancy. The light emitted is the X ray and the orbital energy difference is the energy of the X-ray.
Samples are excited with electron beams, radioactive isotopes or another X-ray beam of shorter wavelength.
X-ray stops in the detector, producing a number of electrons and "holes". The electrons are collected and the current is measured. If each X-ray is separately detected we get pulses-- the amplitude of the pulse is proportional to the x-ray energy.
XFS is typically nondestructive; AA and AES need to dissolve part of the sample. XFS is basically a surface analysis (especially when exciting source is an electron beam.)
9. Calibration-- the use of internal standards for comparison (in AES methods)
- a. What is an internal standard?
- b. Briefly, how would you use an internal standard in such an analysis?
- c. Why is an internal standard more reliable or more convenient than simply measuring a series of standard solutions?
- d. Why are we more likely to use an internal standard in AES than in an AA analysis?
-
An internal standard is another species (another element) added to the sample at a known level.
We measure the relative signal for the IS and the element in question and compute sample concentration from the relative values. We need to base this on results from one or more samples in which both the sample species and the IS species are present at known concentrations.
This typically corrects for instrument variation and many matrix effects of the sample. Sample and IS are subject to the same conditions.
AA may be more difficult since we'd need to change lamps and wavelengths to measure the IS-- too many new variables. AES, especially ICP version, sees al elements (IS and sample) at the same time.
10. Light detectors-- Briefly describe the two light detectors below and their advantages and/or ways they are used in spectrophotometers
- a. Photomultiplier
- b. Diode Array (or a CCD array detector).
A PMT uses the photoelecton effect
light strikes a metal, frees an electron (photocathode)
exiting electron sees a strong electical field (50-100 volts) and is accelerated to another metal suface (dynode)
electron impact with dynode is energetic, kicks several electrons from the surface (amplification)
typical tube has 6-12 such dynodes so tremendous amplification.
micht note that the target area is moderately large (a few mm to a few cm) and the tube housing presvents another detector from being very close. Can only detect one signal at a time (one wavelength.)
A array detector is a solid state light sensor (typically one electron per photon, no amplification)
as an array, 100's-1000's of sensors, side by side, in area of 1-5 cm.
typically charge, wait for light to cause charge to leak away, then measure. (Can improve sensitivity with longer integration periods.)
typically measure light at all wavelengths similtanuously.
11. Lasers
- a. What's the difference between spontaneous and stimulated emission?
- b. Briefly describe an Optical Cavity and its role in a laser design.
- c. What is a Population Inversion?
- d. Briefly describe one way of exciting the medium in a laser.
- e. Describe how one of these lasers achieves a population inversion (He-Ne, Dye, eximer laser)
Spontanous emission-- intensity is proportional to species only; light emitted randomly in time and direction
stimulated emission-- proportional also to the intensity of radiation (can build geometrically in intensity.) Radiation is similtaneous, in phase, in same direction.
Optical cavly typically a pair of parallel mirrors with 99.999+% reflection. Allows field to build up after light passes through medium many times. Takes place of an extreelemly long laser tube.
Population inversion occur when the higher energy state has a higher population than the lower energy state.
Medium excited by
- electrical discharge (in a gas)
- optical pumping
- chemical reaction
- (solid state devices; electrical current)