Exam 2

March 21, 2003

Answer ANY 7 of the 8 questions

1. Beer's Law

• A 0.0045M solution has 37.0% transmission at 450 nm in a standard 1 cm cell.
• Find the concentration of a solution with 85.2% transmission under the same conditions.
  • abs(1)= -log10(0.370) = 0.413
  • molar ext coef = abs/(lc) = 0.413/(1.0 x 0.0045) = 96.0
  • abs(2) = 0.069
  • c = 0.069/96.0 = 0.000725 molar

• The IR spectrum of a molecule is shown in figure 1.
  • It shows vibrational lines at 750, 1500 and 2700 cm^-1
• What might the Raman spectrum look like? Sketch and explain.
  • The exciting light is from a laser, operating at 500.0 nm.
  • (Briefly explain your reply and indicate assumptions you need to make for your answer)
  • (No one answered the question)

• First, what do we mean by allowed and forbidden transitions?
  • Include one example of a selection rule that results in allowed/forbidden
• How is it that we can routinely measure spectral lines corresponding to forbidden transitions?
• What is likely to be the practical difference between observing (measuring) forbidden vs. allowed transitions.

• a. What do we use in IR spectrometers and how does one such detector work?
  • virtually all detectors first convert the radiation to heat (light absorbed by detector and detector temperature rises.) Bolometers, thermocouples (traditional) and solid state detectors now.
• b. What is a good detector for visible light at low intensity
  • (and briefly, why does it work well at low intensities.)
  • The obvious choice is a PMT (Photoelectric effect produces a photon, many dynodes accelerate electron and produce multiple electrons leading to amplification.)
  • Solid state detectors like diode arrays are not really very sensitive (still one electron per photon at best.) However, most integrate signal and if we wait a while, we get a respectable electrical signal for a low rate of photons.
• a. Why won't your visible light detector work in the IR region ?
  • the initial step (convert photon to free electron) doesn't work-- photons in IR do not have enough energy to free electron.
  • (Try to avoid phrases like the light doesn't have much energy, since an intense IR beam has a lot of energy-- but really by having many photons.)

• a. What is ICP (first what do the letters stand for) and what do the words in ICP really mean)
  • Inductively Coupled Plasma-- Inductively Coupled (energy fed in as a radio-frequency field-- no electrodes in the gas.) Strictly speaking, magnetic induction involved. Plasma is a gas with ions, electrons, excited states.
• b. Why is it a preferred source to, say, flames in AA
  • Actually an error in the question-- ICP is generally not used in AA (Absorption) but rather in emission. The plasma torch is clearly at a much higher temperature so we certainly atomize a wider range of materials.
  • For emission, we certainly get a higher concentration of excited states (intensity) and we often get additional useful spectra lines.
  • For absorption, we often depopulate the ground state and ICP might yield lower sensitivity.
• c. Why does AA generally show much lower detection levels for, say Ni, than we can achieve in measuring Ni²⁺ in an aqueous solution.
  • I didn't phrase this correctly and generally ignored reply in grading. I intended this to be a comparison of
    • an AA Ni-atom absorption line (fire/oven) -- remember, even a Ni²⁺ solution produces Ni atoms in the flame or oven. We wouldn't try to get Ni (metal) into the flame-- would dissolve it frst
    • to a Ni2+ aqueous solution, green color, viewed by absorption spectroscopy.
  • In this comparison the Ni²⁺ has additional molecular features so instead of one very intense abs line we get hundreds or thousands of closely spaced spectral lines. Each has only a small fraction of the population so it's weak and we don't resolve individual peaks. Result is a broad spectral peak with greatly reduced extinction coefficients.
  • Even more to the point, visible transitions in Ni²⁺ are d-d forbidden transitions and are expected therefore to be relatively weak. The UV AA transitions tend to be allowed.

• sketch a typical design
• label all the important components.
• briefly describe role of each component and
• what actually is measured at the detector end of the instrument
  • a. FT-IR Spectrometer
  • b. ICP spectrometer with Echelle Grating and Array Detector
  • c. Diode Array Spectrophotometer
  • d. XFS (X-ray Fluorescence Spectrometer)
  • e. SEM (Scanning Electron microscope)

label all the important components.

briefly describe role of each component and

what actually is measured at the detector end of the instrument

a. FT-IR Spectrometer

b. ICP spectrometer with Echelle Grating and Array Detector

c. Diode Array Spectrophotometer

d. XFS (X-ray Fluorescence Spectrometer)

e. SEM (Scanning Electron microscope)