Worksheet: UV–Vis Spectroscopy Methods

In this worksheet, we will practice choosing a UV–vis spectroscopic method and comparing the features and limitations of different spectrometers.

Q1:

In what phase is basic UV–visible spectroscopy designed to analyze samples?

  • APlasma phase
  • BSolid (or suspension) phase
  • CGas phase
  • DSolution (or liquid) phase

Q2:

Which of the following statements about UV–visible spectrophotometers is not true?

  • ADeuterium or tungsten lamps are commonly used as light sources.
  • BSingle-beam and double-beam instruments contain a filter for selecting one wavelength at a time.
  • CA simultaneous UV–vis instrument contains mirrors and a monochromator in order to allow simultaneous detection at various wavelengths.
  • DA diode array detector eliminates the need for a monochromator.

Q3:

Which of the following can not be detected by fluorescence or phosphorescence spectroscopy?

  • APharmaceuticals
  • BVitamins
  • CEnvironmental pollutants
  • DUncomplexed metal ions

Q4:

Which of the following are the analytical methods turbidimetry and nephelometry based on?

  • ALight scattering
  • BPhoton emission
  • CParamagnetism
  • DNuclear repulsion
  • EPhoton absorption

Q5:

How does turbidimetry differ from nephelometry?

  • ANephelometry measures the total metal ion, or inorganic, content; turbidimetry measures total organic content.
  • BTurbidimetry measures the decrease in transmittance of incident radiation; nephelometry measures the intensity of scattered radiation.
  • CNephelometry measures the decrease in transmittance of incident radiation; turbidimetry measures the intensity of scattered radiation.
  • DTurbidimetry measures the total metal ion, or inorganic, content; nephelometry measures total organic content.
  • EThe terms are synonymous; there is no difference.

Q6:

In which phase does flame atomic absorption measure absorption of radiation of analytes?

  • AAqueous phase
  • BGas phase
  • CLiquid (neat) phase
  • DPlasma phase
  • ESolid phase

Q7:

Why are fluorescence and phosphorescence spectra measured at a 90-degree angle to the source?

  • ATo make the overall instrument smaller
  • BBecause the processes of fluorescence and phosphorescence are too intense to observe directly
  • CTo ensure that incident (source) photons are not observed
  • DBecause the sample cell is darkened on two adjacent sides
  • EBecause the monochromator directs the light at a 90-degree angle

Q8:

Why is fluorescence spectroscopy often carried out in a liquid nitrogen environment?

  • APhosphorescent molecules tend to also have explosive properties.
  • BThe detector requires lower temperatures for operation.
  • CThe monochromator slows down the radiation before it hits the sample.
  • DThe source radiation can overheat and destroy the analyte.
  • EPhosphorescence is more likely to occur at low temperatures in a viscous medium.

Q9:

In size exclusion chromatography, what happens to the larger particles?

  • AThey elute first, before smaller particles.
  • BThey remain on the column longer than smaller particles.
  • CThey bind permanently to the stationary phase.
  • DThey are broken down into smaller particles.
  • EThey become oxidized as they move through the column.

Q10:

Which of the following is the correct order in which light passes through a UV–vis spectrometer?

  • ASource, monochromator, sample, detector
  • BDetector, sample, source, monochromator
  • CMonochromator, source, sample, detector
  • DSource, sample, monochromator, detector
  • ESample, source, monochromator, detector

Q11:

What is the purpose of a monochromator?

  • ATo remove stray light from the room
  • BTo allow only light of a certain wavelength to pass from the source to the sample
  • CTo focus light from the sample onto the detector
  • DTo serve as a polychromatic light source
  • ETo interpret the photon signal into a digital readout

Q12:

The UV spectrum of metal complex A has a strong absorption at 412 nm. The UV spectrum of metal complex B has a strong absorption at 654 nm. If equal quantities of the two solutions of metal complexes were mixed, what would the resulting UV spectrum look like?

  • AThe spectrum would show two weak absorption bands, one at 472 nm and the other at 594 nm.
  • BThe spectrum would show two strong absorption bands, one at 472 nm and the other at 594 nm.
  • CThe spectrum would show one very broad absorption band at 533 nm.
  • DThe spectrum would show two strong absorption bands, one at 412 nm and the other at 654 nm.
  • EThe spectrum would show two weak absorption bands, one at 412 nm and the other at 654 nm.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.