AP®ChemistryCollege BoardExam QuestionsUnit 3: Properties of Substances & MixturesSpectroscopy

Spectroscopy (College Board AP® Chemistry): Exam Questions

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Spectroscopy

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1
Sme Calculator1 mark

A colored solution is analyzed by spectrophotometer. The solution has concentration 0.2 M and path length 1.00 cm. When the filter of the spectrophotometer is set to 430 nm, the absorbance is x.

If the experiment is repeated using a cuvette with 0.5 cm path length deduce the new absorbance.

  • x over 4

  • x over 2

  • x

  • 2 x

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11 mark

The wavelength range for ultraviolet radiation is 10-8 m, while that of infrared radiation is 10-5 m. Which type of radiation has more energy, and why?

  • Infrared has more energy because it has a lower frequency.

  • Infrared has more energy because it has a shorter wavelength.

  • Ultraviolet has more energy because it has a higher frequency.

  •  Ultraviolet has more energy because it has a longer wavelength.

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21 mark

If the frequency of a photon of electromagnetic radiation is 5.0 x 1014 Hz, determine the energy of the photon.

  • 1.33 x 10-19 J

  • 2.66 x 10-19 J

  • 3.31 x 10-19 J

  • 6.63 x 10-19 J

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31 mark

Identify the type of energy transition shown in the diagram

Diagram of a water molecule with one hydrogen (H) atom connected to two oxygen (O) atoms, each indicated by arrows pointing to the hydrogen atom.

  • microwave radiation

  • ultraviolet radiation

  • infrared radiation

  • visible light radiation

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41 mark

Which of the following provides the best explanation for what occurs when dye molecules absorb photons of visible light?

  • Certain electrons in the dye molecule transition to a higher energy level, with the energy difference between the two levels matching the energy of the absorbed photons.

  • Certain chemical bonds in the dye molecules start to bend and stretch, with the energy difference between the lower and higher vibrational states equaling the energy of the absorbed photons.

  • The dye molecules begin to rotate more rapidly in specific modes, with the energy difference between the lower and higher rotational states matching the energy of the absorbed photons.

  • Certain covalent bonds in the dye molecules begin to break and re-form, with the bond energies of the bonds being the same as the energy of the absorbed photons.

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1
Sme Calculator1 mark

Determine the amount of energy in joules that a carbon dioxide molecule gains when it absorbs a photon of infrared radiation of wavelength 7 x 10-7 m.

  • 2.26 x 10-19 J

  • 2.47 x 10-19 J

  • 2.62 x 10-19 J

  • 2.84 x 10-19 J

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21 mark
Graph showing a linear relationship between Ni(II) concentration (mol/L) and relative absorbance. Line of best fit is dashed, with data points as black dots.

A student creates a calibration curve of relative absorbance of a fixed wavelength of light for different concentrations of nickel(II) ions.

In a further experiment, they want to use the calibration curve to determine the number of moles of nickel(II) sulfate in a 150 mL sample of a solution of nickel(II) sulfate of unknown concentration. The unknown solution is measured and found to have a relative absorbance of 0.6. Determine the number of moles present in the sample.

  • 0.099 mol

  • 0.110 mol

  • 0.135 mol

  • 0.66 mol

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31 mark
Graph comparing hydrogen and helium electron energy levels, showing peaks in electron numbers relative to energy measured in MJ/mol.

The photoelectron spectra for hydrogen (H) and helium (He) are shown above. Which of the following explanations best accounts for the peak in the He spectrum being positioned farther to the left and higher than the peak in the H spectrum?

  • He has an additional valence electron in a higher energy level than the valence electron in H.

  • He has a greater nuclear charge than H and an additional electron in the same energy level.

  • He has a completely filled valence shell in which the electrons are a greater distance from the nucleus than the distance between the H nucleus and its electron.

  • It takes longer for the electrons in He to be removed due to the higher nuclear mass of He.

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41 mark

Electromagnetic radiation with a maximum wavelength of 540 nm is required to study the photoelectric effect in potassium atoms. What is the approximate frequency associated with this wavelength?

  • 1.8 x 1015 s-1

  • 1.6 x 102 s-1

  • 1.9 x 106 s-1

  • 5.6 x 1014 s-1

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