PHYS 4910 Spectroscopy
Assignment 1

  1. For reasons that we discussed in class the effective optical region for this class is between 300 nm and1000 nm. Within this opticla portion the visible section lies between roughy 400 nm and 700 nm. It is also useful to have an idea of the characteristics in terms of frequency and energy

  2.  
    Wavelength (nm)
    Note: these are all air wavelengths
    Wavelength (Å) Frequency (Hz) Energy (eV) Energy (cm-1)
    257.6 (strong uv line from Hg, used in black lights)
    2576
    1.61 x 1015
    4.82
    38741
    388 (strong line of helium)
    3880
    7.71 x 1014
    3.20
    25721
    515.4 (green line for the Ar+ laser
    5154
    5.49 x 1014
    2.41
    18308
    656.2 (Balmer alpha line of H)
    6562
    4.56 x 1014
    1.89
    15208
    486.1 (Balmer beta line of H)
    4861
    6.15 x 1014
    2.55
    20530
    434 (Balmer gamma line of H)
    4340
    6.89 x 1014
    2.86
    22995
    410.1 (Balmer delta line of H)
    4101
    7.29 x 1014
    3.02
    24335
  3. The (air) wavelengths of the two D lines of sodium are 589.0 and 589.6 nm. Both terminate on the ground state (E=0). From this information, what are the energies (in cm-1 and in eV) of their respective energy levels
    1. For the 589.0 line, the vacuum wavelelngth is 589.0 * 1.0028 = 590.6 nm
    2. Energy = 1/vacuum = 16930.5 cm-1 = 2.099 eV
    3. For the 589.6 line, the vacuum wavelelngth is 589.6 * 1.0028 = 591.3 nm
    4. Energy = 1/vacuum = 16913.1 cm-1 = 2.097 eV
  4. The potassium atom has a pair of D lines, just as sodium does. They also start on the first two excited energy levels and terminate on the ground state. Look up the energies of these two excited levels (see Internet resources), and calculate the D line wavelengths.
    1. The first excited level has an energy of 12985.18 cm-1
      1. This corresponds to a vacuum wavelength of 1 / 12985.18 cm-1 = 770.1 nm
      2. The air wavelength would then be 770.1 / 1.0028 = 767.96 nm
    2. The first excited level has an energy of 13042.90 cm-1
      1. This corresponds to a vacuum wavelength of 1 / 13042.90 cm-1 = 766.70 nm
      2. The air wavelength would then be 766.70 / 1.0028 = 764.56 nm

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