Max Diem

Quantum Mechanical Foundations of Molecular Spectroscopy


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      where the subscript f and i denote, respectively, the final and initial (energy) state of the atom (or molecule). Such a process is referred to as a “emission” of a photon. Similarly, an absorption process is one in which the atom undergoes a transition from a lower to a higher energy state, the energy difference being provided by a photon that is annihilated in the process. Absorption and emission processes are collectively referred to as “transitions” between stationary states and are directly related to the annihilation and creation, respectively, of a photon.

      The wavelengths or energies from the hydrogen emission or absorption experiments were fit by an empirical equation known as the Rydberg equation, which gave the energy “states” of the hydrogen atom as

      In this equation, n is an integer (>0) “quantum” number, and Ry is the Rydberg constant, (Ry = 2.179 × 10−18 J). This equation implies that the energy of the hydrogen atom cannot assume arbitrary energy values, but only “quantized” levels, E(n). This observation led to the ideas of electrons in stationary planetary orbits around the nucleus, which – however – was in contradiction with existing knowledge of electrodynamics, as discussed in the beginning of this chapter.

Schematic illustration of the energy level diagram of the hydrogen atom. Transitions between the energy levels are indicated by vertical lines.

      Equation (1.19) provided a background framework to explain the hydrogen atom emission spectrum. According to Eq. (1.19), the energy of a photon, or the energy difference of the atomic energy levels, between any two states nf and ni can be written as

      (1.20)upper E Subscript photon Baseline equals normal upper Delta upper E Subscript atom Baseline equals upper E Subscript f Baseline minus upper E Subscript i Baseline equals minus upper R Subscript y Baseline left-bracket StartFraction 1 Over n Subscript f Superscript 2 Baseline EndFraction minus StartFraction 1 Over n Subscript i Superscript 2 Baseline EndFraction right-bracket

      At this point, an example may be appropriate to demonstrate how this empirically derived equation predicts the energy, wavelength, and wavenumber of light emitted by hydrogen atoms. This example also introduces a common problem, namely, that of units. Although there is an international agreement about what units (the système international, or SI units) are to be used to describe spectral transitions, the problem is that few people are using them. In this book, all efforts will be made to use SI units, or at least give the conversion to other units.

      Example 1.2 Calculation of the energy, frequency, wavelength, and wavenumber of a photon emitted by a hydrogen atom undergoing a transition from n = 6 to n = 2.

      Answer:

      The energy difference between the two states of the hydrogen atom is given by

      (E1.2.1)normal upper Delta upper E Subscript atom Baseline equals upper E Subscript f Baseline minus upper E Subscript i Baseline equals minus upper R Subscript y Baseline left-bracket StartFraction 1 Over 6 squared EndFraction minus StartFraction 1 Over 2 squared EndFraction right-bracket equals upper R Subscript y Baseline left-bracket one fourth minus one thirty-sixth right-bracket equals 0.22222 upper R Subscript y

      Using the value of the Rydberg constant given above, Ry = 2.179 × 10−18 J, the energy difference is

      (E1.2.2)normal upper Delta upper E Subscript atom Baseline equals 4.842 times 10 Superscript negative 19 Baseline upper J period

      (E1.2.3)normal nu equals StartFraction normal upper Delta upper E Over h EndFraction equals StartFraction 4.842 times 10 Superscript negative 19 Baseline Over 6.626 times 10 Superscript negative 34 Baseline EndFraction equals 7.308 times 10 Superscript 15 Baseline left-bracket StartFraction upper J Over italic upper J s EndFraction equals s Superscript negative 1 Baseline right-bracket equals left-bracket italic upper H z right-bracket

      The wavelength of such a photon is given by Eq. (1.7) as

      (E1.2.4)normal lamda equals StartFraction h c Over upper E EndFraction equals StartFraction 6.626 times 10 Superscript negative 34 Baseline times 2.998 times 10 Superscript 8 Baseline Over 4.842 times 10 Superscript negative 19 Baseline EndFraction equals 4.102 times 10 Superscript negative 7 Baseline left-bracket StartFraction italic upper J s m Over upper J s EndFraction equals normal m right-bracket equals 410.2 left-bracket n m right-bracket

      that is, a photon in the ultraviolet wavelength range. Finally, the wavenumber of this photon is

      (E1.2.5)normal nu overTilde equals StartFraction 1 Over normal lamda EndFraction equals StartFraction 1 Over 4.102 times 10 Superscript negative 7 Baseline EndFraction equals 2.438 times 10 Superscript 6 Baseline left-bracket normal m Superscript negative 1 Baseline right-bracket

      This is a case where the SI units are used infrequently, and the results for the wavenumber are usually given by spectroscopists in units of cm−1, where 1 m−1 = 10−2 cm−1. Accordingly, the results in Eq. E1.5 is written as