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== 9.2#h(0.6em)Molecular Spectra


  Molecular energy levels are more complicated than atomic energy levels because molecules can also vibrate and rotate. The energies associated with such motions lie in different ranges and can therefore be studied separately. Electronic transitions are of order 1 eV, vibrational transitions are of order #math.equation(block: false, alt: "10 to the power −2 eV ,")[$10^(−2) #h(0.2em) "eV" ,$] and rotational transitions are of order #math.equation(block: false, alt: "10 to the power −3 eV .")[$10^(−3) #h(0.2em) "eV" "."$] For complex molecules, these energy changes are difficult to characterize, so we begin with the simple case of a diatomic molecule.


  According to classical mechanics, the energy of rotation of a diatomic molecule is given by


  #math.equation(block: true, alt: "E sub r equals the fraction L squared over 2 I ,")[$E_(r) = frac(L^(2), 2 I) ,$]
  where #emph[I] is the moment of inertia and #emph[L] is the angular momentum. According to quantum mechanics, the rotational angular momentum is quantized:


  #math.equation(block: true, alt: "L equals the square root of l open parenthesis l plus 1 close parenthesis ℏ open parenthesis l equals 0 , 1 , 2 , 3 ,... close parenthesis ,")[$L = sqrt(l ( l + 1 )) ℏ #h(0.2em) ( l = 0 , 1 , 2 , 3 ",..." ) ,$]
  where #emph[l] is the orbital angular quantum number. The allowed #strong[rotational energy level] of a diatomic molecule is therefore


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "E sub r equals l open parenthesis l plus 1 close parenthesis the fraction ℏ squared over 2 I equals l open parenthesis l plus 1 close parenthesis E sub 0 r open parenthesis l equals 0 , 1 , 2 , 3 ,... close parenthesis ,")[$E_(r) = l ( l + 1 ) frac(ℏ^(2), 2 I) = l ( l + 1 ) E_(0 r) #h(0.5em) ( l = 0 , 1 , 2 , 3 ",..." ) ,$]
]


  where the characteristic rotational energy of a molecule is defined as


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "E sub 0 r equals the fraction ℏ squared over 2 I .")[$E_(0 r) = frac(ℏ^(2), 2 I) .$]
]


  For a diatomic molecule, the moment of inertia with reduced mass #math.equation(block: false, alt: "μ")[$μ$] is


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "I equals μ r 0 2 ,")[$I = μ r_(0)^(2) ,$]
]


  where #math.equation(block: false, alt: "r sub 0")[$r_(0)$] is the total distance between the atoms. The energy difference between rotational levels is therefore


  #math.equation(block: true, alt: "Δ E sub r equals E sub l plus 1 minus E sub l equals 2 open parenthesis l plus 1 close parenthesis E sub 0 r .")[$"Δ" E_(r) = E_(l + 1) − E_(l) = 2 ( l + 1 ) #h(0.2em) E_(0 r) .$]
  A detailed study of transitions between rotational energy levels brought about by the absorption or emission of radiation (a so-called #strong[electric dipole transition]) requires that


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "Δ l equals plus or minus 1 .")[$"Δ" l = ± 1 .$]
]


  This rule, known as a #strong[selection rule], limits the possible transitions from one quantum state to another.  is the selection rule for rotational energy transitions. It applies only to diatomic molecules that have an electric dipole moment. For this reason, symmetric molecules such as #math.equation(block: false, alt: "H sub 2")[$"H"_(2)$] and #math.equation(block: false, alt: "N sub 2")[$"N"_(2)$] do not experience rotational energy transitions due to the absorption or emission of electromagnetic radiation.


  #examplebox("Example 1")[The Rotational Energy of HCl][
Determine the lowest three rotational energy levels of a hydrogen chloride (HCl) molecule.


    Strategy
Hydrogen chloride (HCl) is a diatomic molecule with an equilibrium separation distance of 0.127 nm. Rotational energy levels depend only on the momentum of inertia #emph[I] and the orbital angular momentum quantum number #emph[l] (in this case, #math.equation(block: false, alt: "l equals 0")[$l = 0$], 1, and 2). The momentum of inertia depends, in turn, on the equilibrium separation distance (which is given) and the reduced mass, which depends on the masses of the H and Cl atoms.


    Solution
First, we compute the reduced mass. If Particle 1 is hydrogen and Particle 2 is chloride, we have


    #math.equation(block: true, alt: "μ equals the fraction m sub 1 m sub 2 over m sub 1 plus m sub 2 equals the fraction open parenthesis 1.0 u close parenthesis open parenthesis 35.4 u close parenthesis over 1.0 u plus 35.4 u equals 0.97 u equals 0.97 u open parenthesis the fraction 931.5 the fraction MeV over c squared over 1 u close parenthesis equals 906 the fraction MeV over c squared .")[$μ = frac(m_(1) m_(2), m_(1) + m_(2)) = frac(( 1.0 #h(0.2em) "u" ) ( 35.4 #h(0.2em) "u" ), 1.0 #h(0.2em) "u" + 35.4 #h(0.2em) "u") = 0.97 #h(0.2em) "u" = 0.97 #h(0.2em) "u" ( frac(931.5 frac("MeV", c^(2)), 1 #h(0.2em) "u") ) = 906 frac("MeV", c^(2)) .$]
    The corresponding rest mass energy is therefore


    #math.equation(block: true, alt: "μ c squared equals 9.06 times 10 to the power 8 eV .")[$μ c^(2) = 9.06 #h(0.2em) × #h(0.2em) 10^(8) #h(0.2em) "eV" .$]
    This allows us to calculate the characteristic energy:


    #math.equation(block: true, alt: "E sub 0 r equals the fraction ℏ squared over 2 I equals the fraction ℏ squared over 2 open parenthesis μ r 0 2 close parenthesis equals the fraction open parenthesis ℏ c close parenthesis squared over 2 open parenthesis μ c squared close parenthesis r 0 2 equals the fraction open parenthesis 197.3 eV times nm close parenthesis squared over 2 open parenthesis 9.06 times 10 to the power 8 eV close parenthesis open parenthesis 0.127 nm close parenthesis squared equals 1.33 times 10 to the power −3 eV .")[$E_(0 r) = frac(ℏ^(2), 2 I) = frac(ℏ^(2), 2 ( μ r_(0)^(2) )) = frac(attach(( ℏ c ), t: 2), 2 ( μ c^(2) ) r_(0)^(2)) = frac(attach(( 197.3 #h(0.2em) "eV" · "nm" ), t: 2), 2 ( 9.06 #h(0.2em) × #h(0.2em) 10^(8) #h(0.2em) "eV" ) attach(( 0.127 #h(0.2em) "nm" ), t: 2)) = 1.33 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "eV" "."$]
    (Notice how this expression is written in terms of the rest mass energy. This technique is common in modern physics calculations.) The rotational energy levels are given by


    #math.equation(block: true, alt: "E sub r equals l open parenthesis l plus 1 close parenthesis the fraction ℏ squared over 2 I equals l open parenthesis l plus 1 close parenthesis E sub 0 r ,")[$E_(r) = l ( l + 1 ) frac(ℏ^(2), 2 I) = l ( l + 1 ) E_(0 r) ,$]
    where #emph[l] is the orbital quantum number. The three lowest rotational energy levels of an HCl molecule are therefore


    #math.equation(block: true, alt: "l equals 0 ; E sub r equals 0 eV open parenthesis no rotation close parenthesis ,")[$l = 0 ; E_(r) = 0 #h(0.2em) "eV" #h(0.2em) ( "no rotation" ) ,$]
    #math.equation(block: true, alt: "l equals 1 ; E sub r equals 2 E sub 0 r equals 2.66 times 10 to the power −3 eV ,")[$l = 1 ; E_(r) = 2 #h(0.2em) E_(0 r) = 2.66 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "eV" ,$]
    #math.equation(block: true, alt: "l equals 2 ; E sub r equals 6 E sub 0 r equals 7.99 times 10 to the power −3 eV .")[$l = 2 ; E_(r) = 6 #h(0.2em) E_(0 r) = 7.99 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "eV" .$]
    Significance
The rotational spectrum is associated with weak transitions (1/1000 to 1/100 of an eV). By comparison, the energy of an electron in the ground state of hydrogen is #math.equation(block: false, alt: "−13.6 eV")[$−13.6 #h(0.2em) "eV"$].
]


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
What does the energy separation between absorption lines in a rotational spectrum of a diatomic molecule tell you?


    
    #solutionbox[
the moment of inertia
]
]


  The #strong[vibrational energy level], which is the energy level associated with the vibrational energy of a molecule, is more difficult to estimate than the rotational energy level. However, we can estimate these levels by assuming that the two atoms in the diatomic molecule are connected by an ideal spring of spring constant #emph[k]. The potential energy of this spring system is


  #math.equation(block: true, alt: "U sub osc equals the fraction 1 over 2 k Δ r squared ,")[$U_("osc") = frac(1, 2) k #h(0.2em) "Δ" r^(2) ,$]
  Where #math.equation(block: false, alt: "Δ r")[$"Δ" r$] is a change in the “natural length” of the molecule along a line that connects the atoms. Solving Schrödinger’s equation for this potential gives


  #math.equation(block: true, alt: "E sub n equals open parenthesis n plus the fraction 1 over 2 close parenthesis ℏ ω open parenthesis n equals 0 , 1 , 2 , … close parenthesis ,")[$E_(n) = ( n + frac(1, 2) ) ℏ ω #h(0.2em) ( n = 0 , 1 , 2 , "…" ) ,$]
  Where #math.equation(block: false, alt: "ω")[$ω$] is the natural angular frequency of vibration and #emph[n] is the vibrational quantum number. The prediction that vibrational energy levels are evenly spaced #math.equation(block: false, alt: "open parenthesis Δ E equals ℏ ω close parenthesis")[$( "Δ" E = ℏ ω )$] turns out to be good at lower energies.


  A detailed study of transitions between vibrational energy levels induced by the absorption or emission of radiation (and the specifically so-called electric dipole transition) requires that


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "Δ n equals plus or minus 1 .")[$"Δ" n = ± 1 .$]
]


  represents the selection rule for vibrational energy transitions. As mentioned before, this rule applies only to diatomic molecules that have an electric dipole moment. Symmetric molecules do not experience such transitions.


  Due to the selection rules, the absorption or emission of radiation by a diatomic molecule involves a transition in vibrational and rotational states. Specifically, if the vibrational quantum number (#emph[n]) changes by one unit, then the rotational quantum number (#emph[l]) changes by one unit. An energy-level diagram of a possible transition is given in . The absorption spectrum for such transitions in hydrogen chloride (HCl) is shown. The absorption peaks are due to transitions from the #math.equation(block: false, alt: "n equals 0")[$n = 0$] to #math.equation(block: false, alt: "n equals 1")[$n = 1$] vibrational states. Energy differences for the band of peaks at the left and right are, respectively, #math.equation(block: false, alt: "Δ E sub l → l plus 1 equals ℏ ω plus 2 open parenthesis l plus 1 close parenthesis E sub 0 r equals ℏ ω plus 2 E sub 0 r , ℏ ω plus 4 E sub 0 r , ℏ ω plus 6 E sub 0 r , … open parenthesis right band close parenthesis")[$"Δ" E_(l → l + 1) = ℏ ω + 2 ( l + 1 ) E_(0 r) = ℏ ω + 2 E_(0 r) , #h(0.2em) ℏ ω + 4 E_(0 r) , #h(0.2em) ℏ ω + 6 E_(0 r) , "…" #h(0.2em) ( "right band" )$] and #math.equation(block: false, alt: "Δ E sub l → l −1 equals ℏ ω minus 2 l E sub 0 r equals ℏ ω minus 2 E sub 0 r , ℏ ω minus 4 E sub 0 r , ℏ ω minus 6 E sub 0 r , … open parenthesis left band close parenthesis .")[$"Δ" E_(l → l −1) = ℏ ω − 2 l E_(0 r) = ℏ ω − 2 E_(0 r) , #h(0.2em) ℏ ω − 4 E_(0 r) , #h(0.2em) ℏ ω − 6 E_(0 r) , "…" #h(0.2em) ( "left band" ) "."$]


  The moment of inertia can then be determined from the energy spacing between individual peaks #math.equation(block: false, alt: "open parenthesis 2 E sub 0 r close parenthesis")[$( 2 E_(0 r) )$] or from the gap between the left and right bands #math.equation(block: false, alt: "open parenthesis 4 E sub 0 r close parenthesis")[$( 4 E_(0 r) )$]. The frequency at the center of this gap is the frequency of vibration.


  #figure(figph[Figure shows a graph of energy versus internuclear separation. There are two curves on the graph. The curve at the bottom is labeled ground state and the one at the top is labeled excited electronic state. Both are similar in shape, with a sharp dip to a trough, followed by a slow rise till the curve evens out. The ground state curve has five horizontal blue lines bounded by the curve, which look like rungs of a ladder. These are labeled vibrational energy level. Between two blue rungs are smaller purple rungs labeled rotational level. There are four such purple rungs each, between the first and second blue rungs, the second and third blue rungs and the third and fourth blue rungs. There is an arrow pointing up from the center of the trough. To the left of this arrow is a smaller arrow pointing up. This extends from the first purple rung of the first blue rung to the second purple rung of the second blue rung. The excited state curve has four blue rungs.], alt: "Figure shows a graph of energy versus internuclear separation. There are two curves on the graph. The curve at the bottom is labeled ground state and the one at the top is labeled excited electronic state. Both are similar in shape, with a sharp dip to a trough, followed by a slow rise till the curve evens out. The ground state curve has five horizontal blue lines bounded by the curve, which look like rungs of a ladder. These are labeled vibrational energy level. Between two blue rungs are smaller purple rungs labeled rotational level. There are four such purple rungs each, between the first and second blue rungs, the second and third blue rungs and the third and fourth blue rungs. There is an arrow pointing up from the center of the trough. To the left of this arrow is a smaller arrow pointing up. This extends from the first purple rung of the first blue rung to the second purple rung of the second blue rung. The excited state curve has four blue rungs.", caption: [Three types of energy levels in a diatomic molecule: electronic, vibrational, and rotational. If the vibrational quantum number (#emph[n]) changes by one unit, then the rotational quantum number (#emph[l]) changes by one unit.])


  #figure(figph[Graph of intensity versus frequency in Hertz. The curve consists of several pairs of spikes. The spikes have low intensity at the beginning of the curve and also at the end of the curve at 9.2 into 10 to the power 13 hertz. The spikes are longer near the middle but dip at the center. The center frequency for n equal to 0 to n equal to 1 is approximately 8.65 into 10 to the power 13 Hertz. The left side of the graph is labeled transitions where the vibrational energy increases, n=0 to 1 and the rotational angular momentum decreases, j to j minus 1. The right side of the graph is labeled transitions where the vibrational energy increases, n=0 to 1 and the rotational angular momentum increases, j to j plus 1.], alt: "Graph of intensity versus frequency in Hertz. The curve consists of several pairs of spikes. The spikes have low intensity at the beginning of the curve and also at the end of the curve at 9.2 into 10 to the power 13 hertz. The spikes are longer near the middle but dip at the center. The center frequency for n equal to 0 to n equal to 1 is approximately 8.65 into 10 to the power 13 Hertz. The left side of the graph is labeled transitions where the vibrational energy increases, n=0 to 1 and the rotational angular momentum decreases, j to j minus 1. The right side of the graph is labeled transitions where the vibrational energy increases, n=0 to 1 and the rotational angular momentum increases, j to j plus 1.", caption: [Absorption spectrum of hydrogen chloride (HCl) from the #math.equation(block: false, alt: "n equals 0 to n equals 1")[$n = 0 #h(0.2em) "to" #h(0.2em) n = 1$] vibrational levels. The discrete peaks indicate a quantization of the angular momentum of the molecule. The bands to the left indicate a decrease in angular momentum, whereas those to the right indicate an increase in angular momentum.])


  
    
=== Summary


    - Molecules possess vibrational and rotational energy.
- Energy differences between adjacent vibrational energy levels are larger than those between rotational energy levels.
- Separation between peaks in an absorption spectrum is inversely related to the moment of inertia.
- Transitions between vibrational and rotational energy levels follow selection rules.


  
  
    
=== Conceptual Questions


    
      Does the absorption spectrum of the diatomic molecule HCl depend on the isotope of chlorine contained in the molecule? Explain your reasoning.


    
    
      Rank the energy spacing #math.equation(block: false, alt: "open parenthesis Δ E close parenthesis")[$( "Δ" E )$] of the following transitions from least to greatest: an electron energy transition in an atom (atomic energy), the rotational energy of a molecule, or the vibrational energy of a molecule?


    
    #solutionbox[
rotational energy, vibrational energy, and atomic energy
]


    
      Explain key features of a vibrational-rotation energy spectrum of the diatomic molecule.


    
  
  
    
=== Problems


    
      In a physics lab, you measure the vibrational-rotational spectrum of HCl. The estimated separation between absorption peaks is #math.equation(block: false, alt: "Δ f approximately equals 5.5 times 10 to the power 11 Hz")[$"Δ" f ≈ 5.5 #h(0.2em) × #h(0.2em) 10^(11) #h(0.2em) "Hz"$]. The central frequency of the band is #math.equation(block: false, alt: "f sub 0 equals 9.0 times 10 to the power 13 Hz")[$f_(0) = 9.0 #h(0.2em) × #h(0.2em) 10^(13) #h(0.2em) "Hz"$]. (a) What is the moment of inertia (#emph[I])? (b) What is the energy of vibration for the molecule?


    
    
      For the preceding problem, find the equilibrium separation of the H and Cl atoms. Compare this with the actual value.


    
    #solutionbox[
The measured value is 0.484 nm, and the actual value is close to 0.127 nm. The laboratory results are the same order of magnitude, but a factor 4 high.
]


    
      The separation between oxygen atoms in an #math.equation(block: false, alt: "O sub 2")[$"O"_(2)$] molecule is about 0.121 nm. Determine the characteristic energy of rotation in eV.


    
    
      The characteristic energy of the #math.equation(block: false, alt: "N sub 2")[$"N"_(2)$] molecule is #math.equation(block: false, alt: "2.48 times 10 to the power −4 eV")[$2.48 #h(0.2em) × #h(0.2em) 10^(−4) #h(0.2em) "eV"$]. Determine the separation distance between the nitrogen atoms


    
    #solutionbox[
0.110 nm
]


    
      The characteristic energy for KCl is #math.equation(block: false, alt: "1.4 times 10 to the power −5 eV.")[$1.4 #h(0.2em) × #h(0.2em) 10^(−5) #h(0.2em) "eV."$] (a) Determine #math.equation(block: false, alt: "μ")[$μ$] for the KCl molecule. (b) Find the separation distance between the K and Cl atoms.


    
    
      A diatomic #math.equation(block: false, alt: "F sub 2")[$"F"_(2)$] molecule is in the #math.equation(block: false, alt: "l equals 1")[$l = 1$] state. (a) What is the energy of the molecule? (b) How much energy is radiated in a transition from a #math.equation(block: false, alt: "l equals 2")[$l = 2$] to a #math.equation(block: false, alt: "l equals 1")[$l = 1$] state?


    
    #solutionbox[
a. #math.equation(block: false, alt: "E equals 2.2 times 10 to the power −4 eV")[$E = 2.2 #h(0.2em) × #h(0.2em) 10^(−4) #h(0.2em) "eV"$]; b. #math.equation(block: false, alt: "Δ E equals 4.4 times 10 to the power −4 eV")[$"Δ" E = 4.4 #h(0.2em) × #h(0.2em) 10^(−4) #h(0.2em) "eV"$]
]


    
      In a physics lab, you measure the vibrational-rotational spectrum of potassium bromide (KBr). The estimated separation between absorption peaks is #math.equation(block: false, alt: "Δ f approximately equals 5.35 times 10 to the power 10 Hz")[$"Δ" f ≈ 5.35 #h(0.2em) × #h(0.2em) 10^(10) #h(0.2em) "Hz"$]. The central frequency of the band is #math.equation(block: false, alt: "f sub 0 equals 8.75 times 10 to the power 12 Hz")[$f_(0) = 8.75 #h(0.2em) × #h(0.2em) 10^(12) #h(0.2em) "Hz"$]. (a) What is the moment of inertia (#emph[I])? (b) What is the energy of vibration for the molecule?