in what region vibration-rotational spectral lines are observed?
Answers
Answer:
Explanation:
Consider the rotational constant for a rigid rotator B~=h/8π2cμR2e where B~ is the rotational constant, and R2e is the bond length. It is clear from the equation that B~ depends on R2e in a inverse manner, so B~ will decrease as R2e increase, and vice versa.
Vibrational state and the bond length also possess a relationship, and that is bond length, R2e, will increase as the vibrational state, v increases. High vibrational states have large vibrational amplitudes (the amplitude can be visualized as the distance from one side to the other on the harmonic-oscillator parabola, and higher states has greater distance). On a potential graph, imagine a diatomic molecule with one of the atom fixed at the origin, the bond is along the x-axis, and the molecule vibrate toward and away from the origin along x-axis. The range at how much the bond can stretch or squeeze depend on the vibrational amplitude. So, at high vibrational state (thus high amplitude), the bond can squeeze in and stretch out more (just a little more) on the regular harmonic-oscillator potential, but it can stretch out greatly on an anharmonic oscillator. That's why bond length increase with vibrational state.
Anharmonic_oscillator.gif
Figure used with permission from Wikipedia.
So far, there are two relationship established above
B~∝R2e(13.3.1)
(inversely proportional)
and
R2e∝v(13.3.2)
(i.e., directly proportional). Therefore,
B~∝v(13.3.3)
(i.e., inversely proportional). The rotational constant dependent on the vibrational state is denoted as B~v, and the dependence of B~ on v is the vibrational-rotational interaction.
Frequencies of R and P Branches
The energies of the rigid rotator-harmonic oscillator is
E~v,J=G(v)+F(J)=ν~(v+12)+B~J(J+1)(13.3.4)
or in term of B~v
E~v,J=ν~(v+12)+B~vJ(J+1)(13.3.5)
Frequencies of vibrational transition from v=0→1 can be calculated from the equation above as the following:
For R branch, allowed J=0,1,2,...
ν~R(ΔJ=+1)=E~1,J+1−E~0,J(13.3.6)
=32ν~+B~1(J+1)(J+2)−12ν~−B~0J(J+1)(13.3.7)
=ν~+2B~1+(3B~1−B~0)J+(B~1−B~0)J2(13.3.8)
For P branch, allowed J=1,2,3,...
ν~P(ΔJ=−1)=E~1,J−1−E~0,J(13.3.9)
=ν~−(B~1+B~0)J+(B~1−B~0)J2(13.3.10)
According to the vibration-rotation interaction, B~1<B~0 because R2e(v=1)>R2e(v=0).
Decrease in Spacing of Lines in the R Branch with Increasing J
A closer look at
ν~R=ν~+2B~1+(3B~1−B~0)J+(B~1−B~0)J2(13.3.11)
shows that the last term in the parentheses (B~1−B~0) will be always negative because B~1<B~0, and it also multiplies with J2, so the square term (B~1−B~0)J2 will give a larger negative value for increasing J. As a result, ν~R=ν~+(smaller and smaller value) as J increase. Because the rotational frequencies keep getting smaller, the spacing between lines in R branch decreases as J increases.
Increasing in Spacing of Lines in the P Branch with Decreasing J
Using the same analysis as above,
ν~P=ν~−(B~1+B~0)J+(B~1−B~0)J2(13.3.12)
shows that the last term in the parentheses (B~1−B~0) will be always negative because B~1<B~0, and it also multiplies with J2, so the square term (B~1−B~0)J2 will give a larger negative value even for decreasing J. As a result, ν~R=ν~−(smaller and smaller value) as J decreases. Because the rotational frequencies keep getting larger, the spacing between lines in P branch increase as J decreases.