(30) Given the equilibrium bond length of CO is 1.138, explore population fractions (Eq. 2) of…

(30) Given the equilibrium bond length of CO is 1.138, explore population fractions (Eq. 2) of the ground state and first 15 pure rotational excited states relative to the ground state (where {=0). Mathematica is recommended. Do this at 298 K and at 100 K. Comment on how the results differ compared to part a). 1) In the application of quantum mechanical and statistical mechanical principles to samples containing large numbers of species (e.g. macroscopic samples of molecules), there is a celebrated equation due to Boltzmann that quantifies the fraction of species (1) that are in the ih energy level (xi) at a give temperature T (we will hopefully show how this is derived towards the end of the class): fi = liel/kg7 (1) Here, gi refers to the degeneracy of the ih energy level and kg is Boltzmann’s constant. The denominator q is called the partition function. It serves to count up the total number of thermally accessible states at a given temperature. Rather than try to calculate q here (we will need more time to develop this later), lets take a ratio of fractions, thereby eliminating dependence on q (which would be the same for both fractions in the ratio): gjej/kg? _91-(8j-ej)/kgT = 91e-A£ji/kgT figie – €/kg7 (2) This expression is nice because even though we won’t know the fraction of molecules from our macroscopic sample in any one energy level, we will be able to determine how the fraction changes depending on energy level. So for example, if i refers to the ground energy level where e=0 and f refers to an excited energy level (l), we will be able to find out how the excited state population changes relative to the ground state as a function of the energy gap between them as well as changes in degeneracy. Critically, we can also explore the role of temperature. This is exactly the kind of expression that will allow us to understand the overall shape of the CO pure rotational spectrum from last week (how transmission changes as a function of frequency; why it goes up and then goes down). a) (20) Given the force constant for CO is 1860 N/m, calculate the relative population (compared to the ground state) in the first (v=1), second (v=2), and third (v=3) vibrational excited states. Do this at 298 K and at 100 K. Mathematica is recommended. Assume CO is acting as a pure harmonic oscillator.