Question

Estimate the ratio of the number of molecules in the first excited vibrational state of the molecule${\mathbf{\text{N}}}_{\mathbf{2}}$ to the number in the ground state, at a temperature of$\mathbf{450}\mathit{K}$ . The vibrational frequency of${\mathbf{\text{N}}}_{\mathbf{2}}$ is$\mathbf{7}\mathbf{.07}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}\mathbf{}{\mathbf{\text{s}}}^{\mathbf{-}\mathbf{1}}$ .

Short answer

The ratio of the first excited vibrational molecule state to the other in the ground state is equal to $5.25\times {10}^{-4}$.

Step-by-step solution

Given information

Vibrational frequency:$\nu =7.07\times {10}^{13}{\text{s}}^{-1}$

Temperature:$450K$

Concept ofMaxwell–Boltzmann distribution

The Maxwell–Boltzmann distribution is concerned with the energy distribution between identical but distinct particles. It denotes the likelihood of the distribution of states in a system with varying energies.

The so-called Maxwell distribution law of molecular velocities is a specific instance.

Calculate the vibrational frequency

First, find the difference between the energy levels from the vibrational frequency:

$\begin{array}{c}{ϵ}_2-{ϵ}_1=h\nu \\ =(6.63\times {10}^{-34}\text{Js})\left(7.07\times {10}^{1}3{\text{s}}^{-1}\right)\\ =4.69\times {10}^{-20}\text{J}\end{array}$

Calculate the ratio of molecules

The ratio of molecules in the first excited state to the ground state is shown below.

$\begin{array}{c}\frac{P_2}{P_1}=\exp \left(\frac{{ϵ}_1-{ϵ}_2}{K_{b}T}\right)\\ =\exp \left(\frac{-4.69\times {10}^{-20}J}{(1.38\times {10}^{-23}J{K}^{-1})\left(450K\right)}\right)\\ =5.25\times {10}^{-4}\end{array}$