Question

Estimate the ratio of the number of molecules in the first excited vibrational state of the molecule$N_2$ to the number in the ground state, at a temperature of$450K$. The vibrational frequency of${\mathbf{N}}_2$ is.$\mathbf{7}.\mathbf{07}\times {\mathbf{10}}^{13}{\mathbf{s}}^{-\mathbf{1}}$

Short answer

Theratio of the number of molecules in the first excited vibrational state of the molecule$N2$ to the number in the ground state, at a temperature of$450K$ is$\frac{N_1}{N_2}=1.1577\times {10}^{-38}$

Step-by-step solution

Definition of Fundamental state

When vibrational states where only one mode is excited, by one quantum, are called fundamental states.

Relative population of two energy N=N

The relative population of two energy levels is expressed as follows:

$\frac{P\left(\eta 1\right)}{P\left(\eta 2\right)}=exp\left(\frac{-[\mathbf{\epsilon }{\eta }_1-\mathbf{\epsilon }{\eta }_2]}{k_{b}t}\right)$

Here,$P\left(\eta 1\right)$is the probability of finding molecules in energy level$\eta 2$,$P\left(\eta 2\right)$is the probability of finding molecules in energy level$\eta 2$,$\mathbf{\epsilon }\eta$is the energy of state$\eta 1$, $\mathbf{\epsilon }{\eta }_2$is the energy of state$\eta 2$,$k_b$is the Boltzmann constant and $\mathbf{T}$is the temperature of the molecule having value$\mathbf{450}\mathbf{K}.$

Estimate the ratio of the number of molecules

At thermal equilibrium, the ratio $\frac{N_1}{N_2}$is given as follow,

$\frac{N_1}{N_2}=exp[-\frac{E_2-E_1}{kT}]$
The middle of the visible range is taken at$\lambda =550nm$

⇒$\frac{E_2}{E_1}=\frac{h_c}{\lambda }=3.16\times {10}^{-19}J$

⇒$\frac{N_1}{N_2}=exp$$\left(\frac{-3.16\times {10}^{-19}}{(1.38\times {10}^{-23}\frac{1}{2}).\left(450k\right)}\right)$

⇒$\frac{N_1}{N_2}=1.1577\times {10}^{-38}$

Theratio of the number of molecules in the first excited vibrational state of the molecule $N2$to the number in the ground state, at a temperature of$450K$ is$\frac{N_1}{N_2}=1.1577\times {10}^{-38}$