Question

The vibrational frequency of themolecule is$\mathbf{1}.\mathbf{15}\times {\mathbf{10}}^{13}{\mathbf{s}}^{-\mathbf{1}}$. For every million$(\mathbf{1}.\mathbf{00}\times {\mathbf{10}}^6)$molecules in the ground vibrational state, how many will be in the firstexcited vibrational state at a temperature of$\mathbf{300}\mathbf{K}$?

Short answer

The first excited vibrational state at a temperature of $300K$,${\epsilon }_n=24.09\times {10}^{-15}$

Step-by-step solution

Definition of fundamental state

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

Boltzmann distribution

We apply the Boltzmann distribution to describe the probability of finding molecules in each of the vibrational states in a sample of$CO$held at temperature$\mathbf{T}$. We describe the vibrational motions using the harmonic oscillator model, for which the allowed energy levels are

${\mathbf{\epsilon }}_{\eta }=(\eta +\frac{1}{2})h\nu$

Calculation number of first excited vibrational state

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 $T$is the temperature of the molecule having value.$300K$

${\mathbf{\epsilon }}_{\eta }=(\eta +\frac{1}{2})h\nu$

${\epsilon }_n=[1\times {10}^6+\frac{1}{2}]$$6.626\times {10}^{-34}\times 1.15\times {10}^{13}$

${\epsilon }_n=(3.162\times {10}^6).(7.619\times {10}^{-21})$

${\epsilon }_n=24.09\times {10}^{-15}joules$

The first excited vibrational state at a temperature of $300K$,${\epsilon }_n=24.09\times {10}^{-15}$