The bond length of the${\mathbf{\text{N}}}_{\mathbf{2}}$ molecule is$\mathbf{0}\mathbf{.}\mathbf{11}\mathbf{}\mathbf{ }\mathbf{\text{nm}}$ , and its effective spring constant is $\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{\text{N}}\mathbf{/}\mathbf{\text{m}}$at room temperature.

(a) What would be the ratio of molecules with rotational quantum number$\mathit{\ell }\mathbf{=}\mathbf{1}$ to those with$\mathit{\ell }\mathbf{=}\mathbf{0}$ (at the same vibrational level), and

(b) What would be the ratio of molecules with vibrational quantum number$\mathit{n}\mathbf{=}\mathbf{1}$ to those with $\mathit{n}\mathbf{=}\mathbf{0}$(with the same rotational energy)?

The bond length of the molecule is $0.11 \text{nm}$.

Effective spring constant is $2.3\times {10}^3\text{N}/\text{m}$.

Movements of one atom within a molecule in relation to other atoms are known as molecular vibrations. The smallest vibrations are lengthening and shortening of a single link. To understand this, picture the bond as a spring and the molecular vibration as the spring's simple harmonic motion.

The expression for rotational energy is,$E_{rot}=\frac{{\hslash }^{2}l(l+1)}{2\mu a^2}$.

The reduced mass $\mu$ is, $\mu =\frac{m_1{m}_2}{m_1+m_3}$.

The expression for vibrational energy is, $E_{\text{vib}}=\left(n+\frac{1}{2}\right)\hslash \sqrt{\frac{k}{\mu }}$.

Substitute $14.007u$ for $m_1 , m_2$ in the expression $\mu =\frac{m_1{m}_2}{m_1+m_2}$ to find the reduced mass of nitrogen molecule.

$\begin{array}{rcl}\mu & =& \frac{(14.007u)(14.007u)}{(14.007u)+(14.007u)}\\ & =& 7.0035u\\ & =& 7.0035u\left(\frac{1.66\times {10}^{-27}\text{kg}}{1u}\right)\\ & =& 1.16\times {10}^{-26}\text{kg}\\ & & \end{array}$

Use the expression $E=\frac{{\hslash }^{2}l(l+1)}{2\mu a^2}$, calculate the difference of rotational energy.

$\begin{array}{rcl}{E}_{0,1}-E_{0,0}& =& \frac{{\hslash }^2\left(1\right)(1+1)}{2\mu a^2}-\frac{{\hslash }^2\left(0\right)(0+1)}{2\mu a^2}\\ & =& \frac{{\hslash }^2}{\mu a^2}\\ & & \end{array}$

Substitute $1.055\times {10}^{-34}\text{J}.\text{s}$ for$\hslash ,1.16\times {10}^{-26}\text{kg}$ for $\mu$ and $0.11\times {10}^{-9}\text{m}$ for in the above expression.

$\begin{array}{rcl}{E}_{0,1}-E_{0,0}& =& \frac{{\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{\left(1.16\times {10}^{-26}\text{kg}\right){\left(0.11\times {10}^{-9}\text{m}\right)}^2}\\ & =& 7.91\times {10}^{-23}\text{J}\\ & & \end{array}$

According to Boltzmann distribution, the ratio of the number of molecules with rotational quantum number $l=1$to the number of molecules with rotational quantum number is given by, where take three rotational modes for $l=1$ into consideration.

$\begin{array}{rcl}\frac{\left(E_{n=1}\right)}{\left(E_{n=0}\right)}& =& \frac{{\left(NA{e}^{-\frac{E_n}{K_{B}T}}\right)}_{n=1}}{{\left(NA{e}^{-\frac{E_n}{K_{B}T}}\right)}_{n=0}}\\ & =& \frac{NA\left(\exp {\left[\frac{-E_{vib}}{K_{B}T}\right]}_{n=1}\right)\left(\exp {\left[\frac{-E_{rot}}{K_{B}T}\right]}_{n=1}\right)}{NA\left(\exp {\left[\frac{-E_{vib}}{K_{B}T}\right]}_{n=1}\right)\left(\exp {\left[\frac{-E_{rot}}{K_{B}T}\right]}_{n=1}\right)}\\ {\left(E_{rot}\right)}_{n=1}& =& {\left(E_{rot}\right)}_{n=0}\\ \text{ratio}& =& \frac{{\left(\exp \left[\frac{-E_{vib}}{K_{B}T}\right]\right)}_{n=1}}{{\left(\exp \left[\frac{-E_{vib}}{K_{B}T}\right]\right)}_{n=0}}\\ & & \end{array}$

Now substituting values in above equation

$\begin{array}{rcl}\text{ratio}& =& \exp \left(\frac{-2{\hslash }^2}{2\mu a^2}\times \frac{1}{k_{B}T}\right)\\ & =& 2.94\\ \text{ratio}{R}_1& =& 2.94\\ & & \end{array}$

The ratio of the molecules with $l=1$ rotational quantum number to the ones with$l=0$ is $2.94$.

The difference of the vibration energy of the molecules in $n=1$ and the energy of the molecules in $n=0$ is:

$\begin{array}{rcl}{E}_{1,0}-E_{0,0}& =& \left(1+\frac{1}{2}\right)\hslash \sqrt{\frac{k}{\mu }}-\frac{1}{2}\hslash \sqrt{\frac{k}{\mu }}\\ & =& \hslash \sqrt{\frac{k}{\mu }}\\ & & \end{array}$

Substitute $1.055\times {10}^{-34}\text{J}.\text{s}$for $\hslash , 2.3\times {10}^3 \text{N/m}$for K and $1.16\times {10}^{-26}\text{kg}$ for$\mu$ as:

$\begin{array}{rcl}{E}_{1,0}-E_{0,0}& =& \left(1.055\times {10}^{-34}\text{J}.s\right)\sqrt{\frac{2.3\times {10}^3\text{N/m}}{1.16\times {10}^{-26}\text{kg}}}\\ & =& 4.69\times {10}^{-20}\text{J}\\ & & \end{array}$

According to Boltzmann distribution, the ratio of the number of molecules with vibration quantum number n=1 to the number of molecules with vibration quantum number n=0 is given as shown below.

$\begin{array}{rcl}\frac{{\left(E_{\text{vib}}\right)}_{n=1}-{\left(E_{vib}\right)}_{n=0}}{K_{B}T}& =& E_1-E_0\\ E_1-E_0& =& \left(\frac{3}{2}\hslash \sqrt{\frac{\kappa }{\mu }}-\frac{1}{2}\hslash \sqrt{\frac{\kappa }{\mu }}\right)\times \frac{1}{K_{B}T}\\ & =& \frac{1}{K_{B}T}\times \hslash \sqrt{\frac{\kappa }{\mu }}\\ & =& \frac{1}{1.38\times {10}^{-23}\text{J/K}\times 300\text{K}}\times 4.69\times {10}^{-20}\text{J}\\ & =& 11.33\\ \text{ratio}& =& \exp (-11.33)\\ \text{ratio}& =& 1.2\times {10}^{-5}\\ & & \end{array}$

The ratio of molecules with vibration quantum number to the ones with $n=0$ is .$1.2\times {10}^{-5}$