91影视

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

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

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,$E_{rot}=\frac{{\mathrm{鈩�}}^{2}l(l+1)}{2渭a^2}$.

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

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

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

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

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

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

Substitute $1.055脳{10}^{-34}\text{J}.\text{s}$ for$\mathrm{鈩�},1.16脳{10}^{-26}\text{kg}$ for $渭$ and $0.11脳{10}^{-9}\text{m}$ for in the above expression.

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

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

$\begin{array}{rcl}\text{ratio}& =& \exp \left(\frac{-2{\mathrm{鈩�}}^2}{2渭a^2}脳\frac{1}{k_{B}T}\right)\\ & =& 2.94\\ \text{ratio}{R}_1& =& 2.94\\ & & \end{array}$

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:

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

Substitute $1.055脳{10}^{-34}\text{J}.\text{s}$for $\mathrm{鈩�},鈥�鈥�2.3脳{10}^3鈥�\text{N/m}$for K and $1.16脳{10}^{-26}\text{kg}$ for$渭$ as:

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

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.

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

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