Question

The microwave spectrum of ${}^{\mathbf{12}}{\mathbf{C}}^{\mathbf{16}}\mathit{O}$shows that the transition from J = 0 to J = 1 requires electromagnetic radiation with a wavelength of $\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{m}$.

a. Calculate the bond length of CO. See Exercise 60for the atomic mass of ${}^{\mathbf{16}}\mathit{O}$.

b. Calculate the frequency of radiation absorbed in a rotational transition from the second to the third excited state of CO.

Short answer

a) The bond length of CO is $1.45\times {10}^{-10}m$.

b) The frequency of radiation absorbed in a rotational transition from the second to the third excited state of CO is$3.46\times {10}^{11}{s}^{-1}$.

Step-by-step solution

Step 1: The formula used to calculate bond length and frequency

$\mathit{I}\mathbf{=}\mathit{\mu }{\mathit{R}}_{\mathbf{e}}^{\mathbf{2}}$

Where R_e is the bond lengthand$\mu$is the reduced mass of the molecule.

$v=2B\left(J_i+1\right)$

The above formula is used to find the frequency.

Step 2: To find the value of the moment of inertia

We start with the equation below

$$\begin{gathered} \delta E=2hB\left(J_i+1\right)=\frac{hc}{\lambda } \\ \frac{c}{\lambda }=2B\left(J_I+1\right) \\ B=\frac{c}{2\lambda }=\frac{2.998\times {10}^{8}m{s}^{-1}}{\left(2\right)\left(2.60\times {10}^{-3}m\right)} \\ B=5.77\times {10}^{10}{s}^{-1} \end{gathered}$$

Now,

$$\begin{gathered} B=\frac{h}{8{\pi }^{2}I} \\ I=\frac{h}{8{\pi }^{2}B}=\frac{6.626\times {10}^{-34}Js}{8{\pi }^2\left(5.77\times {10}^{10}{s}^{-1}\right)} \\ I=1.45\times {10}^{-46}kg{m}^2 \end{gathered}$$

To calculate the value of the reduced mass of the molecule

Now, we can find the value of 𝜇 by using the following formula

$$\begin{gathered} \mu =\frac{m\left(C\right).m\left(O\right)}{m\left(C\right)+m\left(O\right)} \\ \mu =\frac{\left(12.000\right)\left(15.995\right)}{12.000+15.995} \\ \mu =6.856u \end{gathered}$$

To calculate the bond length

We will use the formula

$$\begin{gathered} I=\mu R_e^2 \\ {R}_e=\sqrt{\frac{I}{\mu }=\frac{1.45\times {10}^{-46}kg{m}^2}{6.856\times {10}^{-10}m}} \\ {R}_e=1.45\times {10}^{-10}m \end{gathered}$$

To calculate the frequency

We will use the formula to find the frequency of transition 2 to 3

$$\begin{gathered} v=2B\left(J_i+1\right) \\ v=\left(2\right)\left(5.77\times {10}^{10}\right)\left(2+1\right) \\ v=3.46\times {10}^{11}{s}^{-1} \end{gathered}$$