Question

The first three absorption lines in the pure rotational spectrum of gaseous${}^{\mathbf{12}}{\mathbf{C}}^{\mathbf{16}}\mathbf{O}$are found to have the frequencies

role="math" localid="1663608192783" $\mathbf{1}.\mathbf{15}\times {\mathbf{10}}^{\mathbf{11}},\mathbf{2}.\mathbf{30}\times {\mathbf{10}}^{\mathbf{11} } \mathbf{and}\mathbf{3}.\mathbf{46}\times {\mathbf{10}}^{\mathbf{11}}{\mathbf{s}}^{-\mathbf{1}}$. Calculate:

1. The moment of inertia Iof role="math" localid="1663608263500" $\mathbf{CO}$(in kg m²)
2. The energies of the role="math" localid="1663608330368" $\mathit{J}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{J}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{and}\mathit{J}\mathbf{=}\mathbf{3}\mathbf{ }$ rotational levels of measured from the $\mathit{J}\mathbf{=}\mathbf{0}$state (in joules)
3. The $\mathbf{CO}$bond length (in angstroms) $\mathbf{CO}$

Short answer

The calculated values are:

1. $I=1.4532\times {10}^{-46}kg m^2$
2. $$\begin{gathered} \Delta E_1=7.65\times {10}^{-23} J \\ \Delta E_2=2.296\times {10}^{-22} J \\ \Delta E_3=4.592\times {10}^{-22} J \end{gathered}$$
   
3. $R_e=1.23$Ǻ

Step-by-step solution

Definition of the rotational spectrum and absorption lines

The rotational spectrum is the spectrum observed when bands arise from measured changes in molecular rotation energy.

Absorption lines are the dark spaces observed in the rotation spectrum that occur due to absorption of the light energy.

Given information

Absorption line frequencies of ${}^{12}{\mathrm{C}}^{16}\mathrm{O}$are$1.15\times {10}^{11},2.30\times {10}^{11 } \mathrm{and}3.46\times {10}^{11}{\mathrm{s}}^{-1}$

 Step 3: Calculation of Inertia(a) 

The average interval between the rotational lines ʋ is the average of the difference between the consecutive frequencies of${}^{\mathbf{12}}{\mathbf{C}}^{\mathbf{16}}\mathbf{O}$in the pure rotational spectrum.$\nu =1.155\times {10}^{-11}{s}^{-1}$

From$\Delta E=h\nu \text{and}\Delta E=\frac{h^2}{4{\pi }^{2}I}$

$\begin{array}{rcl}I& =& \frac{h}{4{\pi }^2\nu }\\ & =& \frac{6.626\times {10}^{-34 } Js}{4{\pi }^2(1.155\times {10}^{11} s^{-1})}\\ & =& 1.4532\times {10}^{-46}kg{m}^2\end{array}$

Determination of energies(b)

The allowed absorption frequencies for transitions are

$$\begin{gathered} \nu =\frac{h}{8{\pi }^{2}I}\left[J\right(J+1)-0]=\frac{h}{8{\pi }^{2}I}\frac{J(J+1)}{2} \\ 0\to 1 \nu =1\times 1.155\times {10}^{11}{s}^{-1} \Delta E_1=7.65\times {10}^{-23} J \\ 1\to 2 \nu =3\times 1.155\times {10}^{11}{s}^{-1} \Delta E_2=2.296\times {10}^{-22} J \\ 2\to 3 \nu =6\times 1.155\times {10}^{11}{s}^{-1} \Delta E_3=4.592\times {10}^{-22} J \end{gathered}$$

Calculation of Bond length(c)

From table 19.1, using nuclidic masses to obtain reduced mass$\mu$

$\begin{array}{rcl}\mu & =& \frac{m_C{m}_O}{m_C+m_O}kg\\ & =& \frac{\left(12\right)(15.9949)u}{\left(12\right)+(15.9949)}\\ & =& 11.385\times {10}^{-27}kg\end{array}$

$$\begin{gathered} R_e=(\frac{I}{\mu }{)}^{\frac{1}{2}} \\ =(\frac{14.532\times {10}^{-47}kg{m}^2}{11.385\times {10}^{-27}kg}{)}^{\frac{1}{2}}=1.23\hat{\text{A}} \end{gathered}$$