4. For the "C" molecule the following millimeter wave pure rotationat transitions has heen ofserwed (in \( \mathrm{MHz} \) ): \begin{tabular}{|c|c|c|c|} \hline Thananthen & \( v=0 \) & \( y-1 \) & \( y=-2 \) \\ \hline\( y=-0 \) & stakn & \( 48\left(673 m^{m}\right. \) & \( 48,280,887 \) \\ \hline\( x=2-\infty \) & \( 9, a m, 9 \times 0 \) & \( 97.2700 x 3 \) & \( 8,5600,30 \) \\ \hline \( 4-3-2 \) & \( 14 n+3 n \) & 145.84 867 & \( 144,818,826 \) \\ \hline\( L=x \) & 18, & 14433431 & 103,113959 \\ \hline \end{tabular} \[ A-A-B \] (a) For each vbrational level derive set of rotational constants by ftting the data. (b) From the results of (a) derve an expression (by fiting) for the vibrational dependence of \( B \). (c) From \( B_{s} \) calculate \( r_{s} \), from \( D_{e} \) calculate \( r_{e} \)
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The rotational constant \( B \) is related to the transition frequency \( \nu \) by the formula: \[ \nu = 2B(v+1) \] where \( v \) is the vibrational level. We can rearrange this equation to solve for \( B \): \[ B = \frac{\nu}{2(v+1)} \] Using the given Show more…
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The vibrational and rotational energies of the CO molecule are given by Eq. (42.9). Calculate the wavelength of the photon absorbed by CO in each of these vibration-rotation transitions: (a) $n=0$ $l=2 \rightarrow n=1, l=3$ (b) $n=0, l=3 \rightarrow n=1, l=2$ (c) $n=0$ $l=4 \rightarrow n=1, l=3$
The vibrational and rotational energies of the CO molecule are given by Eq. $(42.9) .$ Calculate the wavelength of the photon absorbed by $\mathrm{CO}$ in each of the following vibration-rotation transitions: (a) $n=0, l=1 \rightarrow n=1, I=2 ;$ (b) $n=0, I=$ $2 \rightarrow n=1, l=1 ;(c) n=0, l=3 \rightarrow n=1, I=2$
(II) The equilibrium separation of H atoms in the molecule is 0.074 nm (Fig. 29-8). Calculate the energies and wavelengths of photons for the rotational transitions ($a$) $\ell =$ 1 to $\ell =$ 0, ($b$) $\ell =$ 2 to $\ell =$ 1, and ($c$) $\ell =$ 3 to $\ell =$ 2.
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