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Chemistry LibreTexts

Solutions 9

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  • Q1:

    For a given wavelength, using the relations, \( E = hf\) and \(c = \lambda f\) where E is energy in joules, h is Planck's constant, \(lambda\) is wavelength, f is frequency and c is the speed of light in a vacuum. Given \(\lambda\) =  656e-9 m = 656e-7 cm , f = 4.58e14 Hz and \(\tilde{\nu}\) = 1.5232 \(cm^{-1}\).  



    Using the relation, \(\Delta E \Delta t = h\Delta f \Delta t \geq\frac{h}{4\pi} \) and \(c = \lambda f\), we can find frequency,  f = c \ 430e-9 m = 6.972e14 Hz. Given \(\Delta t\) = .5 ns, and rearranging the first relation to \(\Delta f \geq \frac{1}{4} \pi \Delta t \geq\)  1.59e8 Hz. The percent uncertainty in frequency measurement is at least .000022%. 



    Only molecules with dipole moments exhibit microwave spectra. \(O_2\) and \(CCl_4\) do not meet this criteria.


    Given \(B_0\) and the atomic masses of C and O as 12 amu and 16 amu, respectively, we can find the internuclear distance, R, via the relation: \(B = \frac{h}{8 \pi^2 \mu R^2}\). The reduced mass can be found using 12 and 16 amu in the expression, \(\mu = \frac{m_1 m_2}{m_1 + m_2} \), and divided by Avogadro's number to convert to kg. We can rearrange, solving for R:   \(R = \sqrt(\frac{h}{8 \pi^2 \mu B})\) 



    The energy level of a rigid rotor is given by: \(E = BhJ(J+1)\). The frequency of the  \(J = 4 \rightarrow 3\) transition in the pure rotational spectra of \(^{14}N^{16}O\) and \(O_2\) is found by \(E_{J+1} - E_J= \frac{h^2}{4\pi^2 \mu R^2}(J+1)\).  The reduced masses can be found from the atomic masses of \(NO\) and \)O_2\) and the distance R is provided in the bond lengths.