Solutions 9
- Page ID
- 204088
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \(cm^{-1}\).
Q2:
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%.
Q3:
Only molecules with dipole moments exhibit microwave spectra. \(O_2\) and \(CCl_4\) do not meet this criteria.
Q4:
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: \(R = \sqrt(\frac{h}{8 \pi^2 \mu B})\)
Q5:
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.