Solutions 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.