# 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.