Chemistry 2720 Fall 2000 Practice Problems on Quantum Mechanics -- Solutions

1.

$L = \hbar\sqrt{\ell(\ell+1)} = 2\hbar\sqrt{3}$; L_z is one of $\{-3\hbar,-2\hbar,-\hbar,0,\hbar,2\hbar,3\hbar\}$.

2.

To ensure continuity of the wave, we must have $L = n\lambda/2$ or $\lambda = 2L/n$. Since $p=h/\lambda$, we have

$$p = \frac{nh}{2L}.$$

The energy of a photon is related to its momentum by

$$E = cp = \frac{nhc}{2L}.$$

3.

By the usual rules, we would assign the ground state an electronic configuration of [Xe]4f²6s². Assuming that the lanthanides behave somewhat like transition metals, the 3+ ion would then have a ground-state electronic configuration of [Xe]4f¹ while the 4+ ion would simply have a xenon-like electronic configuration.

It turns out that Ce has a ground-state electronic configuration of [Xe]4f¹5d¹6s². The reason why this atom has 3+ and 4+ ions is now a little clearer: The 3+ ion probably has a [Xe]4f¹ configuration while the 4+ ion has a simple [Xe] electronic configuration.

You can see from this example why I didn't make a big deal of the lanthanides and actinides in class: Their chemistry is really quite complex.

4.

The p₀ orbital has two lobes of high electron density along the z axis, above and below the nucleus. If we take the z axis to be the bond axis and imagine adding p₀ orbitals from each atom, we get a $\sigma$ bond, i.e. a bond with enhanced density along the line connecting the two nuclei.¹

The p₁ and p_-1 orbitals are tori. If we add two such orbitals (one from each atom), we get a cylindrical molecular orbital with a nodal line (a region of depressed electron density) along the bond axis. These are $\pi$ orbitals.

5.

Ne₂⁺ would have the ground-state electronic configuration

$$(\sigma_\mathrm{1s})^2(\sigma^*_\mathrm{1s})^2(\sigma_\mathrm... ...ma_\mathrm{2p})^2(\pi^*_\mathrm{2p})^4(\sigma^*_\mathrm{2p})^1.$$

The bond order would be (10-9)/2 = 1/2. The nonzero bond order indicates a stable molecular ion. The unpaired electron in the $\sigma^*_\mathrm{2p}$ orbital would make this ion paramagnetic.

6.

(a)

The reduced mass is

$$\mu = \left(\frac{1}{m_\mathrm{N}}+\frac{1}{m_\mathrm{O}}\right)^{-1} = 7.466\,433\,\mathrm{amu}.$$

Since an amu is a g/mol, we have

$$\mu = \frac{7.466\,433\times 10^{-3}\,\mathrm{kg/mol}}{6.022... ...}\,\mathrm{mol}^{-1}} = 1.239\,83\times 10^{-26}\,\mathrm{kg}.$$

The frequency is

$$\nu_0 = \frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$$

so that

$$k = 4\pi^2\nu_0^2\mu = 1595\,\mathrm{N/m}.$$

(b)

The reduced mass of $\isotope{15}{}{N}\isotope{16}{}{O}$ is 7.740774amu, or $1.285\,385\time 10^{-26}\,\mathrm{kg}$. The fundamental vibration frequency is then

$$\nu_0 = 5.606\times 10^{13}\,\mathrm{Hz}.$$

7.

(a)

The energy of one 680nm photon is

$$E = \frac{hc}{\lambda} = 2.921\times 10^{-19}\,\mathrm{J}.$$

The energy of a mole of such photons is therefore 176kJ. Note that $485\,\mathrm{kJ}/176\,\mathrm{kJ} = 2.76$ so that it takes 3mol of 680nm photons to fix 1mol of carbon.

(b)

The quantum yield is about 1/3.

8.

(a)

The mean rotational splitting is 12.1627GHz which, converted to energy units, is $\Delta_r = 8.059\,10\times 10^{-24}\,\mathrm{J}$. This is an absorption spectrum, so the rotational splitting is $\hbar^2/I$. Thus

$$I = \frac{\hbar^2}{\Delta_r} = 1.379\,96\times 10^{-45}\,\mathrm{kg\,m^2}.$$

(b)

The photon energies are given by

$$\Delta E_J = \frac{\hbar^2}{I}(J+1).$$

Note that the J in this equation is the initial value of the quantum number and that during absorption, this quantum number increases.

$\nu$ (GHz)	$\Delta E_J$ (10-23 J)	$I\Delta E_J/\hbar^2\approx J+1$	transition
24.32592	1.611852	2	$1\rightarrow 2$
36.48882	2.417774	3	$2\rightarrow 3$
48.65164	3.223691	4	$3\rightarrow 4$
60.81408	4.029583	5	$4\rightarrow 5$

(c)

The exciting line has a frequency of 890300GHz. The Raman spectrum has lines at $\pm 3\hbar^2/I, \pm 5\hbar^2/I, \pm 7\hbar^2/I,\ldots$ relative to the exciting radiation. The quantity $\hbar^2/I$, converted to frequency units, is 12.1627GHz. Thus the frequencies of the rotational Raman lines are (in GHz) 890264, 890239, 890215, ...and 890336, 890361, 890385, ...

Footnotes

... nuclei.¹

Conversely, if we subtract the orbitals, we get a $\sigma^*$ orbital with depressed density between the nuclei, but where the maximum electron density is still on the bond axis.