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Determine the lowest three rotational energy levels of ${\text{H}}_{2}.$
${E}_{0r}=7.43\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-3}}\phantom{\rule{0.2em}{0ex}}\text{eV}$ ;
$l=0;{E}_{r}=0\phantom{\rule{0.2em}{0ex}}\text{eV}$ (no rotation);
$l=1;{E}_{r}=1.49\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}\phantom{\rule{0.2em}{0ex}}\text{eV}$ ;
$l=2;{E}_{r}=4.46\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}\phantom{\rule{0.2em}{0ex}}\text{eV}$
A carbon atom can hybridize in the $s{p}^{2}$ configuration. (a) What is the angle between the hybrid orbitals?
List five main characteristics of ionic crystals that result from their high dissociation energy.
Why is bonding in ${\text{H}}_{2}{}^{+}$ favorable? Express your answer in terms of the symmetry of the electron wave function.
Astronomers claim to find evidence of ${\text{He}}_{2}$ from light spectra of a distant star. Do you believe them?
No, He atoms do not contain valence electrons that can be shared in the formation of a chemical bond.
Show that the moment of inertia of a diatomic molecule is $I=\mu {r}_{0}^{2}$ , where $\mu $ is the reduced mass, and ${r}_{0}$ is the distance between the masses.
Show that the average energy of an electron in a one-dimensional metal is related to the Fermi energy by $\stackrel{\text{\u2212}}{E}=\frac{1}{2}{E}_{\text{F}}.$
$\sum}_{1}^{N\text{/}2}{n}^{2}=\frac{1}{3}{\left(\frac{N}{2}\right)}^{3},$ so $\stackrel{\text{\u2212}}{E}=\frac{1}{3}{E}_{F}$
Measurements of a superconductor’s critical magnetic field (in T ) at various temperatures (in K) are given below. Use a line of best fit to determine ${B}_{\text{c}}(0).$ Assume ${T}_{\text{c}}=9.3\phantom{\rule{0.2em}{0ex}}\text{K}\text{.}$
T (in K) | ${B}_{\text{c}}(T)$ |
---|---|
3.0 | 0.18 |
4.0 | 0.16 |
5.0 | 0.14 |
6.0 | 0.12 |
7.0 | 0.09 |
8.0 | 0.05 |
9.0 | 0.01 |
Estimate the fraction of Si atoms that must be replaced by As atoms in order to form an impurity band.
An impurity band will be formed when the density of the donor atoms is high enough that the orbits of the extra electrons overlap. We saw earlier that the orbital radius is about 50 Angstroms, so the maximum distance between the impurities for a band to form is 100 Angstroms. Thus if we use 1 Angstrom as the interatomic distance between the Si atoms, we find that 1 out of 100 atoms along a linear chain must be a donor atom. And in a three-dimensional crystal, roughly 1 out of ${10}^{6}$ atoms must be replaced by a donor atom in order for an impurity band to form.
Transition in the rotation spectrum are observed at ordinary room temperature ( $T=300\phantom{\rule{0.2em}{0ex}}\text{K}$ ). According to your lab partner, a peak in the spectrum corresponds to a transition from the $l=4$ to the $l=1$ state. Is this possible? If so, determine the momentum of inertia of the molecule.
Determine the Fermi energies for (a) Mg, (b) Na, and (c) Zn.
a. ${E}_{F}=7.11\phantom{\rule{0.2em}{0ex}}\text{eV}$ ; b. ${E}_{F}=3.24\phantom{\rule{0.2em}{0ex}}\text{eV}$ ; c. ${E}_{F}=9.46\phantom{\rule{0.2em}{0ex}}\text{eV}$
Find the average energy of an electron in a Zn wire.
What value of the repulsion constant, n , gives the measured dissociation energy of 158 kcal/mol for CsCl?
$9.15\approx 9$
A physical model of a diamond suggests a BCC packing structure. Why is this not possible?
For an electron in a three-dimensional metal, show that the average energy is given by $\stackrel{\text{\u2212}}{E}=\frac{1}{N}{\displaystyle \underset{0}{\overset{{E}_{\text{F}}}{\int}}Eg(E)dE}=\frac{3}{5}{E}_{\text{F}},$
Where N is the total number electrons in the metal.
In three dimensions, the energy of an electron is given by:
$E={R}^{2}{E}_{1},$ where
${R}^{2}={n}_{1}^{2}+{n}_{2}^{2}+{n}_{3}^{2}$ . Each allowed energy state corresponds to node in
N space
$({n}_{1},{n}_{2},{n}_{3})$ . The number of particles corresponds to the number of states (nodes) in the first octant, within a sphere of radius,
R . This number is given by:
$N=2\left(\frac{1}{8}\right)\left(\frac{4}{3}\right)\pi {R}^{3},$ where the factor 2 accounts for two states of spin. The density of states is found by differentiating this expression by energy:
$g\left(E\right)=\frac{\pi V}{2}{\left(\frac{8{m}_{e}}{{h}^{2}}\right)}^{3\text{/}2}\phantom{\rule{0.2em}{0ex}}{E}^{1\text{/}2}$ . Integrating gives:
$\stackrel{\text{\u2212}}{E}=\frac{3}{5}\phantom{\rule{0.2em}{0ex}}{E}_{\text{F}}.$
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