9.8 Superconductivity  (Page 6/12)

$E_{0r}=7.43\times {10}^{\mathrm{-3}}\text{eV}$ ; $l=0;E_r=0\text{eV}$ (no rotation);
$l=1;E_r=1.49\times {10}^{\mathrm{-2}}\text{eV}$ ; $l=2;E_r=4.46\times {10}^{\mathrm{-2}}\text{eV}$

1. They are fairly hard and stable.
2. They vaporize at relatively high temperatures (1000 to 2000 K).
3. They are transparent to visible radiation, because photons in the visible portion of the spectrum are not energetic enough to excite an electron from its ground state to an excited state.
4. They are poor electrical conductors because they contain effectively no free electrons.
5. They are usually soluble in water, because the water molecule has a large dipole moment whose electric field is strong enough to break the electrostatic bonds between the ions.

No, He atoms do not contain valence electrons that can be shared in the formation of a chemical bond.

${\displaystyle \sum }_1^{N\text{/}2}{n}^2=\frac{1}{3}{\left(\frac{N}{2}\right)}^3,$ so $\stackrel{\text{−}}{E}=\frac{1}{3}{E}_F$

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.

a. $E_F=7.11\text{eV}$ ; b. $E_F=3.24\text{eV}$ ; c. $E_F=9.46\text{eV}$

$9.15\approx 9$

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:
$g\left(E\right)=\frac{\pi V}{2}{\left(\frac{8m_e}{h^2}\right)}^{3\text{/}2}{E}^{1\text{/}2}$ . Integrating gives: $\stackrel{\text{−}}{E}=\frac{3}{5}{E}_{\text{F}}.$