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%TCIDATA{Created=Wed Mar 24 13:30:32 1999}
%TCIDATA{LastRevised=Thu Apr 01 12:06:30 1999}
%TCIDATA{Language=American English}

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\begin{document}


\SetTitle{Physics 312 -- First Pledged Problem Set} \SetAuthor{} \Setdate{} %
\TitlePage{}

\vspace{0.5cm}

\noindent\textbf{1}. The ground state of hydrogen is described by the
wavefunction
\begin{eqnarray*}
\psi \left( \mathbf{r} \right) = \frac{1}{\sqrt{\pi}}\left( \frac{1}{a_0}
\right)^{\frac{3}{2}}\,e^{-r/a_0},
\end{eqnarray*}
where $a_{0}$ is the Bohr radius.\vspace{0.5cm}

\noindent (a) Compute the charge density
\begin{eqnarray*}
\rho \left( \mathbf{r} \right) = e\left| \psi\left( \mathbf{r} \right)
\right|^{2}.
\end{eqnarray*}
and plot $4\pi r^{2}\rho$ as a function of the dimensionless variable $r/a_0$%
. Does the plot peak at $r = a_{0}$? \vspace{0.5cm}

We have
\begin{eqnarray*}
\rho \left( \mathbf{r}\right) &=&e\left| \frac{1}{\sqrt{\pi }}\left( \frac{1%
}{a_{0}}\right) ^{\frac{3}{2}}\,e^{-r/a_{0}}\right| ^{2} \\
&=&\frac{1}{\pi }\frac{e}{a_{0}^{3}}\,e^{-2r/a_{0}}.
\end{eqnarray*}
Thus
\[
4\pi r^{2}\rho \left( \mathbf{r}\right) =4\frac{e}{a_{0}}\left( \frac{r}{%
a_{0}}\right) ^{2}\,e^{-2\frac{r}{a_{0}}}.
\]
A plot of this function is below.

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\noindent We see that the plot does peak at $r=a_{0}$. \vspace{0.5cm}

\noindent (b) Compute the average value of $r$. Is it equal to $a_0$?
\vspace{0.5cm}

We have
\begin{eqnarray*}
\left< r \right> &=& \int \psi^{\dagger}\left( \mathbf{r} \right) r
\psi\left( \mathbf{r} \right)\,d^{3}r \\
&=& \int r\left| \psi\left( \mathbf{r} \right) \right|^{2}\,d^{3}r \\
&=& \int_{0}^{\infty}r\frac{1}{\pi}\left( \frac{1}{a_0}
\right)^{3}e^{-2r/a_0}4\pi r^2\,dr \\
&=& 4a_{0}\int_{0}^{\infty}\beta^{3}e^{-2\beta}\,d\beta,
\end{eqnarray*}
where $\beta = r/a_0$. The remaining integral computed by hand, looked up,
or done by computer, and is found to be equal to $3!/2^{4} = 3/8$. Thus
\begin{eqnarray*}
\left< r \right> = \frac{3}{2}a_0.
\end{eqnarray*}
We see that $\left< r \right>$ is \emph{not} equal to $a_0$. \vspace{0.5cm}

\noindent (c) Does the electron have a definite value of $r$? Does it have a
definite speed? Does it have a definite angular momentum? Does it have a
definite energy? If the answer to any of these questions is yes, give the
value of the quantity. \vspace{0.5cm}

The Hamiltonian that describes the electron is
\begin{eqnarray*}
H &=&\frac{\mathbf{p}^{2}}{2m_{e}}+V\left( \mathbf{r}\right) \\
&=&-\frac{\hbar ^{2}}{2m_{e}}\mathbf{\nabla }^{2}-\frac{e^{2}}{4\pi \epsilon
_{0}}\frac{1}{r}.
\end{eqnarray*}
We know from basic quantum mechanics that an eigenfunction of a system
described by a Hamiltonian $H$ will also be an eigenfunction of an operator $%
\mathcal{O}$ if and only if $\mathcal{O}$ commutes with $H$. One can check
that $r$ and $v=p/m$ do not commute with $H$, while $%
L^{2}=L_{x}^{2}+L_{y}^{2}+L_{z}^{2}$ and $H$ itself do commute with $H$.
Therefore the electron has a definite energy and angular momentum, but not a
definite radius or speed. The value of the angular momentum is $L=\sqrt{%
l\left( l+1\right) }\hbar =0$ and the value of the energy is $E=-13.6$ eV.

\vspace{0.5cm}

\noindent (d) Compute the electric field due to this charge density
everywhere in space. (Hint: use Gauss's theorem.) Add it to the field of the
proton. Plot the total (radial) field as a function of $r/a_0$ for $r >
a_{0}/2$ to check that it decreases exponentially. \vspace{0.5cm}

We have from Gauss's law that
\begin{eqnarray*}
\int_{\partial S} \mathbf{E}\left( \mathbf{r} \right) \cdot d\mathbf{a} =
\frac{1}{\epsilon_{0}}\int_{S} \rho\left( r \right)\,d^{3}r,
\end{eqnarray*}
for any closed surface $S$. Because the charge density is spherically
symmetric, it follows that the electric field will be radial. Thus if we
choose for $S$ a sphere of radius $r$ centered at the origin, we find that
\begin{eqnarray*}
E\left( r\right)\,4\pi r^{2} = \frac{1}{\epsilon_{0}}\int_{0}^{r} \rho\left(
r^{\prime}\right)\,4\pi {r^{\prime}}^{2}\,dr^{\prime},
\end{eqnarray*}
so
\begin{eqnarray*}
E\left( r \right) = \frac{1}{\epsilon_{0}}\frac{1}{r^2} \int_{0}^{r}
\rho\left( r^{\prime}\right)\,{r^{\prime}}^{2}\,dr^{\prime}.
\end{eqnarray*}
We have
\begin{eqnarray*}
\frac{1}{\epsilon_{0}}\int_{0}^{r} \rho\left( r^{\prime}\right)\,{r^{\prime}}%
^{2}\,dr^{\prime}&=& \frac{e}{\pi\epsilon_{0}}\int_{0}^{r} \frac{e}{a_{0}^{3}%
}{r^{\prime}}^{2}e^{-2r^{\prime}/a_0}\,dr^{\prime} \\
&=& \frac{e}{\pi\epsilon_{0}}\int_{0}^{r} \left( \frac{r^{\prime}}{a_0}
\right)^{2}e^{-2r^{\prime}/a_0}\,d\left( \frac{r^{\prime}}{a_0}\right) \\
&=& \frac{e}{\pi\epsilon_{0}}\int_{0}^{R} {R^{\prime}}^{2}e^{-2R^{\prime}}%
\,dR^{\prime},
\end{eqnarray*}
where $R = r/a_0$ and $R^{\prime}= r^{\prime}/a_0$. The remaining integral
can be done by \emph{Maple} or the like, but is also easily done by hand.
First we perform a change of variable:
\begin{eqnarray*}
\frac{e}{\pi\epsilon_{0}}\int_{0}^{R} {R^{\prime}}^{2}e^{-2R^{\prime}}\,dR^{%
\prime}= \frac{e}{8\pi\epsilon_{0}}\int_{0}^{2R} \beta^{2}e^{-\beta}\,d\beta,
\end{eqnarray*}
where $\beta$ is simply equal to $2R^{\prime}$. Now we perform two
integrations by parts:
\begin{eqnarray*}
\int_{0}^{2R} \beta^{2}e^{-\beta}\,d\beta &=& -\left.
\beta^{2}e^{-\beta}\right|_{0}^{2R} + \int_{0}^{2R}2\beta e^{-\beta}\,d\beta
\\
&=& -\left( 2R \right)^{2}e^{-2R} - \left. 2\beta e^{-\beta}\right|_{0}^{2R}
+ \int_{0}^{2R}2e^{-\beta} \\
&=& -4R^{2}e^{-2R} - 4Re^{-2R} + 2 - 2e^{-2R} \\
&=& 2 - \left( 1 + 2R + 2R^{2} \right) e^{-2R}.
\end{eqnarray*}
Therefore we have
\begin{eqnarray*}
E\left( r \right) &=& \frac{1}{r^2}\frac{e}{8\pi\epsilon_{0}}\left[ 2 -
\left( 1 + 2R + 2R^{2} \right) e^{-2R} \right] \\
&=& \frac{1}{4\pi\epsilon_{0}}\frac{e}{a_{0}^{2}}\left[ \frac{a_{0}^{2}}{r^2}
- \left( 1 + \frac{a_0}{r} + \frac{a_{0}^{2}}{2r^{2}}\right) e^{-2r/a_0}
\right].
\end{eqnarray*}
The field due to the proton is just the Coulomb field
\begin{eqnarray*}
E\left( r \right) = -\frac{1}{4\pi\epsilon_{0}}\frac{e}{r^{2}},
\end{eqnarray*}
so adding this to our previous result we find that the total field is
\begin{eqnarray*}
E_{\mathrm{tot}}\left( r \right) = -\frac{1}{4\pi\epsilon_{0}}\frac{e}{%
a_{0}^{2}}\left( 1 + \frac{a_0}{r} + \frac{a_{0}^{2}}{2r^{2}}\right)
e^{-2r/a_0}.
\end{eqnarray*}
A plot of this field is below.

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312/Physics 312-Spring 1999/Assignment 7/F9ISU401.wmf';}} \vspace{0.5cm}

\noindent (e) Compute the potential by integrating the field from $r$ to
infinity. \vspace{0.5cm}

Using Maple or the like to do the integration, we find that
\begin{eqnarray*}
V\left( r\right) &=&\int_{r}^{\infty }E_{\mathrm{tot}}\left( r^{\prime
}\right) \,dr^{\prime } \\
&=&-\frac{1}{4\pi \epsilon _{0}}\frac{e}{a_{0}}\left( \frac{1+\frac{r}{a_{0}}%
}{2\frac{r}{a_{0}}}\right) \exp \left( {-2r/a_{0}}\right) .
\end{eqnarray*}

\vspace{0.5cm} \hrule \vspace{0.5cm}

\noindent\textbf{2}. Read the short section on impurity levels on page 368
of \emph{Kittel and Kroemer} and look up the dielectric constants of Si and
Ge, as well as the effective masses for the valence and conduction bands.
Then: \vspace{0.5cm}

\noindent (a) Compute the ``effective Bohr radius'' for donor and acceptor
ground states. The easy way to do this is to scale the usual Bohr radius by
the appropriate factors. \vspace{0.5cm}

From Table 13.1 on page 357 of \emph{Kittel and Kroemer} we obtain the
following values:

\begin{center}
\begin{tabular}{llll}
\textbf{Element} & \textbf{Effective Mass of Holes} & \textbf{Effective Mass
of Electrons} & \textbf{Dielectric Constant} \\
Si & 0.58 $m_{e}$ & 1.06 $m_{e}$ & 11.7 \\
Ge & 0.35 $m_{e}$ & 0.56 $m_{e}$ & 15.8
\end{tabular}
\end{center}

The Bohr radius for hydrogen is given (in MKS units) by
\[
a_{0,\mathrm{hydrogen}}=\frac{4\pi \epsilon _{0}\hbar ^{2}}{m_{e}e^{2}},
\]
which is equal to 0.0529 nm. As a rough approximation, we can treat the
acceptor and donor atoms as hydrogen-like atoms by replacing $m_{e}$ by the
effective electron mass $\left( m_{eff}\right) $ and $\epsilon _{0}$ by $%
\kappa \epsilon _{0}$, where $\kappa $ is the dielectric constant of the
material. Thus we have
\[
a_{0}=\frac{4\pi \kappa \epsilon _{0}\hbar ^{2}}{m_{eff}e^{2}}=\frac{\kappa
}{\left( m_{eff}/m_{e}\right) }a_{0,\mathrm{hydrogen}}.
\]
Substituting, we find the following values for the effective Bohr radii:

\begin{center}
\begin{tabular}{lll}
\textbf{Element} & $a_{0}$ \textbf{for Holes} & $a_{0}$ \textbf{for Electrons%
} \\
Si & 20.2 $a_{0}$ & 11.0 $a_{0}$ \\
Ge & 45.1 $a_{0}$ & 28.2 $a_{0}$%
\end{tabular}
\end{center}
\vspace{0.5cm}

\noindent (b) Compute the ionization energies and compare with Table 13.2.
\vspace{0.5cm}

The ionization energy of the hydrogen atom is given by
\[
E_{ion,\mathrm{hydrogen}}=\frac{m_{e}e^{4}}{2\left( 4\pi \epsilon
_{0}\right) ^{2}\hbar ^{2}},
\]
which is equal to 13.6 eV. Performing the same approximations as before we
find that
\[
E_{ion}=\frac{m_{eff}e^{4}}{2\left( 4\pi \kappa \epsilon _{0}\right)
^{2}\hbar ^{2}}=\frac{\left( m_{eff}/m_{e}\right) }{\kappa ^{2}}E_{ion,%
\mathrm{hydrogen}}.
\]
Thus we find the following table:

\begin{center}
\begin{tabular}{lll}
\textbf{Element} & $E_{ion}$ \textbf{for Holes} & $E_{ion}$ \textbf{for
Electrons} \\
Si & 0.00424 $E_{ion,\mathrm{hydrogen}}$ & 0.00774 $E_{ion,\mathrm{hydrogen}%
} $ \\
Ge & 0.00140 $E_{ion,\mathrm{hydrogen}}$ & 0.00224 $E_{ion,\mathrm{hydrogen}%
} $%
\end{tabular}
\end{center}

We see that this approximation gives the right order of magnitude, but is
not really much more accurate than that.

\vspace{0.5cm} \hrule \vspace{0.5cm}

\noindent \textbf{3}. Compute the bias voltage of a germanium $p$-$n$
junction at $77$ K, at room temperature, and at $500$ K. Comment on the
effect of increased temperature on the performance of such a junction.
\vspace{0.5cm}

We have from page 17 of \emph{Melissinos} that
\[
V_{bi}=\frac{k_{B}T}{e}\ln \left( \frac{N_{D}N_{A}}{n_{i}^{2}}\right) ,
\]
and we have from page 8 that
\[
n_{i}=N_{s}e^{\frac{-E_{g}}{2k_{B}T}}.
\]
Combining these two relations we have that
\begin{eqnarray*}
V_{bi} &=&\frac{k_{B}T}{e}\ln \left( \frac{N_{D}N_{A}}{%
N_{s}^{2}e^{-E_{g}/k_{B}T}}\right) \\
&=&\frac{E_{g}}{e}+\frac{k_{B}T}{e}\ln \left( \frac{N_{D}N_{A}}{N_{s}^{2}}%
\right) .
\end{eqnarray*}
The exact values of $N_{A}$ and $N_{D}$ will depend on the doping, but for
representative values we can use the ones given in Figures 1.8(b) and 1.8(c)
on page 15 of \emph{Melissinos}. These are $N_{A}=10^{16}$ $\mathrm{cm^{-3}}$
and $N_{D}=10^{15}$ $\mathrm{cm^{-3}}$. On page 9 of \emph{Melissinos}, we
learn that for germanium $n_{i}\approx 10^{13}$ cm$^{-3}$ and $E_{g}=0.7$ eV
at room temperature (300 K). From there data we determine that $N_{s}\approx
8$ $10^{18}$ cm$^{-3}$. Plugging these values into the expression for the
bias voltage we find that $V_{bi}=0.596$ V, $0.295$ V, and $0.025$ V at $77$
K, room temperature ($300$ K), and $500$ K, respectively.

Since the bias voltage decreases with temperature, the performance of the
junction becomes worse as the temperature decreases.

\vspace{0.5cm} \hrule \vspace{0.5cm}

\noindent\textbf{4}. Short questions. \vspace{0.5cm}

\noindent (a) Using just the lattice parameter, compute the Fermi energy for
aluminum and compare with the value given, for instance, in \emph{Tipler}.
(Hint: how many valance electrons per atom are there in Al?) If there is a
discrepancy, it indicates that the effective mass of electrons in Al is
different from the free electron mass. \vspace{0.5cm}

Aluminum has a face-centered cubic lattice structure, and so it has four
atoms per unit cell. The density and atomic weight (as found in the \emph{%
CRC Handbook of Chemistry and Physics}) are 2.70 g/$\mathrm{cm}^3$and 26.98
g/mol, respectively. We learn from a periodic table that aluminum has three
valence electron per atom. Therefore the volume occupied by one atom is
(26.98 g/mol)(1 mol/6.022 $10^{23}$ atoms)/(2.70 g/$\mathrm{cm}^3$) = 1.66 $%
10^{-23}$ $\mathrm{cm}^3$/atom. Since there are four atoms per unit cell,
the volume of the unit cell is four time this, or 6.64 $10^{-23}$ $\mathrm{cm%
}^3$, and the lattice parameter is given by 4.05 $10^{-8}$ cm. Since there
are three valence electrons per atom, the number density of valence
electrons is simply $n_e$ = 3/(1.66 $10^{-23}$ $\mathrm{cm}^3$) = 18.1 $%
10^{22}$ electrons/$\mathrm{cm}^3$. From equation 1.4 on page 6 of \emph{%
Melissinos} we see that
\begin{eqnarray*}
E_{F} = \frac{\hbar^{2}}{2m}\left( 3\pi^{2}n_{e} \right)^{2/3}.
\end{eqnarray*}
Plugging in the appropriate quantities, we find that $E_F$ = 7.38 $10^{-21}$
J = 0.046 eV.

In Table 39-1 on page 1295 of \emph{Tipler} (volume 2), the Fermi energy of
aluminum is given as 11.7 eV. Therefore the effective mass of an electron in
aluminum must be considerably different from that of a free electron.
\vspace{0.5cm}

\noindent(b) Using the measured resistivities, given by \emph{Tipler} on
page $722$, compare the resistance (or, if you wish, the conductance) of an
Al wire to that of a Cu wire having the same length and cross section.
\vspace{0.5cm}

From page 722 of \emph{Tipler} we have $\rho_{\mathbf{Al}} = 2.8\times
10^{-8}$ $\Omega/\mathbf{m}$ and $\rho_{\mathbf{Cu}} = 1.7\times 10^{-8}$ $%
\Omega/\mathbf{m}$. On page 721 of \emph{Tipler}, we learn that
\begin{eqnarray*}
R = \rho\frac{L}{A},
\end{eqnarray*}
so if both wires have the same length and cross section then
\begin{eqnarray*}
\frac{R_{\mathbf{Al}}}{R_{\mathbf{Cu}}} &=& \frac{\rho_{\mathbf{Al}}}{\rho_{%
\mathbf{Cu}}} \\
&=& 1.65
\end{eqnarray*}
\vspace{0.5cm}

\noindent (c) Compare the resistance (or, if you wish, the conductance) of
an Al wire to that of a Cu wire having the same length and weight. \vspace{%
0.5cm}

Since the wires have the same weight we have $d_{\mathbf{Al}}A_{\mathbf{Al}%
}L = d_{\mathbf{Cu}}A_{\mathbf{Cu}}L$, or $A_{\mathbf{Al}}/A_{\mathbf{Cu}} =
d_{\mathbf{Cu}}/d_{\mathbf{Al}}$, where $d$ represents the density of the
material. Therefore,
\begin{eqnarray*}
\frac{R_{\mathbf{Al}}}{R_{\mathbf{Cu}}} &=& \frac{\rho_{\mathbf{Al}}}{\rho_{%
\mathbf{Cu}}}\frac{A_{\mathbf{Cu}}}{A_{\mathbf{Al}}} \\
&=& \frac{\rho_{\mathbf{Al}}}{\rho_{\mathbf{Cu}}}\frac{d_{\mathbf{Al}}}{d_{%
\mathbf{Cu}}}.
\end{eqnarray*}
From Table 11-1 on page 333 of \emph{Tipler} we see that $d_{\mathbf{Al}}/d_{%
\mathbf{Cu}} = 0.030$, so we have finally that $R_{\mathbf{Al}}/R_{\mathbf{Cu%
}} = 0.50$. \vspace{0.5cm}

\noindent (d) You should have found from (b) and (c) that, at least by some
criteria, Al is a better conductor than Cu. It is also much cheaper. Why
then is Cu used for most electrical wiring? \vspace{0.5cm}

One reason copper is used rather than aluminum is that aluminum oxide does
not conduct electricity, while copper oxide does. Like most metals, aluminum
and copper quickly develop a thin layer of metal oxide when exposed to air.
This means that if you strip two ends of an aluminum wire and twist them
together, current will not want to flow across the splice. If the same
procedure is repeated with copper, however, one finds that current readily
flows across the splice. Thus, despite the fact that aluminum is cheaper, in
most applications it is more convenient to use copper instead.

Another reason is that aluminum tends to expand more when heated, which
significantly reduces the lifetime of splices. \vspace{0.5cm}

\noindent (e) Why is the filament of an incandescent light bulb made of
tungsten, rather than steel or copper? \vspace{0.5cm}

From pages B-14, B-40, and E-88 of the \emph{CRC Handbook of Chemistry and
Physics, 66th Edition}, we learn that the melting points of copper,
tungsten, and steel are $1083$ ${}^{\circ}\mathrm{C}$, $3410$ ${}^{\circ}%
\mathrm{C}$, and $1500$ ${}^{\circ}\mathrm{C}$, respectively. We also learn
from page 278 of \emph{Bloomfield} that tungsten sublimes very slowly at
temperatures below its melting point. Thus tungsten filaments can be run
much hotter than filaments made of copper and steel before the filament
melts or the rate of sublimation becomes intolerable. Because it can be run
hotter, the tungsten filament also produces a richer, whiter light that is
more similar to the light of the sun. For this reason, tungsten filaments
are used in incandescent light bulbs instead of copper or steel. \vspace{%
0.5cm}

\noindent (f) Why has silicon become the standard semiconductor material for
most purposes? What are possible alternative materials? \vspace{0.5cm}

One reason for the prevalence of silicon in semiconductor devices is that
silicon is a readily available material. Sand (i.e. quartz) is simply
silicon dioxide. Another reason is that when people first started to work
with semiconductors they chose to work with silicon, and over the years
manufacturers have developed tricks to make semiconductor components
smaller, faster, etc. that are specific to silicon. It would cost a great
deal of money to modify these techniques for use with other elements, so
manufacturers continue to work with the traditional silicon.

Alternative materials that are used include germanium and gallium arsenide.
\vspace{0.5cm}

\noindent (g) Why is silicon not suitable for a light emitting diode? What
materials are used for this purpose? \vspace{0.5cm}

A material such as gallium arsenide is much more suitable than silicon for
LEDs because GaAs is \emph{direct gap} semiconductor, while Si is an \emph{%
indirect gap} semiconductor. In both materials the states near the top of
the valence band have zero momentum, but in GaAs the states at the bottom of
the conduction band also have zero momentum, while in Si they have nonzero
momentum. This is because the wavefunction changes sign in going from one Si
atom to the next on the edge of the cubic cell. (See the picture of the
energy bands of Si on page 308 of the \emph{Physicist's Desk Reference}.)
This means that the momentum of an electron in GaAs changes very little when
it emits or absorbs a photon, and thus light emission can occur directly in
this material. In Si, however, photon emission is an indirect process that
cannot occur unless the conduction band electron has a secondary process by
which it can dispose of its momentum. (The terms \emph{direct gap} and \emph{%
indirect gap} refer to whether or not the bottom of the conduction band
lines up directly above the top of the valence band in a plot versus
momentum.)

Gallium arsenide has a bandgap of $1.4$ eV, which corresponds to a red
photon. Thus GaAs is used to make red LEDs. A wider bandgap, resulting in
the emission of yellow or green light, is obtained by replacing some of the
gallium atoms with aluminum, or by replacing some of the arsenic atoms with
phosphorus. \vspace{0.5cm}

\noindent (h) In a $p$-$n$ junction at zero external bias, what is the sign
of the net electrical charge on the $p$ side? On the $n$ side? \vspace{0.5cm}

The net electric charge is negative on the $p$ side and positive on the $n$
side. This is because when the two pieces of semiconductor first make
contact, there is a higher concentration of positive carriers on one side of
the junction and a higher concentration of negative carriers on the other.
Thus there is a diffusive current and positive carriers flow from the $p$
side to the $n$ side and negative carriers flow from the $n$ side to the $p$
side. Once the electrons cross into the $p$ side of the junction they
annihilate with a hole and similarly for the holes that cross into the $n$
side. The new charge distribution produces an electric via Gauss's law, and
the direction of the electric field is such that it \emph{opposes} the
diffusion current. The current continues to flow until the system reaches
equilibrium, where the current due to diffusion is exactly the opposite of
that due to the electric field. \vspace{0.5cm}

\end{document}
