\experiment{Thermal Radiation}

\section{General}

The apparatus for study of the propagation of thermal energy by
radiation is shown schematically in \fref{ThermalRadApparatus}

\begin{figure}[b!]
  \centering
  \igw[4.5in]{ThermalRadApparatus}
  \caption{Thermal radiation apparatus.}
  \label{fig:ThermalRadApparatus}
\end{figure}


Any surface at a temperature above absolute zero radiates energy in
the form of electromagnetic radiation. When the surface is in a
vacuum, as in this case, radiation is the only mechanism for energy
loss. One imagines that the surface of [A] is at some temperature $T$
and asks at what rate $W$ thermal energy is radiated from the surface.
Since [C] is a vacuum, the rate can depend only upon the nature of the
surface and its temperature. The surface has been blackened in this
experiment to eliminate any effect due to the nature of the surface
other than its area $A$. The rate of energy loss may be expressed as
\begin{equation}
	\label{eq:TR1}
	P = \sigma A T^{x}
\end{equation}
The universal constant $\sigma$ is called the Stefan-Boltzmann
constant. We are to determine $\sigma$ and the exponent $x$ relating
$W$ to the surface temperature.

The interior surface of the brass cavity [B] also radiates thermal
energy to [A]. This surface, however, is maintained at a sufficiently
low temperature (77~K) by contact with liquid nitrogen that the
energy received by [A] from [B] is negligible. Further, [A] is
suspended by wires of very small cross-section to reduce conduction of
energy to [A] from the room. In equilibrium, then, the rate at which
energy is radiated through the vacuum from [A] must equal the rate at
which it is generated electrically within [A].

The radiating element [A] in this experiment is a thermistor. Thermal
energy (in watts) is generated within it at a rate given by
\begin{equation}
	\label{eq:TR2}
	P = V I
\end{equation}
where $V$ (in volts) and $I$ (in amperes) are measured by the meters
``V'' and ``I'' shown.
The resistance $R$ of a thermistor varies as a function of its
temperature. Since
\begin{equation}
	\label{eq:TR3}
	R = V/I
\end{equation}
measurement of $V$ and $I$ determines not only $P$ but $R$ as well. 

\section{Thermistor calibration}

The conductivity of a thermistor (semiconductor) is proportional to
the Boltzmann factor, $\exp(-E/kT)$, where $T$ is the temperature (in
kelvins), $k$ the Boltzmann constant ($\val{8.616 \times
  10^{-5}}{eV\,K^{-1}}$) and $E$ the ``band-gap'' energy of the
semiconductor---the energy which must be acquired by an electron for
it to participate in electrical conduction. The resistivity is the
reciprocal of the conductivity so that the thermistor resistance may
be expressed as 
\begin{equation}
	\label{eq:TR4}
	R = a e^{b/T} \qquad(b = E/k)
\end{equation}
The two constants, $a$ and $b$, are determined by measurement of $R$
at two different temperatures. Thus, if $R_{1}$ and $R_{2}$ are the
resistances of the thermistor at temperatures $T_{1}$ and $T_{2}$, then
\begin{equation}
	\label{eq:TR5}
	\ln(R_{1}/R_{2}) = b \left( \frac{1}{T_{1}} - \frac{1}{T_{2}}
    \right) 
    \qquad
    a = R_{1} e^{-b/T_{1}} = R_{2} e^{-b/T_{2}}
\end{equation}

Calibration is performed with atmospheric pressure in the apparatus to
hasten equilibrium. First measure $R$ at room temperature as indicated
by a thermometer. The following circuit is used with a large
(\val{10}{k\Omega}) series decade resistor to limit the power
dissipation in the thermistor to the order of microwatts:
\begin{figure}[h]
  \centering
  \igw[3in]{thermistor}
  \caption{Thermistor circuit.}
  \label{fig:Thermistor}
\end{figure}
The leads of one voltmeter are connected to points [1] and [3] to give
the potential across the thermistor. The second voltmeter reads the
voltage drop across the decade resistor (points [2] and [3]), from
which you can determine the current in the thermistor.\footnote{The
  \val{1}{k\Omega} resistor between the power
  supply and the thermistor is included to protect the thermistor from
  damage when the decade resistance is changed.} Use the potential and
current to determine $R_{1}$ at room temperature. Then surround the
thermistor bath to obtain $R_{2}$ at $T_{2} = \val{273}{K}$. (Record
the potential across the \val{10}{k\Omega} resistor at one- or
two-minute intervals to determine when equilibrium at ice temperature
is reached.) Determine $a$ and $b$ (and $E$) from these data and
equations \eref{TR2}. \eref[e]{TR1} is then used to determine $T$ in
the experiment as written.

As soon
as these data are obtained, have the instructor start the rough
and molecular drag vacuum pumps. Approximately 30 minutes are required
to reach the operating pressure of a few times \val{10^{-5}}{Torr}. Use this
time to analyze your calibration measurements and determine the
constants $a$ and $b$.

\section{Experimental procedure}

The same circuit used to calibrate the thermistor is used to measure
$V$ and $I$ (and thus $R$ and $P$) in the remainder of the experiment
\scream{except that} the series resistor is reduced from
\val{10}{k\Omega} to \val{1}{k\Omega}. 

When a sufficient vacuum is
obtained, establish a current of \val{\sim\! 10}{nA} just before
immersing the brass cavity in liquid nitrogen. (If power is not
applied to the thermistor before cooling, it will rapidly cool after
immersion to a resistance so high that sufficient power cannot be
applied to control its temperature.) Slowly raise the dewar of liquid
nitrogen to cover the brass cavity. The current through the thermistor
will drop a bit as its resistance increases. Wait several minutes for
the temperature of the thermistor to stabilize. Record several voltage
readings during this equilibration period to document the approach to
equilibrium. Determine the thermistor temperature and the power
radiated by the thermistor at equilibrium. 

Vary the current slightly
and again wait for equilibrium. Calculate $T$ and $P$. Proceed in this
way to generate data for $P$ vs.\ $T$ over a temperature range of
approximately \degC{0} to 50--60\degC{}. 

The exponent $x$ is most easily found by using a log-log plot of
equation (1):
\begin{equation}
	\label{eq:TR6}
	\log P = \log(\sigma A) + x \log T
\end{equation}
The slope of the plot $\log P$ vs.\ $\log T$ yields the exponent in
the ``Stefan-Boltzmann radiation law.'' Determine this law from your
data and evaluate the Stefan-Boltzmann constant $\sigma$, given $A =
\val{0.52}{cm^{2}}$. Think carefully about the best way to get a value
for $\sigma$.
