root/trunk/matml/transport/problems/mbe/mbe-solution.tex

\documentclass{article}
\usepackage{fullpage}
\begin{document}
\begin{enumerate}
\item Molecular beam epitaxy deposition rate

  \begin{enumerate}
  \item What I was looking for here was:
    \begin{enumerate}
    \item Use the Clausius-Clapeyron equation to calculate the vapor pressure
      from the source temperature.
    \item \label{meanfreepath} Use the background pressure and temperature to
      estimate the mean free path of sublimating atoms in the chamber.
    \item \label{knudsen} Calculate the Knudsen number, which is the ratio of
      mean free path to chamber size.
    \item If the Knudsen number is greater than one, calculate the sublimation
      rate using the Langmuir equation.
    \end{enumerate}

    Also, given the wording of the problem, another approach which answered the
    stated question was to calculate the sublimation flux using a heat balance.
    Assuming that the difference between flux in from the heat source (electron
    beam) called $q_v$, and heat flux conducted through the solid into the
    cooling water $q_s$, is all used to sublimate the atoms, this gives:
    $$q_v-q_s = \delta H_s\cdot J \Rightarrow J = \frac{q_v-q_s}{\Delta H_s}.$$
    With that approach, steps \ref{meanfreepath} and \ref{knudsen} above were
    still necessary.

  \item This implicitly assumed that electron beam power either went into
    sublimation, or into the cooling water, so the flux balance is again given
    by:
    $$q_v-q_s = \Delta H_s\cdot J.$$
    The net power is then the next flux times the area:
    $$P_{ebeam}-P_{water}=\Delta H_s\cdot J\cdot A.$$

  \item If sublimation flux distribution follows a cosine distribution (like
    radiating photons), and the sublimating atoms follow straight-line
    trajectories through the chamber (like photons), then guess what?  We can
    use the radiation viewfactor to calculate the fraction of sublimating atoms
    which land on the substrate.

    From here it's straightforward: $r_1=2$cm, $r_2=9$cm, $D=30$cm, so
    $D/r_1=15$ (effectively $\infty$ on the viewfactor graph) and $r_2/D=0.3$,
    giving $F_{12}\simeq0.1$.

    The ratio of deposition flux at the substrate to sublimation flux from the
    source is this times the area ratio:
    $$\frac{J_{subs}}{J_{source}}=\frac{A_1F_{12}}{A_2}=
    \frac{\rm\pi(2cm)^2\cdot0.1}{\rm\pi(9cm)^2}\simeq0.005.$$
  \end{enumerate}
\end{enumerate}
\end{document}