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\begin{document} 
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\exnumber{4220}
\heading{Cheimchal equilibrium: H2[3D]==2H[2D]} 
\auname{Edo Hogeg}


{\bf The problem:}

\Dn

An ${H_{2}}$ molecule (mass ${2\mathsf{m}_{H})}$ decomposes into ${H}$ atoms when it
is absorbed upon a certain metallic surface. The ${H_{2}}$ and ${H}$ have an energy gain
${\varepsilon_{2}}$ and ${\varepsilon_{1}}$ per atom respectively, related to each state. The
binding of the ${H}$ atoms is not to particular sites on the surface: the ${H}$ atoms
are free to move on the surface. Consider the ${H_{2}}$ and ${H}$ as an ideal
gas, and express the surface density of the ${H}$ atoms as a
function of the ${H_{2}}$ pressure.


\Dn




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

When the ${H_{2}}$ and ${H}$ come to equilibrium (that is the system's free energy is minimized) their chemical potential are equal to zero.
We'll start by finding the partition function. 

\begin{equation}
Z_{1 H_{2}}=\int\frac{d^{3}xd^{3}p}{(2\pi)^{3}}e^{-\beta(\frac{p^{2}}{2(2m)}+\varepsilon_{2})}=
e^{-\beta\varepsilon_{2}}V\left(\frac{m}{\beta\pi}\right)^\frac{3}{2}
=e^{-\beta\varepsilon_{2}}\frac{2^{\frac{3}{2}}V}{\lambda_{T}^{3}}
\end{equation}

\begin{equation}
Z_{1 H}=\int\frac{d^{2}xd^{2}p}{(2\pi)^{2}}e^{-\beta(\frac{p^{2}}{2m}+\varepsilon_{1})}=
e^{-\beta\varepsilon_{1}}A\frac{m}{2\beta\pi}=
e^{-\beta\varepsilon_{1}}\frac{A}{\lambda_{T}^2}
\end{equation}

Now let's assume there are ${N}$ ${H_{2}}$ molecules to begin with.
With the definition of a macroscopic reaction coordinate ${n}$ (See lecture notes: [5.3] The law of mass action) we know that in equilibrium there are ${N-n}$ ${H_{2}}$ molecules and ${2n}$ ${H}$ atoms. Recalling both ${H_{2}}$ and ${H}$ are ideal gasses we obtain:

\begin{equation}
  Z_{H_{2}}=\frac{1}{(N-n)!}Z_{1 H_{2}}^{N-n}=
  \frac{e^{-\beta\varepsilon_{2}(N-n)}}{(N-n)!}\left(V\left(\frac{m}{\beta\pi}\right)^\frac{3}{2}\right)^{N-n}
  \end{equation}  

\begin{equation}
•Z_{H}=\frac{1}{(2n)!}Z_{1 H}^{2n}=
\frac{e^{-\beta\varepsilon_{1}(2n)}}{(2n)!}\left(A\frac{m}{2\beta\pi}\right)^{2n}
\end{equation}

\begin{equation}
F_{H_{2}}=-\frac{1}{\beta}ln(Z_{H_{2}})=
-\frac{1}{\beta}\left[-\beta\varepsilon_{2}(N-n)-(N-n)ln(N-n)+(N-n)+(N-n)ln(V\left(\frac{m}{\beta\pi}\right)^{\frac{3}{2}})\right]
\end{equation}

\begin{equation}
F_{H}=-\frac{1}{\beta}ln(Z_{H})=
-\frac{1}{\beta}\left[-\beta\varepsilon_{1}(2n)-2nln(2n)+2n+2nln(A\frac{m}{2\beta\pi})\right]
\end{equation}


Note that Stirling's approximation ${ln(N!)\approx Nln(N) -N}$ was used in (5) and (6).
We would like to first find the ${H_{2}}$ pressure, Then we'll equilibrate the chemical potential to find our desired solution.

\begin{equation}
P=-\frac{\partial F_{H_{2}}}{\partial V}=\frac{N-n}{\beta V}
\end{equation}

\begin{equation}
\mu_{H_{2}}+\mu_{H}=0
\end{equation}

\begin{equation}
\frac{\partial F_{H_{2}}}{\partial n}+\frac{\partial F_{H}}{\partial n}=0
\end{equation}

\begin{equation}
-2ln(\frac{2n}{A})-2\beta\varepsilon_{1}+2ln(\frac{m}{2\pi\beta})+
ln(\frac{N-n}{V\beta})+ln(\beta)+\beta\varepsilon_{2}-\frac{3}{2}ln(\frac{m}{\pi \beta})=0
\end{equation}

Denoting the density of the ${H}$ atoms:

\begin{equation}
\mathcal{N}=\frac{2n}{A}
\end{equation}
 
Finally the density ${\mathcal{N}}$ of the ${H}$ atoms is:

\begin{equation}
\mathcal{N}(P)=\frac{e^{-\frac{\beta}{2}(2\varepsilon_{1}-\varepsilon_{2})}}
{2^{\frac{3}{4}}}\sqrt{\frac{\beta P}{\lambda_{T}}}
\end{equation}

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\end{document} 
