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\title{Problem 7.12}
\subtitle{Semiconductors}
\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
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	We're looking at an example of a doped superconductor.
	We have $N - M$ atoms of silicon, and $M$ atoms of phosphorus.
	Each Si atom contributes one electron, and two states (at $\pm \flatfrac{\Delta}{2}$).
	We'll take the energy gap $\Delta = \SI{1.16}{\eV}$ for example.
	
	The phosphorous atoms contribute \emph{two} electrons and two states, at $-\flatfrac{\Delta}{2}$ and $\flatfrac{\Delta}{2} - \epsilon$.
	The impurity energy guy $\epsilon = \SI{0.044}{\eV}$, and is much smaller than the energy gap.
	
	So the ground state here has $N + M$ electrons, so the $N$ valence bands at $-\flatfrac{\Delta}{2}$ and $M$ impurity band states at $-\flatfrac{\Delta}{2} - \epsilon$ are filled, and the $N - M$ conduction band states at $\flatfrac{\Delta}{2}$ are empty.

	\subsection*{(a)}
	Derive a formula for the number of electrons as a function of temperature $T$ and chemical potential $\mu$ for the energy levels of our system.
	
	\section{Solution} \label{sec:solution}
	
	\subsection*{(a)}
	So we have $N + M$ electrons.
	Easy.
	
	Sethna probably wants us to write the right hand side of this too.

	For each energy $E$, the single electron occupation is our Fermi distribution:
	\begin{equation}
		f(T) = \frac{1}{e^{\beta \left(E - \mu \right)} + 1}
	\end{equation}
	So if we add this up by the number of electrons and number of available states, we get
	\begin{equation}
		N + M = \frac{N}{e^{\beta \left(-\flatfrac{\Delta}{2} - \mu \right)} + 1} + \frac{M}{e^{\beta \left(\flatfrac{\Delta}{2} - \epsilon - \mu \right)} + 1} + \frac{N - M}{e^{\beta \left(\flatfrac{\Delta}{2} - \mu \right)} + 1}.
	\end{equation}
	What this really gives us is an implicit relationship for $\mu(T)$, because those are the two undetermined quantities.
	
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