solution tutorial 3

diff --git a/solid_state_physics/tutorial/1_03s.tex b/solid_state_physics/tutorial/1_03s.tex
index 1174731..dd97aa2 100644 (file)
--- a/solid_state_physics/tutorial/1_03s.tex
+++ b/solid_state_physics/tutorial/1_03s.tex
@@ -41,8 +41,10 @@
 \begin{enumerate}
  \item $U=IR \Rightarrow EL=jA\rho\frac{L}{A}
              \Rightarrow E=j\rho$
- \item distance: $v\,dt$\\
-       number of electrons crossing $A$: $n(v\,dt)A$\\
+ \item \begin{itemize}
+        \item distance: $v\,dt$
+        \item number of electrons crossing $A$: $n(v\,dt)A$
+       \end{itemize}
        $\Rightarrow$ $j=\frac{I}{A}=\frac{dQ/dt}{A}=\frac{-e\,n(v\,dt)A/dt}{A}
                        =-nev$
  \item \begin{itemize}
@@ -60,12 +62,30 @@
        \item $j=\left(\frac{ne^2\tau}{m}\right)E$\\
        \item $j=\sigma E \Rightarrow \sigma=\frac{ne^2\tau}{m}$
        \end{itemize}
- \item Energy transfer: $\frac{m}{2}v_{drift}^2$,
-                        $\qquad v_{drift}$:
-                       end drift velocity of the accelerated electron\\
-       $v_{drift} \ne v_{average}$
-
-       
+ \item \begin{itemize}
+        \item Energy transfer: $\frac{m}{2}v_{drift}^2$,
+                               $\quad v_{drift}$:
+                              final drift velocity of the accelerated electron
+       \item $v_{drift}=-\frac{eE}{m}t_0$, $\quad t_0$:
+             free flight time (no collision) of the electron
+        \item $v_{average}=\frac{1}{t_0}\int_{0}^{t_0} v(t) dt
+                         =-\frac{eE}{m}\frac{1}{t_0}[\frac{t^2}{2}]_{0}^{t_0}
+                         =-\frac{eE}{m}\frac{t_0}{2}=:-\frac{eE}{m}\tau$,
+              $\qquad t_0=2\tau$
+       \item Each of the $n$ electrons per unit volume
+             transfer the kinetic energy $\frac{1}{2}mv^2_{drift}$
+             once per $t_0$ to the lattice
+       \end{itemize}
+       \[
+       \Rightarrow \frac{P}{V}=\frac{E_{kin}}{Vt_0}
+                              =\frac{n\frac{1}{2}m\frac{e^2E^2}{m^2}t_0^2}{t_0}
+                             =n\frac{1}{2}\frac{e^2E^2}{m}2\tau
+                             =\sigma E^2=jE=j^2\rho=\frac{I^2}{A^2}\frac{A}{L}R
+                             =\frac{I^2R}{V}
+       \]
+       \[
+       \Rightarrow P=I^2R \textrm{ (Joule heating)}
+       \]
 \end{enumerate}
 
 \end{document}