diff --git a/PS1/doc.tex b/PS1/doc.tex
index 604b74b..52e4a50 100644
--- a/PS1/doc.tex
+++ b/PS1/doc.tex
@@ -541,7 +541,7 @@ ylabel('Current [A]');
             $\mu_1 = 0$ eV. Since $\mu_1 = \mu - qV_S$ and $\mu = 0$ eV, $V_S = 0$ V.
             For $f_2(E + U)$ the step point occurs at $E = \mu_2 - U = -0.25$ eV. Since $U = -0.25$ eV, it follows that 
             $\mu_2 = -0.5$ eV. Since $\mu_1 = \mu - qV_D$ and $\mu = 0$ eV, $V_D = 0.5$ V.
-            
+
             \subsubsection*{(ii)}
 
             For $f_1(E)$ the step point occurs at $E = \mu_1 = 0.25$ eV. Since $\mu_1 = \mu - qV_S$ and $\mu = 0$ eV, $V_S = -0.25$ V.
@@ -549,34 +549,70 @@ ylabel('Current [A]');
             This assumes $U = 0$ eV.
 
         \subsection*{(c)}
-        
+
             \subsubsection*{(i)}
 
-            \begin{figure}
+            \begin{figure}[H]
                 \centering
-                \includegraphics[width=\textwidth]{q4c.jpg}
-                \caption{}
+                \includegraphics[width=0.7\textwidth, angle=90, origin=c]{q4c.jpg}
+                \caption{Visualisation of energy levels.}
                 \label{fig:q4c}
             \end{figure}
 
             \subsubsection*{(ii)}
+                Starting with equation 2 in the assignment we get:
+
+                \begin{equation*}
+                    I = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot \int_{-0.1}^{0.2} [1 - 0] \cdot 10^4 dE \\
+                \end{equation*}
+                \begin{equation*}
+                    I = \frac{q}{\hbar} \cdot \frac{0.005^2}{0.01} \cdot [10^4 \cdot 0.2 - 10^4 \cdot -0.1]
+                \end{equation*}
+                \begin{equation*}
+                    I = 1.8mA
+                \end{equation*}
+
+                (Note that 1)
+
+
             \subsubsection*{(iii)}
 
-        
+
         \subsection*{(d)}
 
         Using equation (5) in the assignment and plugging in the given values we obtain:
 
         \begin{equation*}
-            U = -q[\frac{C_SV_S+C_GV_G+C_DV_D}{C_E}] + \frac{q^2(N-N_0)}{C_E}
-            U = \alpha_SV_S + \alpha_GV_G + \alpha_DV_D + U_0(N - N_0)\:\:[eV]
-            U = 0*0 + 0.5*0\:eV + 0.5*0.6\:eV + 0.25\:eV*(0.325 - 0)\:\:[eV]
-            U = 0.38125 eV$ $
+            \begin{aligned}
+                U = -q[\frac{C_SV_S+C_GV_G+C_DV_D}{C_E}] + \frac{q^2(N-N_0)}{C_E} \\
+                U = -\alpha_SV_S - \alpha_GV_G - \alpha_DV_D + U_0(N - N_0)\:\:[eV] \\
+                U = -0*0 - 0.5*0\:eV - 0.5*0.6\:eV + 0.25\:eV*(0.325 - 0)\:\:[eV] \\
+                U = -0.21875\:eV
+            \end{aligned}
+        \end{equation*}
+
+        Then, starting with Equation \ref{eq:N_new} and plugging in $D(E-U) = \delta(E - \varepsilon)$, we obtain
+
+        \begin{equation*}
+                N = \frac{\gamma_1f_1(\varepsilon + U) + \gamma_2f_2(\varepsilon + U)}{\gamma_1 + \gamma_2}
+        \end{equation*}
+        Recalling that the expressions for the contact Fermi functions are (with $\mu = 0$ eV):
+        \begin{equation*}
+            \begin{aligned}
+                f_1(\varepsilon + U) = \frac{1}{1 + e^{frac{\varepsilon + U + qV_S}{k_BT}}} = \frac{1}{1 + e^{frac{0.2 - 0.21875 + 0}{0.025}}} = 0.679 \\
+                f_2(\varepsilon + U) = \frac{1}{1 + e^{frac{\varepsilon + U + qV_D}{k_BT}}} = \frac{1}{1 + e^{frac{0.2 - 0.21875 + 0.6}{0.025}}} \simeq  0
+            \end{aligned}
+        \end{equation*}
+        Then, solving for $\gamma_2$:
+        \begin{equation*}
+            \begin{aligned}
+                \gamma_2 = \gamma_2\frac{N - f_1(\varepsilon + U)}{f_2(\varepsilon + U) - N} = 5.45 \times 10^-3 \: eV
+            \end{aligned}
         \end{equation*}
 
-        Then, starting with Equation \ref{eq:N_old} and plugging in $D(E-U) = \delta(E - U - \varepsilon)$, we obtain
         
 
+
         \subsection*{(e)}
         \subsection*{(f)}
         \subsection*{(g)}
diff --git a/PS1/q4c.jpg b/PS1/q4c.jpg
index 1c8718b..cb62903 100644
Binary files a/PS1/q4c.jpg and b/PS1/q4c.jpg differ