Update 13_semiconductors.md - fix

src/13_semiconductors.md

@@ -136,9 +136,9 @@ Or in other words
 $$E_e = E_c + \frac{\hbar^2k^2}{2m_e},$$
 $$E_h = E_{v,h} + \frac{\hbar^2k^2}{2m_h} = -E_{v} + \frac{\hbar^2k^2}{2m_h}.$$
 
-Here $E_c$ is the bottom of the conduction band and $E_v$ is the top of the valence band.
+Here $E_c$ is the energy of an electron at the bottom of the conduction band and $E_v$ is the energy of an electron at the top of the valence band.
 
-Observe that because we are describing particles in the valence band as holes, $m_h > 0$ and $E_h > 0$.
+Observe that because we are describing particles in the valence band as holes, $m_h > 0$ and $E_h > -E_v$.
 
 ??? question "a photon gives a single electron enough energy to move from the valence band to the conduction band. How many particles does this process create?"
     Two: one electron and one hole.