# Chemical potential of fermions [closed]

Hey guys I am trying to determine the chemical potential for electrons in metals. I have that:

For the valance band, $$\epsilon\lt\epsilon_\mathrm v$$, $$\rho(\epsilon)=g_\mathrm v$$, while for the conduction band, $$\epsilon\gt\epsilon_\mathrm c$$, $$\rho(\epsilon)=g_\mathrm c$$, where $$g_\mathrm v$$ and $$g_\mathrm c$$ are constants. The Fermi energy $$F=\frac12(\epsilon_\mathrm v+\epsilon_\mathrm c)$$ is located in the middle of the gap. The number of electrons is the same for all $$T$$.

I was able to calculate the average number of electrons in the conduction band $$\langle N_\mathrm c\rangle=2g_\mathrm ck_\mathrm BT\mathrm e^{-(\epsilon_\mathrm c-\mu)/(k_\mathrm BT)}$$ and the average number of holes $$\langle N_\mathrm h\rangle=2g_\mathrm vk_\mathrm BT\mathrm e^{(\epsilon_\mathrm v-mu)/(k_\mathrm BT)}$$ I need to show that $$\mu=\epsilon_\mathrm F-(k_\mathrm BT/2)\ln\left(\frac{g_\mathrm c}{g_\mathrm v}\right)$$ I also know that $$\mu=\epsilon_\mathrm F$$ at $$T=0$$ and that $$\mu=\frac{\mathrm dG}{\mathrm dN}$$.

Since the number of electrons is the same for all T $$N_h = N_c$$ It follows that $$2g_vk_bTe^{(\epsilon_v-mu)/k_BT} = 2g_ck_BTe^{-(\epsilon_c-\mu)/k_BT}$$ Next is just to solve for $$\mu$$ and to use the identity: $$\epsilon_F = \frac{1}{2}(\epsilon_c+\epsilon_v)$$ . To reach the desired expression.