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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 $\mu=\frac{\mathrm dG}{\mathrm dN}$.

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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.

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