How can I calculate the chemical potential and the effective masses in an intrinsic semiconductor? I want to calculate the chemical potential and the effective masses of the germanium charge carriers considering that the relaxation time is the same for the two carriers: electrons and holes. Germanium has an energy gap $E_{\text{gap}}=0,68\, \mathrm{eV}$, its electron mobility being $\mu_{e}=3900\, \mathrm{cm^2/V\cdot s}$ and the mobility of the holes $\mu_{h}=1900\, \mathrm{cm^2/V\cdot s}$ to 300 K. In this case, $\tau_e=\tau_h$. I've been searching the Internet and discovered this expression for the chemical potential:
\begin{equation}
\mu(T)=\frac{1}{2}(E_c+E_v)+(K_BT)\frac{3}{4}\left(\frac{m_{h}*}{m_e*}\right)
\end{equation}
I have already calculated the chemical potential according to the data that I have and obtained a relation between the effective masses. I want to know the effective masses of holes and electrons as being one digit times the mass of the electron. How can I find the masses effective? I'm missing some detail or some equation that I can not remember.
\begin{equation}
\mu_e=\frac{e\tau}{m_e^{*}}
\end{equation}
\begin{equation}
\mu_h=\frac{e\tau}{m_h^{*}}
\end{equation}
Putting together the two equations I get a relation between the effective masses:
\begin{equation}
\frac{\mu_e}{\mu_h}=\frac{3900}{1900}=\frac{m_h^{*}}{m_e^{*}}
\end{equation}
Substituting the relation in the chemical potential I obtain:
\begin{equation}
\mu(T)=\frac{1}{2}\times 0.68 + (8.617\times 10^{-5} \times 300)\times \frac{3}{4} ln\bigg(\frac{3900}{1900}\bigg)=4.74\times 10^{-3}\,eV
\end{equation}
 A: The effective mass can be calculated as 
$$m^* = \frac{\hbar^2}{\left(\frac{\partial^2E(k)}{\partial k^2}\right)}$$

Edited:
Normally, an energy level is identified by some quantum numbers: ${n, \mathcal{l}, s, j, m}$. For example: $n=1; l=0; m=0; s=½$.
When the levels "expand" into bands, those quantum numbers refer to a band. Not a level anymore. Now it is a band.
But, there are many energies inside one band. How to differentiate them?
Well, there is a new quantum number, called "quasi-momentum", $\vec{k}$. This $k$ identifies a concrete energy inside the band (the band is an interval, so we need something else to pick a concrete energy).
So, we would need to know the function $E(k)$. Then, apply the first formula, and you've got it.
BUT. The problem is that we need to solve the Schrödinger's equation for the whole solid if we want $E(k)$. And that is impossible. Solving for one ataom requires approximation. Imagine $1 mol$ of atoms. That's absolutely crazy.
So, we just cannot know $E(k)$ anallytically. However, we can know some things:


*

*$E(k)$ is approximately parabollic near the ends of the band.

*Consequently, the second derivative is constant in the borders of the bands.

*That means that the effective mass is constant in the band borders. The effective mass varies withing the band, but it is constant at the borders.

*And, that constant depends on the material. $Si$ has its values, $Ge$ has its values, $GaAs$ has its values.


In short:
We cannot know $E(k)$, so we cannot know the effective masses; but we can find them experimentally. There's nothing wrong with working experimental data. You do it all the time!
When you use $G=6.67\cdot10^-11 Nm^2/kg^2$, you are using experimental data, and it works fine. Just substitute when you cannot avoid it more, and just that. 
