# How can I calculate the chemical potential and the effective masses in an intrinsic semiconductor? [closed]

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: $$$$\mu(T)=\frac{1}{2}(E_c+E_v)+(K_BT)\frac{3}{4}\left(\frac{m_{h}*}{m_e*}\right)$$$$

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.

$$$$\mu_e=\frac{e\tau}{m_e^{*}}$$$$

$$$$\mu_h=\frac{e\tau}{m_h^{*}}$$$$

Putting together the two equations I get a relation between the effective masses:

$$$$\frac{\mu_e}{\mu_h}=\frac{3900}{1900}=\frac{m_h^{*}}{m_e^{*}}$$$$

Substituting the relation in the chemical potential I obtain: $$$$\mu(T)=\frac{1}{2}\times 0.68 + (8.617\times 10^{-5} \times 300)\times \frac{3}{4} ln\bigg(\frac{3900}{1900}\bigg)=4.74\times 10^{-3}\,eV$$$$

• When this assumption of equality the scattering times of electrons and holes are satisfy? Your proffesor said "do not use the experimental data" but you got the mobility values probably from this site what i put previously in answers. The best fit to your equation of chemical potential are effective masses of density of states of conduction and valence band. – Mateusz Glinkowski Nov 7 '18 at 10:13
• What geometrical assumptions are you allowed to do? Can it be an infinite periodic crystal lattice? Because then I think there are models you can solve for energy levels if you know the lattice vectors. I have only skimmed such books so I don't know any details. It might have been a book on variational physics applied to stat mech. I think I found it when googling that principle where the potential or something is almost parabolic (forgot the name), large deviations or something. – Emil Feb 17 '20 at 19:36
• Was the answer below satisfactory? Or can you copy and paste it on Matter Modeling SE (please)? – user1271772 Jul 23 '20 at 16:43

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:

1. $$E(k)$$ is approximately parabollic near the ends of the band.
2. Consequently, the second derivative is constant in the borders of the bands.
3. 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.
4. 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.

• Thanks for your disponibility. 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 ($m_0=9.11\times10^{−31}\,Kg$). How can I find the masses effective? I'm missing some detail or some equation that I can not remember. P.S: I added in the problem what I have done so far. – Carmen González Nov 6 '18 at 18:13
• Oh, so you have found $\mu(T)$. Good point. But what's the problem then? – FGSUZ Nov 6 '18 at 19:14
• The problem is: I know that: $m_{h}^{*}=(3900/1900)\times m_{e}^{*} \approx 2,05\times m_{e}^{*}$, but I want to know the effective masses as a function of the mass of the electron ($m_0=9,11\times 10^{-31}\,kg$), without resorting to tables (I have to calculate). I just have an equation that relates the two effective masses. How can I know the value of the "absolute" effective masses, that is, as a function of the mass of the electron? – Carmen González Nov 6 '18 at 19:28
• Okay, I edited the answer. I hope it is clearer now. – FGSUZ Nov 6 '18 at 19:47
• This problem was the university professor who asked me to solve it, but she does not want us to use experimental data. It just gave the information of the mobilities, gap energy and relaxation time to be the same for electrons and holes. And she asked to calculate the chemical potential, the effective masses, the number of electrons in the conduction band and the number of holes in the valence band, considering that the semiconductor is intrinsic. – Carmen González Nov 6 '18 at 20:24