2
$\begingroup$

In deriving the density of electrons at the bottom of the valence band in a semiconductor,

$$n(T) = \int^\infty_{\epsilon_C} \frac{g_c(\epsilon)}{e^{\beta(\epsilon - \mu)}+1}$$

we approximate the integral as

$$n(T) = \int^\infty_{\epsilon_C} \frac{g_c(\epsilon)}{e^{\beta(\epsilon - \mu)}}$$

i.e. $$e^{\beta(\epsilon - \mu)} >> 1$$

because that the chemical potential is far below the minimum conduction band energy $\epsilon_C$.

This derivation has reminded me that I never really got to grips with the chemical potential. What is the significance of the chemical potential in this context? Is it simply $\mu(T=0) = \epsilon_F$ or is there more to it? More generally, I'd love to hear a good way to think about the chemical potential, especially in the context of solid state physics.

$\endgroup$
1

1 Answer 1

1
$\begingroup$

I have always found this very difficult to get an intuitive understanding of, this is the way I think about it.

The chemical potential is the energy at which you can add or remove a particle from the semiconductor without changing the entropy of the electron gas.

$$\mu_e = \left(\frac{\partial E}{\partial N_e} \right)_{S}$$

Why? Because of the Exclusion principle.

An electronic gas is a Fermi gas. It has a macrostate with entropy $S$ and temperature $T$ defined by its arrangement of microstates.

For example, if you insert a particle at energy $E$, the ensemble must arrange itself; particles above the energy will all be moved upwards. In general this alteration will have an associated entropy change. The chemical potential is the energy at which there is no entropy penalty to do this.

I would be very interested to hear other peoples intuitive explanations, I’m sure what I’ve written here is not the full story.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks for your response. An expansion on your thought process seems to be found here: aapt.scitation.org/doi/10.1119/1.17844 ... thanks also to @Jon Custer for pointing me in the direction of this resource. $\endgroup$
    – Nik
    Commented Jan 8, 2021 at 14:55
  • $\begingroup$ That’s great, will read! $\endgroup$
    – boyfarrell
    Commented Jan 8, 2021 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.