Free electron density conduction band

[Kara386]

Homework Statement

How many free electrons are there in the CB? Diamond has a bandgap of $5.5$eV.Assume the material is at room temperature and that there are $2 \times 10^{22}$ cm$^{-3}$ electrons in the material. What does this mean for their use in semiconductor devices?

Homework Equations

The Attempt at a Solution

Probably I should use something like this equation:
$n = N_c \exp\left(\frac{E_F - E_C}{kT}\right)$
Where $N_c = \frac{8\pi\sqrt{2}m^{*3/2}}{h^3}\sqrt{E-E_c}$ is the effective density of states, and then multiply by the number of electrons in the material cm$^{-3}$. But there's quantities in there I don't know and can't calculate, like the Fermi energy (although is that half the gap?), and the energy of the conduction band, and the effective mass. I feel like this should be a simple question, but I can't see how to do it. It's possible, although unlikely, that I just have to look these quantities up, but if there's any other way I'd rather not do that. Thanks for any help! :)

 

[mark726]

Hi there!
Yes, you are on the right track. The equation you have mentioned is known as the Fermi-Dirac distribution, and it describes the probability of finding an electron at a given energy level in a semiconductor. To calculate the number of free electrons in the conduction band, you can use the following equation:

$n = N_c \exp\left(\frac{E_F - E_C}{kT}\right)$

As you have correctly pointed out, the Fermi energy (E_F) is typically taken to be halfway between the conduction band minimum (E_C) and the valence band maximum (E_V). In this case, the Fermi energy in diamond will be 2.75 eV. The value of the effective mass (m*) can be found in reference books or online databases, and for diamond, it is approximately 0.2 times the mass of an electron.

Using these values, you can calculate the number of free electrons in the conduction band at room temperature. However, it is important to note that this number may vary depending on the temperature and doping levels of the semiconductor. In general, a higher number of free electrons in the conduction band can lead to better conductivity and improved performance in semiconductor devices. I hope this helps!