Question

If the Fermi level in a piece of silicon is 438meV below the intrinsic Fermi level, then what is the concentration of holes in this piece of silicon per cm³? Assume that the temperature is 300°K, kT = 26meV, and ni = 1 x 10^10/cm³. (Note: For example, to answer 2.54 x 10^17/cm³, enter 2.54e17 into Moodle.)

Answer 1

The concentration of holes in the silicon is 1.56 x 10^9 holes/cm³.

Step-by-step explanation:

To find the concentration of holes in the silicon, we can use the relation:

n = ni * exp(-E/(kT))

Given that the Fermi level is 438meV below the intrinsic Fermi level, we can calculate the energy difference as E = 438meV - 26meV (since kT is given as 26meV). Substituting these values into the equation, we get:

n = (1 x 10^10/cm³) * exp(-412meV/(26meV)) = 1.56 x 10^9 holes/cm³

Since the material is non-degenerate, we can also use the Boltzmann approximation to relate the Fermi level position (EF) relative to the intrinsic level (EFi) with carrier concentrations:

p = ni * exp((EFi - EF)/(kT))

Given EFi - EF = 438 meV and kT = 26 meV,

p = 1 x 10^10 cm-3 * exp(438 meV / 26 meV)

Calculating the exponential factor and computing the concentration:

p ≈ 2.54e17 cm-3