Answer: Fraction of Empty States in GaAs at Room Temp w/E_{F} 0.1eV Below E_{c}

[JoshHolloway]

Homework Statement

The donor state for tellurium (Te) in GaAs is 5.9 meV below the conduction band $$(E=E_{c})$$. At room temperature, what fraction of the states are empty if the Fermi energy lies 0.1 eV below $$E_{c}$$

Homework Equations

$$E_{c}|_{GaAs}= 1.42eV$$
$$E_{d}=$$
$$E_{F}=$$
$$kT|_{(T=300K)}=0.02585 eV$$
$$N_{c}|_{(GaAs)} = 1.04\times10^{19}cm^{-3}$$

Where:
$$E_{d}$$ is the Donor Energy Level
$$E_{f}$$ is the Fermi Energy
$$T$$ is the Temperature
$$k$$ is the Wave Number
$$n_{d}$$ is the Density of Electrons in the Donor Energy Level
$$N_{c}$$ is the Effective Density of States in the Conduction Band$$n_{d} = 1 + \frac{1}{2}exp[\frac{(E_{c}-E_{d})-(E_{c}-E_{F})}{kT}]$$

The Attempt at a Solution

Ok, so here goes:
$$n_{d} = 1 + \frac{1}{2}exp[\frac{( eV - eV) - ( eV - eV)}{0.02585eV}]$$

$$n_{d} = 1 + \frac{1}{2}exp[\frac{ eV - eV}{0.02585eV}]$$

$$n_{d} = 1 + \frac{1}{2}exp[\frac{ eV}{0.02585eV}]$$

$$n_{d} = 1 + \frac{1}{2}exp[-3.64023]$$

$$n_{d} = 1 + \frac{1}{2}(0.026246)$$

$$n_{d} = 1 + (.013123)$$

$$n_{d} = 1.013123 cm^{-3}$$

 

[JoshHolloway]

Disregard this post. Sorry.