- #1

steroidjunkie

- 18

- 1

## Homework Statement

[a) Show that chemical potential in intrinsic semiconductor at T = 0 lies exactly in the middle of the energy gap. At what temperature is the chemical potential equal ##(3/4)(E_c + E_v)##, where ##E_c## and ##E_v## are energies od the bottom of the conduction band and of the top of the valence band?

b) Explain how the chemical potential changes with temperature if we dope the matherial with:

##b_1##) donors;

##b_2##) acceptors?

c) Semiconductor with direct band gap is doped with impurities that give density of ##10^{23}## electron/##m^3##. Calculate density of holes at room temperature if energy band gap is 1 eV, and effective masses of charge carriers in conduction and valence bands are 0,25 and 0,4 mass of free electron, respectively.

## Homework Equations

a) ##n=2(\frac{m_e k_B T}{2 \pi \hbar^2})^{\frac{3}{2}} e^{\frac{\mu-E_C}{k_B T}}##

##p=2(\frac{m_h k_B T}{2 \pi \hbar^2})^{\frac{3}{2}} e^{\frac{E_V-\mu}{k_B T}}##

##n_i=p_i##

c) ##T = 300 K##

## n = 10^{23}/m^3 ##

##E_g = 1 eV ##

## m_e^{*} = 0.25 m_e ##

## m_h^{*} = 0.25 m_e ##

## The Attempt at a Solution

a) Equaling n to p gives out an equation for chemical potential:

## \mu=\frac{E_C + E_V}{2} + \frac{3}{4} k_B T \ln \frac{m_h}{m_e}##

and for ##T=0##:

## \mu=\frac{E_C + E_V}{2}##

because the second term vanishes.

The second part is just putting the ##(3/4)(E_c + E_v)## instead of ##\mu## in equation ## \mu=\frac{E_C + E_V}{2} + \frac{3}{4} k_B T \ln \frac{m_h}{m_e}## and solving equation:

##(3/4)(E_c + E_v) = \frac{E_C + E_V}{2} + \frac{3}{4} k_B T \ln \frac{m_h}{m_e}##

for T gives:

##T = \frac{E_C + E_V}{3 k_B \ln \frac{m_h}{m_e}} ##

b) and c) are the problem.

b) I know what happens with chemical potential if the semiconductor is doped, but I am clueless on proving it.

c) I found this equation in Kittel's "Introduction to solid state physics" (page 210) but I'm not sure if it's applicable to the problem:

## n = (n_0 N_d) e^\frac{E_d}{2 k_B T} ##, ## n_0 = 2 (\frac{m_e^* k_B T}{2 \pi \hbar^2})^{\frac{3}{2}} ##