# Homework Help: Intrinsic semiconductor, carrier concentration

1. Mar 12, 2017

### Incand

1. The problem statement, all variables and given/known data
An intrinsic semiconductor with a direct gap have valance band $\epsilon_k = E_v-b|k|^2$ and conduction band $\epsilon_k = =E_c+a|k|^2$, with $E_v=6.0$eV, $E_c = 5.5eV$, $a=5.0eV\cdot Å^2$, and $b=3.0eV\cdot Å^2$.
Calculate the chemical potential $\mu$ and the number of charge carriers $n+p$.

2. Relevant equations
Effective mass:
$m^*_e = \frac{\hbar^2}{2a}$
Intrinsic semiconductor $n=p$:
$p=n = \sqrt{pn} = \frac{1}{\sqrt{2}}\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}(m_e^* m_n^*)^{3/4}e^{-E_g/(2k_BT)}$
$e^{(2\mu-E_g)/k_BT}= \left( \frac{m_h^*}{m_e^*}\right)^{3/2}$
$\mu = \frac{E_g}{2}+\frac{3}{4}k_BT\ln \frac{m_h^*}{m_e^*}$
At $T\approx 300$, $k_BT \approx 25.7meV$.

3. The attempt at a solution
For the first part I get the correct answer just plugging in the numbers, with $E_g = E_c-E_v = 0.5$eV and then shifting the energy by $E_V$ and noting that $m_h/m_e=a/b$
$\mu = 5.5+0.25+25.7/1000 \cdot ln(5/3) \approx 5.76eV$.

For the second part we can rewrite
$n=\frac{1}{\sqrt{2}}\left( \frac{k_BT}{2\pi \sqrt{ab}}\right)^{3/2}e^{-E_g/(2k_BT)}$
plugging in the numbers here we get $n \approx 1.45\cdot 10^{-9} Å^{-3}= 1.45\cdot 10^{21} m^{-3}$. However the answer claims I should get $3.9\cdot 10^{23}m^{-3}$.

Any obvious error I'm doing or is the answer incorrect here? The factor in front can change a bit from simply changing $k_BT$ and the answer may be by a factor 2 as well if they mean $n+p$ but the factor of 100 is troubling.

Here is the numerical calculation in matlab
Code (Text):

eg=0.5;
kbt = 25.7e-3;
sqab=sqrt(15);
c=1e30*(kbt/(2*pi*sqab))^(3/2)*exp(-eg/kbt/2)/sqrt(2)

2. Mar 12, 2017

### kuruman

Where did the "2 π" in the denominator come from? I got just π. This will change the value from 1.45 to about 4 but I agree with you that the power of 10 is +21 given your numbers.

3. Mar 12, 2017

### Incand

It's from the 2 in the effective masses $\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}(m_e^*m_h^*)^{3/4} =\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}\left(\frac{\hbar^2 \hbar^2}{(2a)(2b)}\right)^{3/4}=\left( \frac{k_BT}{\pi \hbar^2}\right)^{3/2}\left(\frac{\hbar^2}{2\sqrt{ab}}\right)^{3/2}$.
Anyway I take your answer as confirmation that I'm using the correct method so despite the answer being different I think it can be considered solved! Thanks for looking it over!