Average Kinetic Energy of Electron in the Conduction Band

[Teymur]

Homework Statement:

Show that:
$$<\:K.E.>\:=E_c+3/2\:k_B\:T$$

Relevant Equations:

$$<\:K.E.>\:=\frac{\left(total\:K.E.\right)}{\left(no.of\:electrons\right)}$$

$$<\:K.E.>\:=\:\frac{\int \:\left(E-E_c\right)g\left(E\right)f\left(E\right)dE}{\int \:g\left(E\right)f\left(E\right)dE}$$

Hello,
I've seen in a few books on solid state physics that one can deduce an expression for average K.E.:

$$<\:K.E.>\:=E_c+3/2\:k_B\:T$$

from the following:

$$<\:K.E.>\:=\:\frac{\int \:\left(E-E_c\right)g\left(E\right)f\left(E\right)dE}{\int \:g\left(E\right)f\left(E\right)dE}$$

I can't, however, find any work through of how to do so. I've had a go at the bottom part:

where $n=\int g\left(E\right)f\left(E\right)dE$ and $\int \:x^{\frac{1}{2}}exp\left(-x\right)dx=\frac{\pi \:^{\frac{1}{2}}}{2}$

and

$g\left(E\right)=\frac{\left(2m_e\right)^{\frac{3}{2}}\left(E-E_c\right)^{\frac{1}{2}}}{2\pi ^2ℏ^3}$ and $f\left(E\right)\approx exp\left(\frac{\mu -E}{k_B\:T}\right)$

to get:

$$n=2\left(\frac{m_ek_B\:T}{2\pi ℏ^2}\right)^{\frac{3}{2}}\:exp\left(\frac{\mu -E_c}{k_B\:T}\right)$$

But how does one integrate the numerator with the $\left(E\:-E_c\right)$ term and simplify to the desired result?

 

[Teymur]

p.s. I used: $x=\left(\frac{E-E_c}{k_B\:T}\right)$ for the integral: $\int \:g\left(E\right)f\left(E\right)dE \rightarrow \int \:x^{\frac{1}{2}}exp\left(-x\right)dx$

 

[TSny]

But how does one integrate the numerator with the $\left(E\:-E_c\right)$ term and simplify to the desired result?

The numerator integration is very similar to the integration in the denominator. The factor $\left(E\:-E_c\right)$ has a simple relation to $x$.

 

[Teymur]

Aha .. I'm not sure why I didn't spot that.

 

[vanhees71]

Another very important trick used in statistical physics is to calculate the denominator, the socalled "partition sum" and then take a derivative wrt. $\beta=1/(k_{\text{B}} T)$, which is an application of the celebrated Feynman-Hellmann theorem.