Average Kinetic Energy of Electron in the Conduction Band

In summary: This sums up all the contributions to the average KE from particles with energies between ##E_c## and ##E##.
  • #1
Teymur
16
3
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?
 
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  • #2
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##
 
  • #3
Teymur said:
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##.
 
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  • #4
Aha .. I'm not sure why I didn't spot that.
 
  • #5
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.
 
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