
#1
Oct2104, 03:09 AM

P: 161

Hey kids,
The question I'm having trouble with (this time) is as follows: Show that the FermiDirac distribution function, [tex] f_{FD}(E)=\frac{1}{e^{(\frac{EE_f}{kT})}+1} [/tex] Has the following functional form at T= 0K (see attachment) Now, the first thing that screamed at me was the division by T in the exponential bit. If T=0, what is going on!? The obvious things are: E>Ef then f(E) = 0 and E<Ef then f(E) = 1. I'm just really confused at how I can show that the function has that form at T=0K Any ideas? Cheers 



#2
Oct2104, 03:54 AM

Sci Advisor
HW Helper
P: 2,004

I don't understand your question. Didn't you just show the function has that form?
[tex]\lim_{T \rightarrow 0}f_{FD}(E)=\left\{ \begin{array}{ll}1 & \mbox{if} E<E_f\\ \frac{1}{2} & \mbox{if} E=E_f \\0 & \mbox{if} E>E_f[/tex] 



#3
Oct2204, 08:29 AM

P: 161

I'm glad somebody else doesn't understand the question either.
They want me to 'show' that the distribution has the (attached pic) form at T=0. The real problem I'm having is how do I "show" that it has that form? Via two lines of maths? That's it? 



#4
Oct2204, 08:36 AM

Sci Advisor
HW Helper
P: 2,004

FermiDirac Statistics
You HAVE just shown it. By taking the limits.
So yeah, that's it. :) 



#5
Oct2204, 08:36 AM

P: 4,008

Well, just use the definition of the Fermilevel... It is the the maximum energylevel at T = 0 K. Just fill up all the available energylevels with all available electrons. The last electron is placed at the highest energylevel which is called the Fermilevel. Ofcourse all this is done at zero Kelvin. That is why the distribution function has the drawn form. All levels are filled up (probability one) till the Fermilevel. Above this level there are no filled levels since the Fermilevel is the highest. It is just by QMdefinition of the Fermilevel
regards marlon 


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