Hey kids, The question I'm having trouble with (this time) is as follows: Show that the Fermi-Dirac distribution function, [tex] f_{FD}(E)=\frac{1}{e^{(\frac{E-E_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
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]
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?
Well, just use the definition of the Fermi-level... It is the the maximum energy-level at T = 0 K. Just fill up all the available energy-levels with all available electrons. The last electron is placed at the highest energylevel which is called the Fermi-level. 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 Fermi-level. Above this level there are no filled levels since the Fermi-level is the highest. It is just by QM-definition of the Fermi-level regards marlon