Question

The Fermi energy for copper is 7.00 eV. For copper at 1000 K, (a) find the energy of the energy level whose probability of being occupied by an electron is 0.900. For this energy, evaluate (b) the density of states and (c) the density of occupied states

Answer 1

By definition of Fermi energy,
7.00eV is the energy of the energy level whose probability of being occupied is 1/2.

Answer 2

Need to find the energy level whose occupancy probability is 0.9.

Can there be more than one level with the same probability of occupancy ?
How do we know that the occupancy probability function is one-to-one ?

Answer 3

I'm guessing the one-to-one ness has to do with the pauli exclusion principle, but im very rusty with these probability ideas...

Answer 4

I haven't studied this but I've heard about it and I have a vague idea of what it is. So I'll talk just generally about energy states and the Pauli exclusion principle, since electrons are Fermions I'm assuming it's all related.

You can have multiple different energy states with the same energy, these are called degenerate states. p-orbitals are a simple example of this if you imagine the \(p_x\) orbital it can hold 2 electrons and the \(p_y\) orbital can hold 2 electrons but since they're both p orbitals (assuming they have the same quantum number n) then they have exactly the same energy for all 4 electrons.

So you are not excluded from having 20 electrons from having the same energy. You're just excluded from having them be in the same state. This is good news, otherwise every single atom in the universe would have trouble with being stable since they'd be somehow competing for the same energy despite being on entirely different atoms...!

Answer 5

Ohkay, so all the bound electrons with same \(n\) and \(l\) values will have the same energy

Answer 6

The function for probability of occupancy of a specific level could still be one to one as it relates to energy levels, and not to the particles. I see, okay...

Answer 7

Yeah, and even though the two electrons in any given orbital have opposite spins, there's no difference between their energy because you can look at it upside down and it's exactly the same thing by symmetry so they must be the same. However if you put an atom into a magnetic field, then this splits the energy of the electrons but yeah...

I studied mostly organic chemistry while in university so metals and band gaps and this kinda stuff wasn't my interest but maybe I'll start to look into this sorta stuff now.

Answer 8

Yeah ok I think you're right about that, this is starting to sound familiar to me based on studying statistical thermodynamics of the Boltzmann distribution. You have a bunch of different energy states with different populations each. But that's where I stop cause I don't know any of the specifics of fermi energy levels to say.

Answer 9

My textbook gave some random formulas and asked to lookup fermi-dirac statistics if interested in the derivations...

Answer 10

https://en.wikipedia.org/wiki/Fermi%E2%80%93Dirac_statistics#Fermi.E2.80.93Dirac_distribution

Answer 11

Random video seems to be explaining it, https://www.youtube.com/watch?v=1SdCMIKtXt8
it looks like the same idea as the Boltzmann distribution, there are probably better videos though haha.

Answer 12

from that wiki link

\[P(E) = \dfrac{1}{e^{(E-E_f)/kT}+1}\]

I think for part a, i simply need to plugin \(P(E) = 0.9, E_f = 7eV, T=1000\) and solve \(E\)

Answer 13

Yeah that will exactly solve it you're right.

Answer 14

that's exactly the video im looking for... thanks !

Answer 15

Yeah, in case you're curious, this is exactly the same thing as the Boltzmann distribution except a special case of only considering electrons so you end up with binomial coefficients. If you were to do it in more generality you'd end up with multinomial coefficients for the \(w_i\) and it's the same idea with Stirling's approximation and all that.

Answer 16

So these formulas are basically derived/based from thermodynamics is it ? there is another similar formula that gives the relation ship between number of particles and energy difference :

\[N_x = N_0e^{-(E_x-E_0)/kT}\]

Answer 17

No, it's not derived from thermodynamics, this is statistical mechanics and statistical mechanics allows you to derive thermodynamics. This was sorta the reason why Boltzmann's entropy formula \(S=k \ln w\) was so controversial at the time cause it assumes particles exist and was his expression from statistical considerations that gave thermodynamic properties.

Yeah, that formula there is the form of the Boltzmann distribution.

Answer 18

To be honest, @Frostbite knows waaaaaaay more than I do about this haha.

Answer 19

interesting.. statistical mechanics looks similar to the kinetic theory of gases from thermo

Answer 20

ahh alright, thermodynamics is part of statistical mechanics

Answer 21

Haven't worked with fermi-Dirac stats in some time so I need to do a recap first, can we agree that:
\[\large f_{FD}(E)=\frac{ 1 }{ \exp \left( \frac{ E-\mu }{ kT }\right)-1 }\]
where \(\mu\) is the chemical potential, if I recall, earlier also called the Fermi level?

Answer 22

Well I think we pretty much solved it, but I think now we're just sorta conceptually talking about what's going on. I dunno though haha.

Answer 23

Good good :D

Answer 24

is that -1 in the denominator a typo..

Answer 25

Nope

Answer 26

This is not the Boltzmann distribution.

Answer 27

Oh wait.. sorry yes. I meant +1

Answer 28

Else it is bose-einstein <.<

Answer 29

My bad.

Answer 30

i think il be able to work b and c, thank you both :)