2
$\begingroup$

My understanding is that given the DOS of a material we find the fermi level by filling electrons into those energy levels and when we run out of electrons we reach close to the fermi level (or exactly the fermi energy if $T = 0K$)

So theoretically, given the DOS of a material the fermi level $E_f$ is always implicit in it.

Now coming specifically to the problem, given below are the DOS of Pd and PdH

enter image description here

As can be clearly seen, the fermi level when hydrogen is introduced increases and shifts to the right. This is being reasoned in the following ways:

In Pd, although the electronic configuration of the atomic Pd is [Kr]4d105s0, the 4d band of fcc Pd is not filled and the Fermi level is located at the 4d band due to an overlap of 4d and 5s bands (Figure 5a).[78,79] In the  phase, a Pd–H chemical bond is formed, resulting in the formation of bonding and antibonding bands below and above the 4d band (Figure 5b).[80] As a result, the 4d band is filled and the Fermi level shifts to the 5s band, which has been investigated not only theoretically

also by,

Palladium hydride therefore differs from palladium by the strong perturbing influence of hydrogen in the octahedral interstice in the lattice. This hydrogen is strongly attractive for states (which may or may not be filled) having s-like character about this site ; the character (or number) of states about the palladium is of minor importance except as far as they were originally filled or empty. There exists strong hydrogen-hydrogen interactions which can lower states sufficiently to be filled. The filling of these states as well as states characteristic of the "host" palladium raises the Fermi level to a region of low density of states

But I am having a hard time grasping how having more states in lower energies are going to increase the fermi level, shouldnt a high DOS in the lower energies imply that they get filled first and as a result the fermi level decreases ? Why is it being implied that the low hydrogen states ensure that the 4d palladium bands get filled ? An energy level gets filled if starting from the lowest energy if one were to put electrons one by one and upon reaching the said state there actually are electrons left to be filled. Also this whole idea of lowering energy seems confusing because according to band theory and MO, bands combine to create lower (bonding) and higher (anti bonding) states, so what precisely does lowering of energy states by hydrogen mean and what is the guarantee that these lowered states get filled ?

$\endgroup$
4
  • $\begingroup$ Number of available states matter, but so do number of electrons in the unit cell to fill the states. I think they are trying to say that the new lower band is basically filled by the Hydrogen electron already, so doesn't drop the Fermi level. And the (formerly) PD 4d and 5s bands are now not so hybridized anymore so the 4d gets filled and the 5s partially filled. Of course this is all a little "hand-wavey" since band structure is a hand-wavey concept that strictly only applies to independent particles (whereas in real life electrons interact with each other)... $\endgroup$
    – hft
    Commented Apr 4 at 18:51
  • $\begingroup$ I have been wondering that as well, the more atoms you add the more states you create but you also have more electrons to fill, so if no band restructuring were to happen the fermi level would remain unchanged. But what makes the 4d and 5s lose the hybridization ? $\endgroup$ Commented Apr 4 at 19:02
  • $\begingroup$ Two additional things besides my answer: 1) there's likely some post-hoc analysis done by the experimentalists and theorists to get to those statements. I wouldn't give them blind trust. 2) Go to those references to see if those statements can be justified. $\endgroup$
    – Dr. Nate
    Commented Apr 4 at 19:51
  • $\begingroup$ @AjaykrishnanR "But what makes the 4d and 5s lose the hybridization?" I see the opposite. In pure Pd, I see the d-band (I even see 5 bumps) and the s-band clearly. Adding hydrogen makes the s-band less distinct, which means more hybridized. $\endgroup$
    – Dr. Nate
    Commented Apr 4 at 19:55

1 Answer 1

0
$\begingroup$

Start with pure palladium and integrate over the DOS below the Fermi energy. You should get the number of non-core electrons from your atoms.

In the next step, you add hydrogen. New states appear and bands change, causing the DOS to change shape. But when you integrate, you will get the correct new number of electrons for the system.

As another way to look at it, you could look at the equation for a free electron gas: \begin{equation} E_F\propto\left(\frac{N}{V}\right)^{3/2}.\end{equation} Hydrogen will go interstitial into metallic lattices with little perturbation to the volume, so adding electrons is going to increase the Fermi energy.

$\endgroup$
4
  • $\begingroup$ So for hydrogen to be increasing the number of "free" electrons it has to essentially act like a donor because isnt N the number of free electrons ? But the bonding electrons arent really free right ? $\endgroup$ Commented Apr 4 at 19:55
  • $\begingroup$ 1) I think you might be importing some concepts from chemistry that don't apply here. All the electrons contribute to bonding. 2) I'm not going to use some of your terminology because it might be confusing, but, yes, $N$ is the number of non-core electrons from the atoms. ( FYI: DOS diagrams don't show core electrons and their analysis ignores them because those electrons are hundreds of eV below the $E_F$.) $\endgroup$
    – Dr. Nate
    Commented Apr 4 at 20:06
  • $\begingroup$ I think what i meant was that an electron can be non core but still not be free in which case it wouldnt increase N. By core electrons i guess you are referring to inner shell electrons that are for the most part localized to the atoms themselves. $\endgroup$ Commented Apr 4 at 20:16
  • $\begingroup$ Yes, that's what I mean by core electrons. There are (a) core electrons and (b) the electrons in $N$/the DOS we talk about. $\endgroup$
    – Dr. Nate
    Commented Apr 4 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.