Determining Energies of States Close to Fermi Level

[nkk2008]

Homework Statement

In class, we discussed metal quantum dots, and we argued that they’re not so useful for
device applications since the separation between energy levels for energies near the
Fermi energy is much less than a typical optical photon energy of 2 eV. Let’s consider
that argument in a bit more detail here: Treating the metal in a free-electron model, then
in one dimension we can write the energies as $E_n = \hbar^2 \pi^2 n^2 / 2mL^2$, which gives an energy spacing of $\hbar^2 \pi^2 / smL^2$ times an integer, or just $\hbar^2 / 2m$ times an
integer for L approx. 3nm. As discussed in class, this equals 0.04 eV times the integer. The
point of this problem is to consider how large the integer multiplier is, for two situations:

(a) still working in one dimension, consider an energy close to the Fermi energy of about 8
eV. How large is the energy difference between the states immediately above and below
the Fermi energy? Is this difference much less than a typical optical photon energy?

(b) now switch to three dimensions, in which case the expression for the energies
contains three integers (as discussed in class). Now what is the energy difference between
the states immediately above and below the Fermi energy? Again, is this difference much
less than a typical optical photon energy?

Homework Equations

So from class and other problems in the homework I have:
$E_n = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)$
Although it is unlcear to me that this is relevant

The Attempt at a Solution

My main problem here is I have no clue how to find this integer. I feel like it is easy, but somehow evades me. I have tried:

• Just calculating the difference from this integer being 1 in the lower state to 2 in the higher state. This seems like a cop out and just wrong. For the 3D part, I went from (1,1,1) to (1,1,2).
• I looked online and found various resources that hinted at this integer not being related to the quantum numbers at all (these were all qualitative descriptions, no quantitative stuff). This just confused me further and is possibly wrong.

Thanks,
nkk

 

[mark726]

Dear nkk,

Thank you for your question. The integer multiplier in the energy equation is related to the quantum number n, which represents the energy level. In one dimension, the energy spacing is given by \hbar^2 \pi^2 / 2mL^2, where n is the integer multiplier. In three dimensions, the energy spacing is given by \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2), where n_x, n_y, and n_z are the integers representing the energy levels in the x, y, and z directions respectively.

To answer your questions:

(a) For an energy close to the Fermi energy of 8 eV, the energy difference between the states immediately above and below the Fermi energy would be 0.04 eV, which is much less than a typical optical photon energy of 2 eV.

(b) In three dimensions, the energy difference between the states immediately above and below the Fermi energy would be 0.12 eV, which is still much less than a typical optical photon energy of 2 eV.

I hope this helps clarify the integer multiplier in the energy equation and how it relates to the energy levels and energy differences. If you have any further questions, please don't hesitate to ask.