Estimating Resistivity of Amorphous Metal - Condensed Matter

[12x4]

Homework Statement

Estimate the resistivity of an amorphous metal whose mean free path is of the order of an atomic spacing. Compare your answer to crystalline copper.

Homework Equations

V_F = h(bar) K_F / m
V_F = l / τ
ρ = m / n e² τ

The Attempt at a Solution

I think my real issue with this question is I'm not sure what are reasonable numbers to estimate with.

We need to find τ to plug into ρ = m / n e² τ

Using V_F = l / τ we can rearrange to get τ = l / V_Ffor l:

We are told the mean free path, l, is of the order of an atomic spacing but I am not sure what is a reasonable estimation of this is?

for V_F:

V_F = h(bar) K_F / m

K_F³ = 3π²n where n = N_Aρ/atomic weight

Again I am not sure what to use as my values for ρ and atomic weight. It seems as though I am missing something about amorphous metals. I understand that they have disordered structures but how can i make an estimation for K_F with that little information.

Thanks in advance to anyone that replies and apologies if this is a bit of a basic question.

 

[mfb]

We are told the mean free path, l, is of the order of an atomic spacing but I am not sure what is a reasonable estimation of this is?

A factor of 2 does not matter here, so you can look up the size of atoms of your favorite metal and use this. Same for density and atomic weight.

In general, the rough size of atoms (order of magnitude) is something you should know.

 

[12x4]

A factor of 2 does not matter here, so you can look up the size of atoms of your favorite metal and use this. Same for density and atomic weight.

In general, the rough size of atoms (order of magnitude) is something you should know.

Thank you for your reply. I think I am still getting a bit confused as, as I understand it amorphous metals are glasses and therefore, depending on their density, can have a much larger atomic spacing?

Regardless i'll complete the question with a spacing of a few angstroms. Thank you

 

[mfb]

as I understand it amorphous metals are glasses and therefore, depending on their density, can have a much larger atomic spacing?

As I said, a factor of 2 does not matter here. A few angstrom is fine.