# Electrons & Braggs Law

1. Jun 4, 2004

### heardie

Hi - come across a problem I am unsure if I am doing the right way.

At what energy is Bragg's Law satisfied to first order in Cu given that the lattice parameter is a=0.361nm

i) Electrons propogate in the [100] direction
ii) [110] direction

My basic step was
$$2d_{hkl} \sin \theta = n\lambda$$

$$d_{hkl} = \frac{a}{{\sqrt {h^2 + k^2 + l^2 } }}$$

I was unsure what to do with theta, however I seem to recall coming across a similar problem, where we assumed electon-lattic interctions caused a reflection, so theta=90. However this seems a little hazy to me.

So upon finding the wavelength I then used
$$E = \frac{{\hbar ^2 }}{{2m}}\left| k \right|^2 {\rm{, where }}k = \frac{{2\pi }}{\lambda }$$

I think I got about 2 ev. Is this the right approach?

And finally...comparing this to the Fermi energy of Cu (7ev), "comment on the significant of this comparision with respect to teh assumption of free electrons in Cu". Since E < E[f], do I conclude this is a good assumption??

2. Jun 7, 2004

### Gokul43201

Staff Emeritus
This is a little different from the way I'm familiar with using Bragg's Law. If you are doing X-ray or electron diffraction off a crystalline solid, the ratios of the sines of the peak angles tell you what the unit cell structure is (bcc, fcc, etc.), and the values of the angles tell you the plane spacing (or lattice parameter).

Have not thought about how to use Bragg's Law to deal with electronic transport within the lattice. Clearly, as you've indicated, "what is theta ?" is a good question that need resolving. Hmmm...will think about this.

Last edited: Jun 9, 2004
3. Jun 7, 2004

### Gokul43201

Staff Emeritus
I think you want to say that $$sin(\theta) <= 1, so \lambda <= d$$

This cannot be wrong...but may not be sufficient.

4. Jun 9, 2004

### heardie

If electrons are propogating in the [100] direction they are travelling perpendicular to a plane though the x-axis. Thus when they interact with that plane, they are reflected at 90 deg. What about all other planes? this is bothering me!

5. Jun 14, 2004

### clint

Using a=0.361nm and theta=90 (i.e. normal incidence and 180 change in direction or back scattering) gives Lamda=a, and Bragg's law is satisfied for Ea=11.5eV (23eV) for i (ii). Try your numbers again, and see if you agree.

So this is the opposite. Ea>Ef=7eV, which makes Cu free electron like, and is thus a good assumption. If Ea=Ef, then one has the extreme case of an insulator, ie NOT a free electron. Internal electrons keep getting backscattered because of the lattice periodicity, and therefore do not move, hence an insulator.

6. Jun 14, 2004

### heardie

Well I have the 'official' answersr now: 2.89eV and 1.45 eV

In both cases the opening of the enegry gap due to the lattic interaction or Bragg reflection, is well away from the Cu Fermi energy of 7eV. The condunction band of Cu is always partially filled, and Cu maintains its metallica nature.

7. Jun 15, 2004

### clint

Right, the Bragg law is

2*d_hkl*sin(theta) = n*Lamda

I was using the what you wrote without the factor 2. Still, as long as the energy isn't in the vicinty of Ef, free electron model is fine I think.

Last edited: Jun 15, 2004
8. Aug 4, 2004

### broomfieldjay

Use a different form of Bragg's law

Hopefully, I won't mess up the Latex typsetting too much, I'll try to come back and fix it later.

a = .361 nm
Braggs law using the Reciprocal lattice vector G is

k = (1/2)G

Then the energy is

E = $${hbar^2}{k^2}$$/(2m)
Ignore the first "/" in this equation. I can't edit it out!

$${k^2}= (1/4){G^2}$$
which is needed in the equation for the energy that was given in a post above.

G = $$h b _1 + kb_2 + l b_3$$
where h,k,l are the miller indices and the b's are the primitive lattice vectors for the fcc recpirical lattice.

So for the first part in the [100] direction,
$${G^2} = b^2_1$$

The result for the energy is
E = 3 $${hbar^2}{pi^2}$$/$$({a^2}{2m})$$
= 3 * 2.89 eV = 8.6 eV

For the [110] direction,

$$G = b_1 + b_2$$

$${G^2} = b^2_2 + 2b_1 b_2 + b^2_1$$

and the energy is

E = 4 $${hbar^2}{pi^2}/{a^2}$$2m

= 4 * 2.89 eV
= 11.5 eV

The fermi energy for Copper is 7 eV. The fermi energy is the energy of the uppermost electrons. Since this energy is below the top of the bands (which are at the Bragg planes) the band is only partially filled and Copper is a metal. In previous posts, the assumption was made that theta = 90, this assumption cannot be made.

According to the nearly free electron model, the major effect of the lattice is at the Bagg planes. Since Copper's fermi energy is well below the Bragg plane energy, the free electron model works.

Last edited: Aug 5, 2004