Electrons & Braggs Law

**heardie** · Jun4-04, 04:28 AM

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
$LaTeX Code: $2d_{hkl} \\sin \\theta = n\\lambda $<BR>$

$LaTeX Code: $d_{hkl} = \\frac{a}{{\\sqrt {h^2 + k^2 + l^2 } }}$<BR>$

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
$LaTeX Code: $E = \\frac{{\\hbar ^2 }}{{2m}}\\left| k \\right|^2 {\\rm{, where }}k = \\frac{{2\\pi }}{\\lambda }$<BR>$

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??

Thanks in advance

**Gokul43201** · Jun7-04, 03:50 PM

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.

**Gokul43201** · Jun7-04, 04:11 PM

I think you want to say that $LaTeX Code: sin(\\theta) <= 1, so<BR>\\lambda <= d$

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

**heardie** · Jun9-04, 07:41 AM

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!

**clint** · Jun14-04, 12:00 PM

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.

**heardie** · Jun14-04, 09:55 PM

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.

**clint** · Jun15-04, 07:58 AM

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.

**broomfieldjay** · Aug4-04, 02:49 PM

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

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

G =

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,
$LaTeX Code: {G^2} = b^2_1$

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

For the [110] direction,

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

and the energy is

E = 4 $LaTeX Code: {hbar^2}{pi^2}/{a^2}$ 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.