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06-26-2012   #1 (permalink)
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Default Material Science!

Bragg's Law and stuff. i do not know the value for a (edge length of a crystal structure) for a cubic BaTiO3 and the only thing given is Lambda and Teta. it is also stated that it was done using Cu K-alpha-x rays :/
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06-26-2012   #2 (permalink)
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I don't study material science, so I am useless.

Okay, I did some researching on what this is for the past 30 minutes:

Bragg's Law
n * lambda = 2d sin theta
d = (2sine theta)/(n*lambda)
Do you know what n is suppose to mean?
n is suppose to be the order? And usually it is suppose to be 1?

I looked up the copper k-alpha x rays and it's wavelength is about 1.5418..
I don't know if that is helpful or even related to this problem...

Uh..

Er.. I don't know... sorry... hopefully somebody who studies this can help you way more than I can.

I should stop trying to answer questions that I don't know.
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06-26-2012   #3 (permalink)
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Default

Quote:
Originally Posted by iFarted
I don't study material science, so I am useless.

Okay, I did some researching on what this is for the past 30 minutes:

Bragg's Law
n * lambda = 2d sin theta
d = (2sine theta)/(n*lambda)
Do you know what n is suppose to mean?
n is suppose to be the order? And usually it is suppose to be 1?

I looked up the copper k-alpha x rays and it's wavelength is about 1.5418..
I don't know if that is helpful or even related to this problem...

Uh..

Er.. I don't know... sorry... hopefully somebody who studies this can help you way more than I can.

I should stop trying to answer questions that I don't know.
I believe you are on the right track.
The only thing useful about k-alpha xrays is the 1.5418 (nm?) wavelength which is lambda. I assume these xrays have a very well known wavelength. This would be analogous to mercury vapour that has a number of very discrete wavelengths that it emits.
Many objects do not emit wavelengths of light like this, they'd emit all wavelengths instead of a select few.
I would let n=1
Bragg's law is based on when constructive intereference occurs. You are shining some EM radiation and it is bouncing off like a mirror, however some will penetrate the first layer of the lattice. The light bouncing from two points of the lattice will then interfere either constructively or destructively.
From what I can tell n has to do with what layer the em radiation is bouncing off. If it is bouncing off 1 and 2 n=1, if it is bouncing off 1 and 3 n=2. Do not quote me on this.



Check your answer here.
Bragg's Law
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06-26-2012   #4 (permalink)
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Default thanks!

i'm done! i have two answers though, one using summation of atomic radii and the other using density and manipulating its units to get a side length for the fcc latice :> thanks for the help last night LadyEl and thanks for the help too iFarted ^^ hopefully i have done this right :>
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