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Topic: x-ray diffraction  (Read 11917 times)

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Offline bogger

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x-ray diffraction
« on: September 28, 2007, 02:14:58 AM »
hi, im new here and i got a question.

in bragg's equation n*lambda=2*d*sin(theta)

how do you determine the value of n, the index of reflection?

Offline AWK

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Re: x-ray diffraction
« Reply #1 on: September 28, 2007, 02:19:39 AM »
n is any nonzero integer

Offline Alpha-Omega

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Re: x-ray diffraction
« Reply #2 on: January 04, 2008, 08:55:01 PM »
In the Bragg Equation n denotes the order of reflection.  It is indeed a whole number integer. One wavelength may have 1st, 2nd, and 3rd orders of reflection thru different angles (theta 1, theta 2, theta 3).

The total reflection is made up of the various reflection orders (1, 2, ... n). The
higher the reflection order, the lower the intensity of the reflected proportion of radiation.

How great the maximum detectable order is depends on the wavelength, the type of crystal used and the angular range of the spectrometer.

From Bragg's equation that the product of reflection order n = 1, 2, ... and wavelength lambda'
for greater orders, and shorter wavelengths lambda *< lambda" that satisfy the condition lambda* = lambda/n', give the same result.

Accordingly, radiation with one half, one third, one quarter etc. of the appropriate wavelength (using the same type of crystal) is reflected through an identical angle "theta":
1 lambda = 2(lambda/2) = 3(lambda/3) = 4(lambda/4) = ...


lambda is the wavelength, d is the lattice plane distance and theta is half the diffraction angle. This relation is used for the structure analysis of crystals.

The easiest way to understand where n comes from is to derive Bragg's Law (attached diagrams 1 and 2):

Figure 1: 

Bragg's Law can easily be derived by considering the conditions necessary to make the phases of the beams coincide when the incident angle equals and reflecting angle.

The rays of the incident beam are always in phase and parallel up to the point at which the top beam strikes the top layer at atom z .

The second beam continues to the next layer where it is scattered by atom B. The second beam must travel the extra distance AB + BC if the two beams are to continue traveling adjacent and parallel.

This extra distance must be an integral (n) multiple of the wavelength (lambda) for the phases of the two beams to be the same:

n lambda = AB +BC eq. (2).

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