X-ray diffraction (1 Viewer)

[inasero]

this is basically an offshoot from the superconductor thread i posted earlier, but can someone please explain to me the essence of the Braggs' X-ray diffraction experiment and how they derived the formula n,\=dsintheta

thanks alot

 

[Halo]

Braggs' X-Ray diffraction experiment was to determine the structure of a crystal, by analysing the diffraction patterns when X-Rays bombard it.

To determine the formula (where d = interatomic spacing of the crystal), suppose two synchronised X-Ray waves are fired at the crystal, and each reflect on a consecutive 'atomic layer'. The X-Ray which travelled further in the crystal (ie. the X-Ray wave that reflected on the deeper atomic layer) would need an integral number of wavelengths to remain synchronised with the other wave when both waves exit the crystal. Using trigonometry, you can calculate the extra distance travelled by the longer wave to be 2dsin(theta). Putting this together gives you the formula n(lander) = 2dsin(theta). Looking at the book's diagram would make this much clearer.

I think this is how it goes...

[Edit: Wrong formula Quoted supplied formula]

 

[Giant Lobster]

jacaranda has a diagram that geometrically proves the result.

Im prolly as lost as u r in this part of superconductors, i tried asking my phys teacher and he was kinda lost as well but he goes theres jack all about X-ray diffraction in the hsc, cept maybe a brief brief outline of braggs experiments.

In which case i think all u have to mention is that EMR at a particular wavelength, when shone on a crystalline structure, will diffract in a way characteristic of the spacing between the lattice layers of the crystalline sample. [e.g. NaCl is widely used as an example]

 

[Giant Lobster]

forgot to mention
the formula is nlamda = 2dsintheta
careful there.

 

[inasero]

yeah it has the diagram but i dont understand it!! wahhhhhhh
also why is it 2dsintheta?

 

[Constip8edSkunk]

theta(@) is the angle between the ray and a line perpendicular 2 the layers and the hypotenus is the perpendicular distance between the layers(d), then use trig 2 find dsin@.

u'll notice that this length * 2 is the difference in the distance the 2 rays(1 'reflected' on the top layer, 1 'reflected' from the 2nd) travels as the rays r ||.

As @ is when max. constructive interference occurs, the 2 rays peaks and troughs corresponds 2 eachother. therefor teh difference btween the 2 rays is nlamda(whole wavelengths)

therefor
2dsin@ = nlamda