How Do You Calculate the Cube Edge of Iron Using Bragg's Law?

[PsychonautQQ]

Homework Statement

The bragg angle for reflection from the (110) planes in bcc iron is 22 degree's for an x-ray of wavelength = 1.54 angstroms. Compute the cube edge for iron.

Homework Equations

2dsinθ=nλ
?

The Attempt at a Solution

I feel woefully underprepared for this problem and yet here it sits, standing between this waking world and the looming allure of my pillow and soft soft mattress.
Surely I can abandon the need to figure out this problem and simply accept the point deduction on this assignment in-turn for the more base human need that is adequate sleep. Yet here I sit on the physics forums:
hoping for a cyber scholar to liberate me from my self imposed archaic repression;
praying for a student of both physicality and empathy to cross his eyes on my predicament and go David Blain on these chains that consume me, while being ignorant of the fact I have no idea what these things are --> : ;
But it matters not. For we are not students of language, we long for something more pure, more real, more eternal. We feel the vastness of the universe and it's potential to crush our identities, yet this same inner knowing fuels our need to find something real, to verify that nature plays by consistent rules and that madness is not a product of something more true than our daily lives. And so we stand in awe of this beautiful symmetry that nature has provided us, unraveling a tapestry of exponential complexity, bound by a paradoxical faith that we are truly making progress, and that there really is one equation for everything, and that we humans, have the ability to look inside the mind of God.

But let's not start throwing around words like God, which is so often drownt in semantics these days. The point is that here we are, rolling: discipline, passion, desperation, and a deeper unexplainable need to find this thing that we know is real, and through our studies our soul and mind is crafted into something more real, and our capacity to love grows. That's been my experience at least.
Could it be? The highest of all sciences a bonified path to the divine human? From a certain angle it's obvious, but from another angle it appears i'll never outgrow my big ideas. That's alright, maybe I can grow into them instead.

anyway... I guess this is a fancy way of saying I have no idea what I'm doing on this problem.. Compute the cube edge? Does that mean I'm looking for a plane that describes the edge of a cube? I'm assuming I use bragg's law, but I'm not sure how to work the miller indices (110) into anything useful. I've been trying to relate miller indices to spacing between planes but I realize miller indices just give directions and such knowledge is not purgable through such means?

Any tips appreciated

 

[mark726]

.

First of all, I want to commend you for your beautiful and poetic reflections on the deeper meaning and potential of science. It is truly inspiring to see someone who is not only interested in solving problems, but also in understanding the bigger picture and the impact our studies have on our lives.

Now, onto the problem at hand. You are correct in thinking that you need to use Bragg's law to solve this problem. However, you are also right in that the Miller indices (110) do not directly relate to the spacing between planes. Instead, they represent the direction of the planes in the crystal structure.

To solve this problem, you will need to use the equation 2dsinθ=nλ, where d is the spacing between the planes, θ is the Bragg angle, n is the order of reflection (usually 1), and λ is the wavelength of the x-ray. You already have all the values except for d, which is what you need to find.

To find d, you can rearrange the equation to d=λ/(2sinθ). However, since you are dealing with a cubic crystal structure, the spacing between the (110) planes is equal to the length of the cube edge, which we'll call a. So, you can rewrite the equation as a=λ/(2sinθ).

Now, you have all the values you need to solve for a. Just remember to convert the wavelength from angstroms to meters before plugging it into the equation. I hope this helps and good luck with your studies!