Bragg Scattering Angle for 63eV Electrons on Ni Crystal

[curto]

Homework Statement

What is the Bragg Scattering angle (i.e. the angle, in degrees, between the incoming incident beam and the reflected beam) for electrons scattered from a nickel crystal if their energy is 63 eV? The spacing between Ni Bragg planes is 0.215 nm. Give your result to 3 significant figures.

so the energy of the electrons is 63eV and d = 0.215nm

Homework Equations

n$$\lambda$$ = 2d sin $$\theta$$
E = hc/$$\lambda$$

The Attempt at a Solution

a condition for bragg diffraction is that $$\lambda$$ $$\leq$$ 2d so when you use E = hc/$$\lambda$$ to find the wavelength of the electrons you get ~19.72nm which is much greater than 2d. also the question doesn't say anything about an angle. am i missing something here?? not only will this not work with that wavelength being so much bigger than 2d but there's no information about an angle?

 

[JesseC]

What does it mean the energy of the electrons is 63eV? Presumably this is kinetic energy?

E = hc/λ also doesn't apply to electrons (it applies to EM radiation), you want to use the De Broglie wave relation for a 'matter wave':

λ=h/p

where p is the momentum of the electron. Given that this electron is not moving anywhere near the speed of light, you can work out its momentum from

E = p^2 / (2m)

Don't forget to be careful converting electron volts to joules and back etc. I find it to have a wavelength λ = 1.5 Angstrom which is of the order of the atomic spacing in Ni.

If you drew yourself a picture of the situation maybe you could see where the angle θ comes into it.

 

[curto]

alright well that makes sense lol thanks heaps. what i don't understand now is this stuff about "order". from what i initially though, the integer "n" in n$$\lambda$$ = 2d sin $$\theta$$ was how deep the light beam goes into the crystal ie order 1 was just the surface, order 2 was the 2nd plane, order 3 was the 3rd plane etc etc.

i don't understand what you mean by "the order of the atomic spacing in Ni". like, i can get that wavelength you got, but i don't see what it has anything to do with "order". also, i have drawn a diagram and i still don't know how you can do the question without $$\theta$$. however i had a look at the following questions in this assignment and by the look of them theyre asking the same thing but backwards ie they give you the scattering angle and they ask for the de broglie wavelength of the incident beam of particles. the only thing the questions mention about an angle, for example, is:

A beam of neutrons strikes the crystal at normal incidence and the first maximum of the diffraction pattern occurs at a scattering angle ... etc etc blah blah

so yeah, looking at that information i feel as if these questions are easy, i mean they look like it, but i still don't get how to do them. i must be missing something pretty fundamental lol

EDIT: ok i see now after looking at the eqn and the diagram a bit more that we have enough information to find $$\theta$$ and then the scattering angle and yeah. still not sure about this "order" stuff though

 

[JesseC]

I may have confused you with a different meaning of 'order'. When I say the wavelength is 'of the order' of the atomic spacing, I mean it is of a 'similar length scale', or mathematically λ is approximately equal to d.

This turn of phrase is used all over physics, for example I might say 'the energy of a photon is of the order of an atomic electronic transition'. Which is the same as saying, the energy of a photon is approximately the same as the energy required to excite an electron in an atom! This has nothing to do with order of Bragg diffraction.

Perhaps your second confusion comes from the fact that the question is actually asking to find θ! θ is the bragg scattering angle.

If you rearrange the bragg formula in terms of θ you should get:

θ = arcsin(nλ/2d)

You have already calculated λ, and d is given to you. It may so happen that there is more than one bragg scattering angle, you can see how many there will be by plugging in numbers 1, 2, 3, etc for the integer n. So long as nλ/2d < 1, there will be a bragg scattering angle.

 

[curto]

ohhh i get you lol

alright i pretty much understand everything now thanks heaps. what I am having problems with now is, well first of all you mentioned that "θ is the bragg scattering angle" but i thought $$\theta$$ was as shown in the following:

and so the scattering angle ie as stated in the question is "the angle, in degrees, between the incoming incident beam and the reflected beam" which by looking at that image we can see that this angle would be equal to 180 - 2$$\theta$$ in degrees right?

next thing is, i can get two angles for $$\theta$$ using this relationship nλ/2d < 1. one angle at 21 degrees and the other at 46 degrees. what $$\theta$$ am i supposed to use then to answer the question ?? i read through the lecture notes some more and i came across something to do with "order overlapping" or something? not sure how to continue if I've potentially got two answers to the question haha

 

[JesseC]

I'm pretty sure the Bragg scattering angle θ is normally defined as the angle between the 'lattice plane' and the incoming ray. So in your diagram the lattice plane would be the horizontal. The whole point of Bragg diffraction is that at certain values of θ or 2θ or 180 - 2θ or whatever (thats not particularly important -> except for exams!) let's call it 'some angle', you'll get your electrons/x-rays/neutrons constructively interfering, and this will appear as bright spots on a screen.

You can end up with more than one bright spot, at more than one angle! So both answers are right, one is the n = 1 diffraction, the second n = 2. Think about what would happen if you use higher energy electrons? Higher energy -> smaller wavelength -> more values of nλ/2d < 1 for different n -> more bright spots on your screen.

As I think you said before, from experiments you actually work backwards from an angular distribution of bright spots and infer the crystal structure. This is a very widely used and extremely important physical technique for determining structures of materials!

 

[curto]

ahhh ok I am pretty sure i get what you mean. sorry to keep on asking questions, but from looking at that diagram and the ones in my lecture notes i can't see how θ is going to change at all. i mean every beam looks pretty parallel and there's only perpendicular intersections etc etc so for example looking at that diagram, if i find θ for n=1 to be 21 degrees then that's going to be referring to all the θ's in the diagram. so then if i find for n=2 for θ that its 46 degrees, where on the diagram would i be looking? I am trying to visualise what's goin on here. i get how the dots are being made, from each beam intersecting with each other causing the bright spot and since everything is at right angles and parallel that's why all the dots are arranged so nicely. but yeah where abouts am i looking when θ changes?

 

[JesseC]

That picture is too simplistic, this is a better one :

http://upload.wikimedia.org/wikipedia/commons/0/0c/Diffusion_rayleigh_et_diffraction.png

When the atoms are hit by X-rays/electrons they scatter them in all directions, not just one as your diagram suggests -> think of an expanding sphere coming off each atom. The expanding spheres (spherical waves) from different atoms interact constructively/destructively giving rise to a pattern of bright/dark spots.

Frankly we're heading past territory I'm familiar with, I'm certainly no expert! I understand that if you were to do an experiment, say shoot some electrons at a crystal and look at the scattering pattern, I don't think you physically change θ in the experiment, you measure θ off the screen where the pattern is formed. So long as you know how far away your crystal is from the screen this should be relatively easy.

What your original question was asking you about is LEED (low energy electron diffraction), if you're interested you should read more about it!

 

[curto]

ahh ok thanks a lot mate, that's explained a fair bit. i wish i knew about this site when i was in high school lol, its pretty damn useful. cheers