Finding angles with Bragg's Law

[baseballfan_ny]

Homework Statement

A beam of electrons of mass m and speed v is incident on the surface of a 2-dimensional square crystal of lattice spacing a. It makes an angle $\theta = 45^{o}$ with the surface. The Bragg diffracted beam leaves at an angle of $\theta^{'}$ with the surface, as shown.

(a) When $\theta^{'} = 45^{o}$, what is the smallest possible value of v?
(b) [This question requires some real physical thought. It is not in any way a plug-and-chug question.] Give two other possible values for the angle $\theta^{'}$, and give the smallest possible value of v for each of these.

Relevant Equations

$\theta_{incident} = \theta_{reflected}$
$n\lambda$ = 2d\sin(\theta)##
$p = \frac {h} {\lambda}$

Part (a)
Ok so for (a) $\theta_{incident} = \theta_{reflected}$, so I assume I could just consider the horizontal planes in these atoms.

$n\lambda = 2a\sin(\theta)$
$p = \frac {h} {\lambda}$

$\frac {nh} {2amv} = \sin(\theta)$
$v = \frac {nh} {2am\sin(\theta)}$

I suppose the smallest value of v would be when n = 1, so $v = \frac {h} {2am\sin(\theta)}$

Part (b)
Ok so here $\theta_{incident} \neq \theta_{reflected}$, so I suppose I need to look at a different plane. Not sure if this is the right approach, but I took an arbitrary plane at an arbitrary angle $\alpha$ from the horizontal like so:

I've shown the original beams at angles $\theta$ and $\theta^{'}$ from the horizontal, the arbitrary plane is at an angle $\alpha$ from the horizontal. $\beta_{i}$ and $\beta_{r}$ are the angles of the beams with respect to the arbitrary plane.

So for this arbitrary plane, we need $\beta_{i} = \beta_{f}$
So $\theta + \alpha = \theta^{'} - \alpha$

So I've got this with
$n\lambda = 2a\sin(\theta)$
$p = \frac {h} {\lambda}$

Not sure I have enough equations because I think the most I can do here is
$\frac {nh} {mv} = 2a\sin(\theta + \alpha) = 2a\sin(\theta^{'} - \alpha)$

And I don't think I have enough equations to solve for $\theta^{'}$ or $\alpha.$ Or I took the wrong approach?

Thanks in advance.

 

[haruspex]

Don't you need to choose a plane that has some relationship to the lattice, like two nodes at a knight's move apart?

 

[baseballfan_ny]

Don't you need to choose a plane that has some relationship to the lattice, like two nodes at a knight's move apart?

Thanks for the reply.

Maybe? Not sure I completely understand the analogy (idk chess by the way if this is a chess thing).

" Two cells, not in the same row or column, are said to be diagonally adjacent when their distance apart is 2 and a knight's move apart when it is 3. "
https://core.ac.uk/download/pdf/82762799.pdf

Two nodes at a knight's move apart would be those with a distance of 3 from each other? Is this a diagonal distance?

 

[haruspex]

Thanks for the reply.

Maybe? Not sure I completely understand the analogy (idk chess by the way if this is a chess thing).

" Two cells, not in the same row or column, are said to be diagonally adjacent when their distance apart is 2 and a knight's move apart when it is 3. "
https://core.ac.uk/download/pdf/82762799.pdf

Two nodes at a knight's move apart would be those with a distance of 3 from each other? Is this a diagonal distance?

A knight's move is one step one way and two steps perpendicularly, for a distance of √5. More simply, you could have one and one, for a distance of √2.

 

[baseballfan_ny]

A knight's move is one step one way and two steps perpendicularly, for a distance of √5. More simply, you could have one and one, for a distance of √2.

Ok, that makes sense.

Don't you need to choose a plane that has some relationship to the lattice, like two nodes at a knight's move apart?

Ok, so what I'm getting it here is that I shouldn't be making $\alpha$, the angle between this "new plane" and the horizontal plane arbitrary. I suppose -- now looking at the wording of the question again and seeing it asks for two "possible" values of $\theta^{'}$ -- I could choose any possible value for this angle $\alpha$ that relates to the lattice and solve for $\theta^{'}$.

Not sure if I'm jumping the gun here, but here's my attempt for fixing $\alpha$ to 45 degrees and 63.435 degrees, which should correspond to a distance of $\sqrt2$ and $\sqrt5$ respectively.

Here's my interpretation of fixing it to 45 degrees:

So now using the condition
$\theta_{incident} = \theta_{reflected}$
$\theta + 45 = \theta^{'} - 45$
$\theta^{'} = \theta + 90 = 130$

And then doing the same thing for when I fix the angle between the two planes as 63.435 degrees,
$\theta^{'} = 166.87$ degrees

And I suppose finding the smallest possible value of each would just use the formula I got in part a (if that looks okay)

So for fixed angle = 45 deg and $\theta^{'}$ = 130 degrees
$v = \frac {h} {2am\sin(130)}$

So for fixed angle = 63.435 deg and $\theta^{'}$ = 166.87 degrees
$v = \frac {h} {2am\sin(166.87)}$

Is this right?

 

[haruspex]

$\theta^{'} = \theta + 90 = 130$

Check your arithmetic.
That might be valid, but it means the electron bounces straight back where it came from.
I would look for pairs of lattice points that make a shallower angle to the horizontal.

 

[baseballfan_ny]

Check your arithmetic.
That might be valid, but it means the electron bounces straight back where it came from.
I would look for pairs of lattice points that make a shallower angle to the horizontal.

Oh gosh, you're right! I used $\theta$ = 40 for some reason instead of $\theta$ = 45. That's what I get for staying up too late :)

Anyways that should mean $\theta^{'}$ = 135, which would bounce straight where it came from. I suppose, as you suggested, 45 degrees is too large.

Here's my attempt for some shallower angles:

The first angle I took is 26.565 degrees below the horizontal (two nodes left and one down)

$\theta$ + 26.565 = $\theta^{'}$ - 26.565
$\theta^{'}$ = 45 + 2(26.565) = 98.13

And then the smallest v should be $v = \frac {nh} {2a\sin(26.565)*m*\sin(98.13 - 26.565)}$

The next angle I took is 18.4349 degrees below the horizontal (three nodes left and one down)

$\theta$ + 18.4349 = $\theta^{'}$ - 18.4349
$\theta^{'}$ = 45 + 2(18.4349) = 81.8698

And then the smallest v should be $v = \frac {nh} {2a\sin(18.4349)*m*\sin(81.8698 - 18.4349)}$

I also noticed that in my last post when solving for the smallest possible v, I neglected to consider that the perpendicular distance between the "new set of planes" that I've chosen would not be a, but actually $a\sin(\alpha)$ where $\alpha$ is the fixed angle. I adjusted for that in this attempt. Hopefully this looks okay.

 

[haruspex]

$n\lambda = 2a\sin(\theta)$

I didn't pay enough attention to that line before. How do you get that? Think what it means when the angle is small.

 

[baseballfan_ny]

I didn't pay enough attention to that line before. How do you get that? Think what it means when the angle is small.

I got that from the Bragg condition for constructive interference -- the path difference between two consecutive beams ($2a\sin(\theta)$) should be an integer multiple of $\lambda$.

Okay so when $\theta$ is small $\sin(\theta) \approx \theta$ but I'm not sure the linear approximation is necessary here.

It could also be that $\sin(\theta) = \frac {n\lambda} {2a}$ so when the angle is really small $\sin(\theta)$ is really small and the index n is small? I guess maybe we're getting at that if the angle is large n would be large... is that why we had to choose a shallower angle?

 

[haruspex]

I got that from the Bragg condition for constructive interference -- the path difference between two consecutive beams ($2a\sin(\theta)$) should be an integer multiple of $\lambda$.

Okay so when $\theta$ is small $\sin(\theta) \approx \theta$ but I'm not sure the linear approximation is necessary here.

It could also be that $\sin(\theta) = \frac {n\lambda} {2a}$ so when the angle is really small $\sin(\theta)$ is really small and the index n is small? I guess maybe we're getting at that if the angle is large n would be large... is that why we had to choose a shallower angle?

ok - just me displaying my ignorance of the subject. Ignore post #8.

Moving on to

And then the smallest v should be
$v = \frac {nh} {2a\sin(26.565)*m*\sin(98.13 - 26.565)}$

why are you dividing by two sines? Previously you only divided by one. Shouldn't you just divide by sin(θ+α)?

 

[baseballfan_ny]

why are you dividing by two sines? Previously you only divided by one. Shouldn't you just divide by sin(θ+α)?

In Post 7 I ended up with a second sine, whereas when I did it in Post 5 I only had one sine. I realized that in the formula I obtained in part (a) from Post 1: $v = \frac {nh} {2am\sin(\theta)}$ only applies when considering horizontal planes when the perpendicular distance between planes is a. When changing sets of planes, the perpendicular distance should also change. I attached a diagram in Post 7 that shows that the perpendicular distance between a set of planes an angle $\alpha$ from the horizontal is $a\sin(\alpha)$.

In this formula: $$v = \frac {nh} {2a\sin(26.565)*m*\sin(98.13 - 26.565)}$$

the first sine is $\sin(\alpha)$, the second is, as you stated, $sin(\theta + \alpha)$ or equivalently $\sin(\theta^{'} - \alpha)$.

 

[haruspex]

In Post 7 I ended up with a second sine, whereas when I did it in Post 5 I only had one sine. I realized that in the formula I obtained in part (a) from Post 1: $v = \frac {nh} {2am\sin(\theta)}$ only applies when considering horizontal planes when the perpendicular distance between planes is a. When changing sets of planes, the perpendicular distance should also change. I attached a diagram in Post 7 that shows that the perpendicular distance between a set of planes an angle $\alpha$ from the horizontal is $a\sin(\alpha)$.

In this formula: $$v = \frac {nh} {2a\sin(26.565)*m*\sin(98.13 - 26.565)}$$

the first sine is $\sin(\alpha)$, the second is, as you stated, $sin(\theta + \alpha)$ or equivalently $\sin(\theta^{'} - \alpha)$.

I need a better diagram to get my head around this.

I have two beams reflecting off 'planes" AA' and BB'.
The two other fine dashed lines are normals to these.
The dash-dot lines are normal to the beams and reflections.
The extra path length of the right hand beam is the sum of the lengths of the two double-ended arrows, yes?

 

[baseballfan_ny]

I need a better diagram to get my head around this.
View attachment 277935
I have two beams reflecting off 'planes" AA' and BB'.
The two other fine dashed lines are normals to these.
The dash-dot lines are normal to the beams and reflections.
The extra path length of the right hand beam is the sum of the lengths of the two double-ended arrows, yes?

Before I check that, let me confirm to make sure I understand this correctly: are the fine dashed lines the ones bounded by the double arrows and are the dash-dot lines the ones that make up planes AA' and BB'? Or is it the other way around?

 

[haruspex]

Before I check that, let me confirm to make sure I understand this correctly: are the fine dashed lines the ones bounded by the double arrows and are the dash-dot lines the ones that make up planes AA' and BB'? Or is it the other way around?

Two fine dashed lines are the two planes of reflection, AA' and BB', and the other two are normals to these at the points of reflection.
The four dash-dot lines are normals to the incoming beams and their reflections.
The solid double ended arrows are distances between parallel pairs of dash-dot lines.

 

[baseballfan_ny]

The extra path length of the right hand beam is the sum of the lengths of the two double-ended arrows, yes?

Ok, my understanding of the system seems to indicate, yes, that is right. I'm not 100% sure (maybe confirm with someone else) and the drawing I'm attaching below (a representation of your diagram above) may not be that clear but this is how I'm viewing it:

I've defined the interplanar distance between AA' and BB' as d. I've also defined the angle between the incident beam and AA' as $\theta$. I've translated the second beam over to make things easier to visualize. That translated beam forms and angle $\theta$ with the interplanar distance d. The path difference will be $d\sin(\theta)$ plus another $d\sin(\theta)$ to account for the extra distance the beam has to travel enter and exit an extra plane deeper. So the total path difference would be $2d\sin(\theta)$, or as you stated, the sum of those two-double ended arrows.

Hopefully this helps in the visualization.

 

[haruspex]

Ok, my understanding of the system seems to indicate, yes, that is right. I'm not 100% sure (maybe confirm with someone else) and the drawing I'm attaching below (a representation of your diagram above) may not be that clear but this is how I'm viewing it:

View attachment 277942

I've defined the interplanar distance between AA' and BB' as d. I've also defined the angle between the incident beam and AA' as $\theta$. I've translated the second beam over to make things easier to visualize. That translated beam forms and angle $\theta$ with the interplanar distance d. The path difference will be $d\sin(\theta)$ plus another $d\sin(\theta)$ to account for the extra distance the beam has to travel enter and exit an extra plane deeper. So the total path difference would be $2d\sin(\theta)$, or as you stated, the sum of those two-double ended arrows.

Hopefully this helps in the visualization.

I wish you had not redefined θ.
Your diagram marks two angles as θ, one being the complement of the other?
I'll define φ as the angle the incident beams make to the tilted plane, so in terms of the original variables that's θ+α.
Also, it seems to me you have the relationship backwards. Aren't the extra distances d/sin(φ)?

 

[baseballfan_ny]

I wish you had not redefined θ.
Your diagram marks two angles as θ, one being the complement of the other?
I'll define φ as the angle the incident beams make to the tilted plane, so in terms of the original variables that's θ+α.
Also, it seems to me you have the relationship backwards. Aren't the extra distances d/sin(φ)?

Let me change it to $\varphi$:

Ok what I head earlier seemed to have confused myself too, so let's forget about that version.

I translated the second beam so its a continuation of the first one, hopefully that's easier to visualize. $\varphi$ is the angle between the plane and the continuation of the beam, so the extra path difference is, as you said $\frac {d} {\sin(\varphi)}$

 

[haruspex]

Let me change it to $\varphi$:

View attachment 277955

Ok what I head earlier seemed to have confused myself too, so let's forget about that version.

I translated the second beam so its a continuation of the first one, hopefully that's easier to visualize. $\varphi$ is the angle between the plane and the continuation of the beam, so the extra path difference is, as you said $\frac {d} {\sin(\varphi)}$

Ok. So what is the relationship between d and a?

 

[baseballfan_ny]

Ok. So what is the relationship between d and a?

Ok now I'm confused again.

I drew the lattice of the original problem:

I have the two parallel planes at angle alpha from the horizontal. The planes are those that go two nodes across and one down. The interplanar distance between the original horizontal planes is a, and that between the new planes is d.

I get two right triangles. The one one the top right suggests d = $a\cos(\alpha)$. The one on the bottom left suggests d = $2a\sin(\alpha)$.

 

[haruspex]

I get two right triangles. The one one the top right suggests d = $a\cos(\alpha)$. The one on the bottom left suggests d = $2a\sin(\alpha)$.

Is that necessarily a contradiction?

 

[baseballfan_ny]

Is that necessarily a contradiction?

Oh wait. If I plug in the numbers for that angle those are the same thing. Haha.