# Homework Help: Complex problem from crystallography - rotating the crystal

1. Jul 27, 2013

### 71GA

The problem statement:

What I have managed to do:

This problem seems a bit tricky at first because it is talking about rotating a crystal while also changing the voltage - changing two variables at the same time makes no sense to me. This is why I assumed that while we change the voltage we do it at a constant angle $\vartheta$! Distance $d$ between crystal planes which is a material property is also constant so I can write two equations which I can equate and solve for a number $N_1$ (which I don't know yet):

\begin{align}
\left.
\begin{aligned}
\substack{\text{Brag's law for voltage $U_1$}}\longrightarrow\quad2d\sin\vartheta &= N_1 \lambda_1\\
\substack{\text{Brag's law for voltage $U_2$}}\longrightarrow\quad2d\sin\vartheta&=\!\!\!\! \smash{\underbrace{(N_1+1)}_{\substack{\text{At voltage $U_2$}\\\text{I get an extra}\\\text{mirror reflection}}}}\!\!\!\lambda_1\\
\end{aligned}
\right\}
N_1 \lambda_1 &= (N_1+1) \lambda_2\\
\frac{N_1+1}{N_1} &= \frac{\lambda_1}{\lambda_2}\\
1+\frac{1}{N_1} &= \frac{\lambda_1}{\lambda_2}\\
\frac{1}{N_1} &= \frac{\lambda_1}{\lambda_2} - 1\\
N_1 &= \frac{1}{\lambda_1/\lambda_2 -1}
\end{align}

Ok so I can calculate $N_1$ if i know the ratio $\lambda_1/\lambda_2$ which I can get from voltages (with a litle help of Lorentz invariant):

\begin{align}
\substack{\text{Lorentz invariance}}\longrightarrow p^2c^2 &= E^2 - {E_0}^2\\
p^2c^2 &= (E_k + E_0)^2 - {E_0}^2\\
p^2c^2 &= {E_k}^2 + 2E_kE_0 + {E_0}^2 - {E_0}^2\\
p^2c^2 &= {E_k}^2 + 2E_kE_0\\
p &= \frac{\sqrt{{E_k}^2 + 2E_kE_0}}{c}\\
~\\
~\\
~\\
\substack{\text{I use Lorentz invariance}\\\text{and De Broglie's hypothesis}\\\text{to calculate ratio $\lambda_1/\lambda_2$}} \longrightarrow \frac{\lambda_1}{\lambda_2} &= \frac{h p_2}{p_1 h} = \frac{p_2}{p_1} = \frac{\sqrt{{E_{k2}}^2 + 2E_{k2}E_0}}{\sqrt{{E_{k1}}^2 + 2E_{k1}E_0}} = \sqrt{\frac{e^2{U_{2}}^2 + 2eU_{2}E_0}{e^2{U_{1}}^2 + 2eU_{1}E_0}}= \\
&= \sqrt{\frac{e^2\cdot 214^2V^2 + 2e\cdot 214V\cdot0.51\times10^6eV}{e^2\cdot 137^2V^2 + 2e\cdot 137V\cdot0.51\times10^6eV}} = 1.25
\end{align}

If I use this ratio I can calculate that $\boxed{N_1=4}$.

What I haven't managed to figure out:

Question 1:
Does my result $N_1 = 4$ mean that at voltage $U_1=137V$ we get the fourth maximum or does it mean that If I change angle $\vartheta$ I will get maximums at $4$ different angles? Is it both? Please explain!

Question 2:
I used De Broglie relation and Lorentz invariant to calculate
\begin{align}
\lambda_1&=1.04\times10^{-10}m\\
\lambda_2&=8.37\times10^{-11}m
\end{align}
but I have absolutely no idea on how to calculate the distance between the crystal planes $d$. Can anyone give me a hint?

Last edited: Jul 27, 2013
2. Jul 27, 2013

### TSny

Hello.

Think about the value of the angle $\vartheta$ at which a new reflection just begins to occur as the voltage just reaches a value to produce the new reflection.

[Aside: You can safely use nonrelativistic expressions for the energy to make the calculation easier. You'll still get a ratio of 1.25 for the wavelengths.]

3. Jul 27, 2013

### 71GA

I don't think I understand this approach. Can you be a litle more speciffic?

I know yes - in nonrelativistic approach I could just neglect the $e^2U^2$ part inside the square root, but it was fairly simple to calculate ratio relativistically and i didn't do that. It was easy because I expressed momentum in $eV/c$.

In this case It wasn't so simple and I was forced to use the nonrelativistic way. I still wish i could solve that problem relativistically =)

4. Jul 27, 2013

### TSny

From $2d\sin\vartheta = n \lambda$ you see that the angles at which you get a reflection maximum depend on the wavelength, which in turn depends on the voltage.

Suppose the voltage $U$ is set to a value between $U_1$ and $U_2$. As you vary $\vartheta$ between 0 and $\pi/2$, how many reflection maxima will occur?

If $U$ is now increased somewhat, but still lies between $U_1$ and $U_2$, what happens to the angles $\vartheta$ at which you see the maxima? Do the angles get bigger or smaller?

Just as the voltage gets to $U_2$, an extra maximum appears. At what angle $\vartheta$ does the new maximum occur?

Knowing the value of $\vartheta$ for which the extra reflection occurs will allow you to find $d$.

Regarding using nonrelativistic expressions for the energy, you can show that $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{U_2}{U_1}}$.

Last edited: Jul 27, 2013
5. Jul 27, 2013

### 71GA

I can see this yes.

This is what i don't know. and is hardest to understand to me. I would guess that $4$ but I have no clue why. I would need a further explaination on this.

I know that increase in $U$ decreases the $\lambda$ so in order to maintain equality $2d\sin\vartheta = n \lambda$ we can:
• decrease the $\vartheta$
• increase $N$ (Is this allowed? I guess not untill voltage exceeds or is equal to $U_2$.) Please confirm!
If i would chose to decrease the $\vartheta$ the anwser to your questions would be: "The angles between maxima should get smaller and smaller the more we increase the voltage $U$".

The new maximum must appear at angle $90^\circ$ or $0^\circ$ right? Because the central maximum will always be at $2\vartheta=0^\circ$. Is my thinking correct? Please confirm!

This will be easy once I understand the above. So lets leave it standing.

Last edited: Jul 27, 2013
6. Jul 27, 2013

### TSny

Yes. It's 4. You already found $(N_1+1)/N_1 = 1.25 = 5/4$. So, $N_1 = 4$.

Yes, $\vartheta$ decreases as $U$ increases. So, if $U$ is set at some value between $U_1$ and $U_2$, and you rotate the crystal, you will find 4 values of $\vartheta$ that will produce a maximum reflection. That is, for that value of $U$, values of $\vartheta$ can be found such that Bragg's law $2d\ sin\vartheta = n \lambda$ will be satisfied for n equal to 1, 2, 3, or 4. But it can't be satisfied for n = 5.

If you then increase $U$ a little bit, you will find that the 4 values of $\vartheta$ that satisfy Bragg's law for the new $\lambda$ will have decreased a bit. As you continue to increase $U$, you will finally reach the voltage $U_2$ where you can rotate the crystal and find 5 different angles that produce maxima. So there now exist values of $\vartheta$ such that $2d \ sin\vartheta = n \lambda$ for n = 1, 2, 3, 4, and 5.

The key is to see what the value of $\vartheta$ will be for the n = 5 case when the voltage is right at $U_2$.

I don't understand what you mean by "central maximum" in the context of Bragg reflection. Make sure you understand the geometrical setup and the meaning of $\vartheta$.

For a nice diagram, see http://www.britannica.com/EBchecked/media/17859/Bragg-diffraction [Broken].

From the diagram, you can see that $\vartheta = 0$ means that the incoming electrons are moving essentially parallel to the crystal planes and $\vartheta = 90^\circ$ means the electrons are moving perpendicular to the planes.

Last edited by a moderator: May 6, 2017
7. Jul 28, 2013

### 71GA

[Broken]

After drawing this image for the angles at voltage $U_1=137V$ it occured to me that the angle $\vartheta_4$ for $N=4$ will be the biggest (among the angles for maxima which occur at $U_1$). So i can write down the Brag's law like this:
\begin{align}
2d\sin\vartheta_4 = 4\lambda_1
\end{align}
Still i cannot calculate the $\vartheta_4$ so i need a new equation which I could write for $N=1$. This time it is:
\begin{align}
2d\sin\vartheta_1 = 1\lambda_1
\end{align}
So if i could find the relation between $\vartheta_1$ and $\vartheta_4$ i could calculate $\vartheta_4$ and then all the other angles. Is it possible that for the $U_1$ i write these four equations:
\begin{align}
&
\left.
\begin{aligned}
2d\sin\vartheta_4&=4\lambda_1\\
2d\tfrac{3}{4}\sin\vartheta_4&=3\lambda_1\\
2d\tfrac{2}{4}\sin\vartheta_4&=2\lambda_1\\
2d\tfrac{1}{4}\sin\vartheta_4&=1\lambda_1
\end{aligned}
\right\} \quad \text{at voltage $U_1=137V$}
& &
\left.
\begin{aligned}
2d\sin\vartheta_5&=5\lambda_2\\
2d\tfrac{4}{5}\sin\vartheta_5&=4\lambda_1\\
2d\tfrac{3}{5}\sin\vartheta_5&=3\lambda_1\\
2d\tfrac{2}{5}\sin\vartheta_5&=2\lambda_1\\
2d\tfrac{1}{5}\sin\vartheta_5&=1\lambda_1
\end{aligned}
\right\} \quad \text{at voltage $U_2=214V$}
\end{align}
I am a bit confused at the moment and don't know how exactly to get an angle out of these equations (wel generaly speaking it is the same equation with 2 variables... $d$ and $\vartheta_4$ OR $d$ and $\vartheta_5$...

ls it possible that the $\vartheta_5(U_2)$ is the same as $\vartheta_4(U_1)$?

Last edited by a moderator: May 6, 2017
8. Jul 28, 2013

### TSny

That's a nice drawing! There's a key correction that should be made in the drawing. You know that increasing U will cause $\vartheta$ to decrease for each reflection maximum. Likewise, decreasing U will increase the angles. So, if you were to imagine U to decrease a bit below U1, then all of the angles in your drawing would increase a little and you would still have 4 maxima. But that contradicts the fact that U1 is the special voltage at which you just get the 4th maximum (n = 4). In other words, the n = 4 reflection at U = U1 must be at an angle that cannot increase further, otherwise you could decrease U below U1 and still have 4 maxima.

[Edited an incorrect statement on my part here.]

Yes!

Last edited: Jul 28, 2013
9. Jul 28, 2013

### 71GA

I think what you are saying here is that $\vartheta_4(U_1)$ should be $90^\circ$??? But wouldn't i get a maximum at $\vartheta_4=90^\circ$ only if distance $d=N\lambda_1$... But if $d \neq N\lambda_1$ the maximums shuld start dissapearing before the angle $\vartheta = 90^{\circ}$ like this:

[Broken]

Should i somehow use this to calculate my missing results $\vartheta_4$ and $d$? How? I always end up with a ratio $\lambda_1/\lambda_2$ which i already know...

Last edited by a moderator: May 6, 2017
10. Jul 28, 2013

### TSny

Yes, at U = U1, the n = 4 maximum will occur at $\vartheta = 90^\circ$.

So, what does the Bragg law become when you let n = 4 and $\vartheta = 90^\circ$?

Similarly, consider the Bragg law for n = 5 when U = U2.

11. Jul 28, 2013

### 71GA

I have been thinking about this all the time, but wasn't certain that the new maximums always occur at $90^\circ$. Are you sure about this. Is there any good demonstration / animation etc about this?

I get:
$$d=2\lambda_1$$

12. Jul 28, 2013

### TSny

For U = U1, you just start to get a new maximum. This new maximum must occur at $\vartheta = 90^\circ$. The reason is as follows. Suppose the new maximum that appears at U1 occurs at an angle less than 90o, say at 85o. Then, according to Bragg's law, you could decrease U a little below U1 and that would cause the angle of the new maximum to increase a little, say to 87o. Thus, you would have found a value of U less than U1 that still has the same number of maxima as U1. This contradicts the statement that U1 is the lowest value of U for which the new maximum occurs.

So, whenever a new maximum starts to occur, it always starts to occur at $\vartheta = 90^\circ$.

With that in mind, I think it would be good to recap the problem from the beginning.

(1) You have shown how to determine the values of $\lambda_1$ and $\lambda_2$ corresponding to the voltages $U_1$ and $U_2$.

(2) It is stated in the problem that new maxima appear when $U = U_1$ and $U = U_2$. These new maxima will first appear at $\vartheta = 90^\circ$.

(3) If $N_1$ is the order of the new maximum that occurs at $U = U_1$, then the Bragg law for this maximum becomes $2d \ \sin90^\circ = N_1 \lambda_1$. Likewise, $2d \ \sin90^\circ = N_2 \lambda_2$, where $N_2$ corresponds to the order of the maximum that first appears when $U = U_2$. Simplify these equations using the known value of $\sin90^\circ$.

(4) Dividing the two equations in (3) and using the known values of $\lambda_1$ and $\lambda_2$ gives the ratio $N_2/N_1$ = 1.25 = 5/4. Since you know that $N_2 = N_1+1$, you conclude that $N_1 = 4$ and $N_2 = 5$. So, you find that 4 Bragg reflections will occur for $U = U_1$ and 5 will occur for $U = U_2$.

(5) All that's left to do is to find $d$. (Hint: use one of the equations from (3) above.)

Last edited: Jul 28, 2013
13. Jul 28, 2013

### 71GA

Thank you. This problem is solved and i think i understand the physics behind it.

14. Aug 8, 2013

### mcodesmart

The scattering in general is given as follows:

$I(K) \propto S(K) = 1 + N/V \int g(r) e^{iK\bullet r} dr$