Complex problem from crystallography - rotating the crystal

[71GA]

The problem statement:

We are observing an electron diffraction on a "Ni" crystall. We point
the narrow beam of electrons on a crystall which can be rotated so that
we are changing the angle ##\vartheta## between an incomming beam and a crystal
planes.

We are also increasing the the voltage which accelerates the
electrons. By increasing the voltage the number of angles at which we
get mirror reflection also increases.

An extra mirror reflection on a same set of crystal planes appears
when we accelerate electrons with a voltage ##U_1=137V##, while next one
appears at voltage ##U_2=214V##.

How many angles, at which we get a mirror reflection, we get when
using voltage ##U_1=137V##? What is the distance ##d## between crystal planes?

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}
\quad
\right\}
\qquad
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?

 

[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.]

 

[71GA]

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.

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

[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.]

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 =)

 

[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}}##.

 

[71GA]

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.

I can see this yes.

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?

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 explanation on this.

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?

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 until 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##".

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

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!

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

This will be easy once I understand the above. So let's leave it standing.

 

[TSny]

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 explanation on this.

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

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 until 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##"

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##.

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!

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 .

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.

 

[71GA]

After drawing this image for the angles at voltage ##U_1=137V## it occurred 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##...

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##.

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

 

[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.]

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

Yes!

 

[71GA]

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.

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:

Yes!

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...

 

[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.

 

[71GA]

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

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?

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

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

 

[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 90^o, say at 85^o. 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 87^o. 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.)

 

[71GA]

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

 

[mcodesmart]

The scattering in general is given as follows:

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