Solid State electron diffraction

[Brewer]

Homework Statement

A crystal with unknown lattice spacing is studied using electron diffraction.
The electron wavelength is 3.7pm.

a) assuming a simple cubic lattice, what is its size if the smallest diffraction peak is at 0.5 degrees?

b) At what angle is the nxt peak?

c) At what angle is the [122] peak?

d) The [122] and [310] planes will be situated close to each other. Without calculating the angle at which the [310] peak scatters electrons, can the two peaks be resolved if the accuracy to which the angle can be determined is 0.03 degrees?

e) Comment on the locations of the [220] peak and the second order peak of [110].

Homework Equations

$$n\lambda = 2dsin\theta$$

The Attempt at a Solution

a) Rearranged the above equation to give d=2.12*10^-10 m=a

b) said that the next peak will be at [110] and so d=$$\sqrt{2}$$a, put that into the equation above to give theta = 0.71 deg.

c) said that d=3a and put this into the above equation to give theta = 0.83 deg.

d) I don't really understand how to go about this question. Any hints would be appreciated.

e) I haven't fully looked at this part yet (stuck on the previous bit), but first off I would say that the locations of the peaks are in the same place. I think anything else would be a bonus!

Any help/comments would be appreciated.

Thanks

Brewer

 

[Jhero]

Dear Brewer,

Thank you for your post. I understand your interest in studying this crystal with unknown lattice spacing using electron diffraction. I will try my best to help you with your questions.

a) Your calculation for the size of the crystal assuming a simple cubic lattice is correct. The size of the crystal is 2.12*10^-10 m or a= 212 pm.

b) The next peak will indeed be at [110]. To find the angle, we can use the same equation n\lambda = 2dsin\theta. Since we already know the value of d, we can rearrange the equation to solve for theta. This will give us theta = 0.71 degrees, which is the same as your answer.

c) Similarly, for the [122] peak, we can use the equation n\lambda = 2dsin\theta and plug in the value of d=3a. This will give us theta=0.83 degrees, which is also the same as your answer.

d) To determine if the two peaks can be resolved with an accuracy of 0.03 degrees, we need to consider the spacing between the two peaks. The spacing between the [122] and [310] planes is given by d= a/\sqrt{5}. If we assume that the angle at which the [310] peak scatters electrons is approximately the same as the angle for the [110] peak, then we can calculate the value of d for the [310] peak using d=\sqrt{2}a. We then find the difference in angles between the two peaks, which is 0.71 degrees. This difference is greater than the accuracy of 0.03 degrees, so the two peaks can be resolved.

e) As you mentioned, the locations of the [220] peak and the second order peak of [110] will be in the same place. This is because the [220] peak is the second order of the [110] peak, and the spacing between the [220] and [110] planes is the same as the spacing between the [110] and [100] planes. Therefore, the two peaks will overlap.

I hope this helps. If you have any further questions, please let me know.