GR8677 #91


Problem



Optics}Bragg Reflection
Bragg diffraction is basically the wavelength change of a wave impinging two adjacent layers of the crystal. More specifically, it relates the wavelength difference of the incident wave on the top layer to that of the reflected wave on the lower layer. The distance between layers (``lattice planes", in SolidStateSpeak) is , and the angle of incidence is . The change in wavelength is thus a simple geometry problem, the result being the celebrated Bragg Reflection Relation,
where corresponds to the order of the reflection.
The problem supplies the following numbers
and if one recalls the de Broglie relation, , one has, in general, . Plugging in some numbers, one finds that , since , and where the approximation scheme to battle the nocalculatorsallow requirement is shown. Thus, one has , which is very close to choice (D).
The problem .


Alternate Solutions 
chrisfizzix 20081003 14:32:17  I feel that the posted solution lacks a bit of transparency; this is the same solution but reformatted.
Bragg Scattering gives the result
and the DeBroglie relation gives
.
Firstorder reflection lets us set , also the relation usually fails if n > 1 anyway, so we can rearrange to write
The answer in E is greater than , so we can get rid of that. All of the other answers are nonrelativistic (even for D, ) so we can just use .
Solving this, we get
Plugging in values gives a result closest to D.  

Comments 
anmuhich 20090321 15:03:39  I don't know hardly any optics, but one can eliminate A, B and E because they are completely unreasonable speeds for an electron. D seems the most plausible out of the two left.   student2008 20081014 06:07:56  Since the orders of magnitude in answers differ a lot, we can even don't get use of & Bragg condition. All we use is , since (actually, in this problem , so we even obtain the exact result).
wittensdog 20090728 08:06:08 
Just out of curiosity, I wanted to test how well this strategy would work in general. For 30 degrees the answer is indeed pretty much exactly on. To have the 2 * sin(theta) term introduce as much as a factor of ten, the incident angle would have to be as small as about 2.86 degrees, just grazing the surface. Even with a factor of ten, you should be able to discriminate between answers enough to get the right one.
Sounds like a powerful trick to me! I'll have to keep it in mind if I ever get stuck on a problem with wave phenomenon in general.

wittensdog 20090728 08:10:07 
Also, if you wanted to see if it could introduce a factor of ten the other way, as in to create a decrease from the correct answer, you'd end up with theta = arcsin( 5 ). So need to worry about that!

rahat90 20140911 03:43:11 
This is a useful trick!

  chrisfizzix 20081003 14:32:17  I feel that the posted solution lacks a bit of transparency; this is the same solution but reformatted.
Bragg Scattering gives the result
and the DeBroglie relation gives
.
Firstorder reflection lets us set , also the relation usually fails if n > 1 anyway, so we can rearrange to write
The answer in E is greater than , so we can get rid of that. All of the other answers are nonrelativistic (even for D, ) so we can just use .
Solving this, we get
Plugging in values gives a result closest to D.
luwei0917 20140402 08:08:41 
thanks.

  madfish 20071102 19:04:53  if one knows electrons travel very fast to give off xrays, but not faster than the speed of light...one can elminate all but C & D
chrisfizzix 20081003 14:13:43 
The electron isn't radiating Xrays  the spacing between lattice planes in the crystal has been previously measured using Xrays. Thus, the threshold for Xray emission by an electron has nothing to do with this problem.
Also, electrons need both high energy and some kind of accelerating force to emit Xrays, such as the magnetic field from a bending magnet in a synchrotron.

 

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