De Broglie wavelength neutrons

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  • #1
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Question-A certain crystal has a set of planes spaced 0.376 nm apart. A beam of neutrons strikes the crystal at normal incidence and the first maximum of the diffraction pattern occurs at a scattering angle (i.e angle between incoming neutrons and reflected beam) of 41.4 degrees. What is the de Broglie wavelength of the neutrons in units nm and to 3 decimal places.

Usefull equations -
2d sin theata = nλ
where d = plane spaceing, n is the order in this case =1

My Attempt-
useing above eqn,
2*0.0376E-9 (m)*sin(41.4)=4.973E-11 m

I have been told that this solution is wrong as the question states "at normal incidence" which apparantly complicates things and makes a need to change the angle theata. Any suggestions? thanks in advance.
 

Answers and Replies

  • #2
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Right, real simple. Try drawing a picture and then revisit what they mean by "angle between incoming neutrons and reflected beam" and make sure you drew it correctly.

You're almost there.
 
  • #3
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yeah its the drawing which is what is stuffing me up. when i draw a picture i have the neutrons comeing in at 90 degrees to the plane ie normal and the reflecting out at 41.4 degrees. but i dont think this is right?
 
  • #4
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diagram.png
 
  • #5
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Perform an internet search for "Bragg Diffraction."
 
  • #6
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Bragg_diffraction.png
 
  • #7
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Well done. Now think on that for a bit.
 
  • #8
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Looking at the question again it states that the neutrons are hitting the plane normal ie at theata = 90 degrees which is the dotted line on that diagram right. Also we know the first maximum of the diffraction pattern occurs at a scattering angle (i.e angle between incoming neutrons and reflected beam) of 41.4 degrees.
So does this mean that as theata is measured with respect to the plane, theata is 90-41.4?
 
  • #9
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Excellent! That's the second step. We're missing the first one. Nearly there. Think about where the "angle between incoming neutrons and reflected beam" is.
 
  • #10
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diff 2.png


well the angle (phi) is the angle between incomming neutrons and reflected beem, and in our case phi= 41.4 degrees. So what am i missing? thank you very much for your patience.
 
  • #11
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No troubles. You've done quite well. Great picture. I believe you're a good student for being so patient yourself.

Answer: We need to halve it.

Then do what you were saying.
 
  • #12
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I see, so we have 90-1/2phi
=90-20.7
=69.3 degress
and from this we can plug theata into the 2dsintheata=nlambda
correct?
 
  • #13
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Time to celebrate.
 
  • #14
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thank you very much.
 

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