Diffraction grating problem?

  • Thread starter kraphysics
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  • #1
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Homework Statement


Electrons are accelerated to an energy of 2.40 keV and then passed through a thin graphite film. The distance between the carbon atoms is 1.42 * 10^-10 m.

Homework Equations


b) Describe the diffraction pattern on a screen 5.00 cm from the graphite film. We're looking for x.

The Attempt at a Solution


Well first I found the wavelength. I'm sure I got this part right.
E = 1/2mv^2
then using v, I found lambda in mv = h/lambda
Now to solve the above question for x, I used ,
lambda = xd/nl
but I get a slightly different result from the supposed answer. The answer is 8.84 * 10^-3m yet I keep getting 8.83 * 10^-3 m. I kept all the digits and I keep getting this. Can anyone help?
Does the angle matter?
 

Answers and Replies

  • #2
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I don't know why the third digit is a concern, but what rest mass do you use for the electron and to how many significant figures?

You might try to the use the relativistic momentum, and it might give a different numeral in the third significant figure after round-off.
 
  • #3
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I don't know why the third digit is a concern, but what rest mass do you use for the electron?

You might try the use the relativistic momentum, and it might give a different numeral in the third significant figure.

I'm in high school and we haven;t studied relativity yet. This is supposed to be solved simply. I use 9.11 * 10^-31kg as my rest mass. The thing is that I remember the teacher saying something about using tan angle instead of sin or something like that. I don't really understand it.
 
  • #4
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I don't know why the tangent comes up, but you could try the following.

[tex]\theta = sin(\lambda/d)[/tex]

[tex]x/l = arctan(\theta)[/tex]

If this doesn't work change sin to tan and arctan to arcsin to get the proscribed result.
 
  • #5
4,239
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OK, now I see where the tangent operator comes from, and your teacher is not so nuts in this particular case. If using the two equations above doesn't do it, there are two more higher order corrections you can use.
 
  • #6
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I don't know why the tangent comes up, but you could try the following.

[tex]\theta = sin(\lambda/d)[/tex]

[tex]x/l = arctan(\theta)[/tex]

If this doesn't work change sin to tan and arctan to arcsin to get the proscribed result.

Well I tried this another way.
sin theta = lambda /d
theta = 10.18

then tan 10.18 = x/l
tan 10.18 * 0.05 m = x
I get an answer that's even more off. I get 8.98 * 10^-3 m. What am I doing wrong?
 
  • #7
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Oh wait! I just got the right answer. But I don't know why it works:
tan theta = lambda/d
tan theta = 0.1767605634
tan theta = x/l
0.1767605634 * 0.05m = x
x = 8.84 * 10^-3 m

I dont get how tan equals both lambda/d and x/l. Can somebody explain?
 
  • #8
4,239
1
How do you know it's the right answer?

You seem to have manage to utilize your calculator's round off error for trigonometric functions to obtain the desired result.

Shall we call your activity science, or numerology and deference to authority? How do you know which answer is really correct to three figures? Maybe you were right the first time. Maybe both are wrong to 3 figures.
 
Last edited:
  • #9
34
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How do you know it's the right answer?

You seem to have manage to utilize your calculator's round off error for trigonometric functions to obtain the desired result.

Shall we call your activity science, or numerology and deference to authority? How do you know which answer is really correct to three figures? Maybe you were right the first time. Maybe both are wrong to 3 figures.


No. I did not do that. Well it's a practice problem and my teacher gave us the answers to the problems. I did not round off anything in my calculator. Maybe if you try the calculations on your calculator, you will see too.
My question still remains. Why does what I did work?
 
Last edited:

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