Chemistry

You are conducting experiments with your x-ray diffractometer.

(a) A specimen of molybdenum is exposed to a beam of monochromatic x-rays of wavelength set by the Kα line of silver. Calculate the value of the smallest Bragg angle, θhkl, at which you can expect to observe reflections from the molybdenum specimen.

DATA:
λKα of Mo = 0.721 Å
lattice constant of Mo, a = 3.15 Å
λKα of Ag = 0.561 Å
lattice constant of Ag, a = 4.09 Å

(b) If you wish to repeat the experiment described in part (a) but using an electron diffractometer, across what vale of potential difference must electrons be accelerated from rest in order to give the identical value for the smallest Bragg angle, θhkl?

asked by Odesa
  1. Someone please answer this ;(

    posted by Anonymous
  2. 3.93

    posted by Anonymus
  3. 3.93 is correct?

    posted by Anonymus
  4. no ...

    posted by Chemistry ;)
  5. a 7.23 is correct

    posted by anonim
  6. and (b)?

    posted by anonymous
  7. How did you calculate it?

    posted by anonymous
  8. What is b) ?

    posted by Ling
  9. and b) is a voltage, needed to accelerate electrons to get the same.. :)

    posted by alex
  10. What is the answer to B??

    posted by Steve
  11. ok, b)

    eV=E
    To get electrons behave like a waves with the same lenght as 0.561A, we need to give them some impulse p according to de Broglie eq.
    and E = p^2/2m..

    posted by alex
  12. to be honest i don't understand b)
    it would be great to get a full detailed info

    posted by babol
  13. b) 475 ev

    posted by babalovic
  14. babol, for diffraction we need a wave, and according to de Broglie (and later experiments) any matter behave like a waves with different lengths ( depends on their energy). So actually we make a diffraction of electrons (as a waves) on a Mo surface.

    posted by alex
  15. Thank you alex and babalovic for B)and anonim for A)

    posted by Odesa

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