Chemistry

posted by Odesa

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?

  1. Anonymous

    Someone please answer this ;(

  2. Anonymus

    3.93

  3. Anonymus

    3.93 is correct?

  4. Chemistry ;)

    no ...

  5. anonim

    a 7.23 is correct

  6. anonymous

    and (b)?

  7. anonymous

    How did you calculate it?

  8. Ling

    What is b) ?

  9. alex

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

  10. Steve

    What is the answer to B??

  11. alex

    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..

  12. babol

    to be honest i don't understand b)
    it would be great to get a full detailed info

  13. babalovic

    b) 475 ev

  14. alex

    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.

  15. Odesa

    Thank you alex and babalovic for B)and anonim for A)

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