Asked 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?
Answered by Anonymous
Someone please answer this ;(
Answered by Anonymus
3.93
Answered by Anonymus
3.93 is correct?
Answered by Chemistry ;)
no ...
Answered by anonim
a 7.23 is correct
Answered by anonymous
and (b)?
Answered by anonymous
How did you calculate it?
Answered by Ling
What is b) ?
Answered by alex
and b) is a voltage, needed to accelerate electrons to get the same.. :)
Answered by Steve
What is the answer to B??
Answered by 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..
Answered by babol
to be honest i don't understand b)
it would be great to get a full detailed info
Answered by babalovic
b) 475 ev
Answered by 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.
Answered by Odesa
Thank you alex and babalovic for B)and anonim for A)
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