Chemistry
posted by Odesa .
You are conducting experiments with your xray diffractometer.
(a) A specimen of molybdenum is exposed to a beam of monochromatic xrays 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?

Someone please answer this ;(

3.93

3.93 is correct?

no ...

a 7.23 is correct

and (b)?

How did you calculate it?

What is b) ?

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

What is the answer to B??

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.. 
to be honest i don't understand b)
it would be great to get a full detailed info 
b) 475 ev

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.

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