find the Edge (geometry) and the Volume of Unit cell
the question syas: if the radios of Niobium is 1.43 Ångström what will be the length Edge and what will be the Volume of Unit cell, if is know that Niobium has bcc structure.
what I know is:in bcc structure there 2 atoms.
so to find the Volume I just need to use this formula:4/3*pi*r^3 when r=1.43*10^-8cm?
I really need help here.
thanks.
Comment on Unknown008's post
No there is only 2 (NET) atoms that involved in bcc structure.
And in fcc there are 4 atoms (NET) involved.
In both I mean number of atoms per unit cell.
http://www.virginia.edu/bohr/mse209/chapter3.htm
Comment on Unknown008's post
to find a I have to use this formula: a=4r/sqer3 and is give us that a=3.3 Ångström
now to find the volume because we speak about cube is will be a^3.
tell me what you think about this way?
and remember we speak about Volume of Unit cell, :)
thanks a lot .
Comment on Unknown008's post
in the Volume of unit cell=2.45/0.68=36 from where you taken that 0.68?
I found it by using (3.3)^3 Ångström
Comment on Unknown008's post
OK thanks.
Last thing what do say about my way?
Comment on Unknown008's post
lol forgot about this formula.
I just speak about the formula for length of the Edge :4r/sqrt3 is give us a
and for the volume for unit cell =a^3 .
you can see it in my former post up hree :)
Comment on Unknown008's post
"The volume (V) of the unit cell is equal to the cell-edge length (a) cubed."
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch13/unitcell.php
Comment on Unknown008's post
Yes I know in this formula I did mistake
Comment on Unknown008's post
Well this formulas I got in my book.
But anyway I solve it :)
Thanks you :)