pop000 Posts: 352, Reputation: 6 Full Member #1 May 7, 2011, 02:07 AM
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.

 pop000 Posts: 352, Reputation: 6 Full Member #2 May 7, 2011, 02:49 AM
is will be correct if I use this formula d=m/v to find the Volume of Unit cell.
if I know that the unit cell in this case is bcc so I can find d by this formula:d=(3^3/2M)/(32Nar^3)

this what I thinking to do:v=m/d when m=xMw/Na when Mw is molar mass, and x is the NO of atoms in unit cell in this case I have 2, and the molar mass of Niobium is 92.91 g mol-1.

so I get v=(2*92.91 g mol-1/6.022*10^23)/(3^3/2M)/(32Nar^3).

what are you think this is a correct way to find the Volume of Unit cell,

sorry about the mess :)
thanks
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #3 May 8, 2011, 02:46 AM

BCC is body centered cube, right? Then, isn't there 5 atoms involved? :confused:

I never did this chapter, but this seems strange to me...
 pop000 Posts: 352, Reputation: 6 Full Member #4 May 8, 2011, 06:03 AM
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
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #5 May 8, 2011, 09:35 AM

Okay, if I understand well, the atoms along a diagonal touch each other, but along not along an edge, like this:

The diagonal in this case is equal to 4 radii, which is 5.72 Ångström (black lines, any one of them). This means that the diagonal along the base is 5.72/\sqrt2 = 4.04 Ångström (that diagonal is the hypotenuse of an isosceles right angled triangle, the triangle formed by the two red lines and one black line I picked). This base is now another hypotenuse of an isosceles right angled triangle, and is equal to 4.04/\sqrt2 = 2.86 Ångström. (in the triangle with blue lines and one red line)

The volume is then that value to the cubed, which is 23.4 Ångström cube.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #6 May 8, 2011, 09:39 AM

If I understood your site, you could this too:

Yu use the APF, which would give you:

Volume of atoms = $2\times \frac43 \pi(1.43)^3 = 2.45$ Ångström cube

Volume of unit cell = 2.45/0.68 = 36.0 anmstrom cube

Edge of unit cell = $\sqrt[3]{36.0} = 3.30$ Ångström

That's quite far =/

But I guess this is better and shorter, since I'm not sure that the atoms even touch each other.
 pop000 Posts: 352, Reputation: 6 Full Member #7 May 8, 2011, 09:57 AM
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.
and remember we speak about Volume of Unit cell, :)
thanks a lot .
 pop000 Posts: 352, Reputation: 6 Full Member #8 May 8, 2011, 10:06 AM
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
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #9 May 8, 2011, 10:24 AM

Um... I actually used the Volume to get the length of the Edge in my latest post :o

And I got 36.0 Ångström.

But when it comes to your formula, I'm having trouble understanding how

Density =(3^3/2M) / (32Na r^3)

Mass = 3^3/2M ?
Volume = 32 Na r^3 ?

:confused:
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #10 May 8, 2011, 10:26 AM

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
Ooh, I got it from the site you linked me to. For BCC, the APF is 0.68.
 pop000 Posts: 352, Reputation: 6 Full Member #11 May 8, 2011, 10:35 AM
Comment on Unknown008's post
OK thanks.
Last thing what do say about my way?
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #12 May 8, 2011, 10:46 AM

I don't know where these values come from :(
 pop000 Posts: 352, Reputation: 6 Full Member #13 May 8, 2011, 10:49 AM
Comment on Unknown008's post
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 :)
 pop000 Posts: 352, Reputation: 6 Full Member #14 May 8, 2011, 11:07 AM
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
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #15 May 8, 2011, 11:08 AM

Hm... this very closely reminds me of the one I used, but I honestly don't know :(

If you get the answer required, then I guess it's fine :)

Also, the site you gave me said that in the denominator, you have nA, where A is the mass in amu. 92.91 is not in the units of amu as far as I see, since 1 amu is about 12 g/mol
 pop000 Posts: 352, Reputation: 6 Full Member #16 May 8, 2011, 11:10 AM
Comment on Unknown008's post
Yes I know in this formula I did mistake
 pop000 Posts: 352, Reputation: 6 Full Member #17 May 9, 2011, 03:18 AM
OK I did it but:
look on the picture I want to calculate it and I get as a result 21.2

but I know the correct answer is 8.17g/cm^-3

what you get?

thank you.
 pop000 Posts: 352, Reputation: 6 Full Member #18 May 9, 2011, 03:20 AM
Comment on pop000's post
Is speak about Cu with radios 128pm
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #19 May 9, 2011, 09:32 AM

I told you I don't understand your formula at all :(

Why the $3^{3/2}$, I don't know, why the 32 in the denominator, I don't know :(

Plus the site you gave me earlier doesn't have it...
 pop000 Posts: 352, Reputation: 6 Full Member #20 May 10, 2011, 08:18 AM
Comment on Unknown008's post
Well this formulas I got in my book.
But anyway I solve it :)

Thanks you :)

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