[vanalm]

Hello. I am doing my homework and I am stuck on this question:

Calculate the percent volume that is actually occupied by spheres in a body-centered cubic lattice of identical spheres. You can do this by relating the radius of a sphere, r, to the length of an edge of a unit cell, 1. The volume of a sphere is given as v=4/3 pi r³.

"The edge of a unit cell, 1"... does that mean the length is 1? If so, does that mean the radius is 1/2? Or is finding the radius more complicated than that?
Once I figure out the radius I just plug that into the volume formula, correct?

[DevaDevil]

The unit cell; as you stated yourself is cubical; therefor has no radius.

What you have to do is draw the bcc - unit cell, and then just look at the atoms in it.

- How many atoms make up the unit cell;
- and for what amount are they actually IN the unit cell (For example, corner-atoms are present for 1/8 in the unit cell, as the form the corner of 8 unit cells)

That gives you the amount of spheres in the unit cell.
The volume of that is (of course) given by: Amount * Volume per sphere.
Look at your drawing again to see what the ratio between radius of the atoms and length of the unit cell is.
The length (and width, and depth; since it's a cube) of the unit cell is given as 1; the ratio will give you the value of r.

then calculate the volume of the unit cell itself;

Finally you have 2 volumes; then see what the % of the volume of the spheres is in comparison to the volume of the unit cell

[vanalm]

I found a formula in my notes that I think may help. Would you look over my work and let me know if this looks right?

the formula says:
a=length of a side
b=diagonal of a side
c=diagonal
b²=2a²
c²=3a²=4r

So I have:
c²=3, c=square root of 3
r=(square root of 3)/4= 0.433
volume of a sphere =0.340 times 2 spheres per cube= 0.680

volume of cube=l*w*h=1

percent volume=68%

Did I do this correctly?

[AWK]

http://mathworld.wolfram.com/CubicClosePacking.html

[vanalm]

Thanks alot for the link, that looks like it will be much more helpful than my book. My book pictures aren't very clear.

By the way, does that mean that my answer was wrong?

[DevaDevil]

c²=3a²=4r

c = 4r; not c²

c²=3, c=square root of 3
r=(square root of 3)/4= 0.433

But here you do it right

percent volume=68%
Did I do this correctly?

Yes

[DevaDevil]

http://mathworld.wolfram.com/CubicClosePacking.html

Nice website, even if it gives the calculation for the fcc unit cell

[AWK]

http://www.seas.upenn.edu/~chem101/sschem/metallicsolids.html


68 % is OK