Titanium Edge
Titanium metal has a body-centered cubic unit cell. The density of titanism is 4.50 g/cm^3.?
Calculate the edge length of the unit cell and a value for the atomic radius of titanium. (Hint: In a body-centered arrangement of spheres, the spheres tough across the body diagonal.
I know how to get the length factor label and then cubing when I get the answer in cm from the density. How do I then find the radius? Can I use the pythagorean theorem as in the face-cenetered cubic cell? If not, what equation should I use instead. I was told to do it this way but I don’t understand. Please explain.
length=328 pm.
Take the square root of 328^2 + 328^2 = x.
x= 464
328^2 + 464 ^2= 4r^2
32280 = 16r^2
Divide by 16.
r= 45 pm
Is this right?
(volume / g) x (g / mole) x (mole / atoms) x (atoms / cell) = (volume / cell)
do you see that? everything cancels except volume and cell. and since this is a cube, the cube root of the volume = the edge length.
volume / g = 1 cm^3 / 4.50g
g / mole = the molar mass of titanium from a periodic table
moles / atoms = 1 mole / 6.022×10^23 atoms
atoms / cell = 2
fyi.. atoms / cell = 2 because there are 8 corner atoms. each of those exists in 8 adjacent cells. so only 1/8th of each corner atom is in any given cell.. 8 x 1/8 = 1.. and 1 center atom makes 2 per cell.
see this picture
http://www.google.com/imgres?imgurl=http://mrsec.wisc.edu/Edetc/SlideShow/images/unit_cells/body_centered_cubic2.jpg&imgrefurl=http://www.mrsec.wisc.edu/Edetc/SlideShow/slides/unit_cells/body_centered_cubic.html&h=336&w=320&sz=32&tbnid=d4zt8lnJ5tYkYM:&tbnh=119&tbnw=113&prev=/images%3Fq%3Dbody%2Bcentered%2Bcubic%2Bunit%2Bcell&hl=en&usg=__qmlkHDPuNfByyCqvwlzI4te3GzU=&ei=9o9DS7GMMJKENv_fnYIJ&sa=X&oi=image_result&resnum=4&ct=image&ved=0CA8Q9QEwAw
**********
As to the radius….
this is complicated. look at this picture
http://www.chem.lsu.edu/htdocs/people/sfwatkins/merlot/lattice/03bcc.html
the diagonal of the cube = line from the upper left front corner through the center atom to the lower right back atom. ok? those atoms touch. and since the corner atoms are only partially in a unit cell.. from the center of the atom on, the length of that diagonal = 1 radius from the upper left atom + 2 radius from the center atom + 1 radius for the lower right atom. got it?
and that diagonal forms a right triangle with 1 leg being the edge from the upper left front atom to the lower left front atom and the other leg being the diagonal of the bottom FACE of the cube. and that diagonal has 2 legs both being edge lengths. lower left front to lower right front and lower right front to lower right back.. got that?
thanks to pythagoreus…
(FACE diagonal length)² = (edge length)² + (edge length)²
(CUBE diagonal length)² = (edge length)² + (FACE diagonal length)²
(CUBE diagonal length)² = (edge length)² + ((edge length)² + (edge length)²
(CUBE diagonal length)² = 3 x (edge length)²
CUBE diagonal length = √3 x edge length..
4r = √3 x edge length..
**************
can you finish?
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