Question

Titanium is bcc above 882 °C. The atomic radius increases 2% when the bcc structure changes to hcp during cooling. Determine the percentage volume change. Hint: there will be a change in atomic packing factor. What ramifications might this have on forming processes used to make biomedical implants?

Answer 1

Answer: according to the given informations , In FCC structure the number of atoms per unit cell are 4 an atomic radius r = a[2]^1/2 / 4

n HCP unit cell, the corner atoms are touching the centre atom on top and bottom faces. Therefore a = 2r or r = a/2

Volume of an FCC unit cell (remember that in the FCC structure atoms are in contact along the face diagonals): (4r/(2^0.5))^3=22.62*r^3

and for the HCP :   = 24 x √2 x r³

  now radius increases 2% measn volume of HCP becomes = 192 *   √2 10^-6 r³

now percentage = 271.529 * 10^-6 / 33.94 * 100

= 12.49 %

Hence the Percentage change in volume is comes out to be 12.49 %

Thank you :)