4. Above 882°C, titanium has a BCC crystal structure, with a = 0.332 nm. Below t

Question

4. Above 882°C, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperture, titanium has a HCP structure, with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC titanium transforms to HCP titanium. Is this a contraction or expansion? 5. Aluminum foil used to package food is approximately 0.001 inch thick. Assume that all of the unit cells of the aluminum are arranged so that a0 is perpendicular to the foil surface. For a 4 in. x 4 in. square of the foil, determine (a) the total number of unit cells in the foil and (b) the thickness of the foil in number of unit cells cm

Explanation / Answer

For BCC crystal structure

a = 0.332 nm

Volume of BCC = 0.332^3 = 0.03659 nm3

For HCP structure

a = 0.2978 nm

c = 0.4735 nm

Volume of HCP = 3*a^2*c x Sin 40

= 3*0.2978^2 * 0.4735 x Sin 40

= 1.0911 nm3

Equivalent volume of 2 atoms /unit cell = 1.0911 nm3/3 = 0.03637 nm3

% change in volume = (V HCP - V BCC) *100/V BCC

=( 0.03637 - 0.03659) *100/0.03659

= - 0.60%

This is contraction

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