Mats 2001 Summer 2012 Assigned 06/25/12 Problem Set 3: 1) At 20 C, Molybdenum is
ID: 705057 • Letter: M
Question
Mats 2001 Summer 2012 Assigned 06/25/12 Problem Set 3: 1) At 20 C, Molybdenum is BCC. Calculate the lattice parameter for 2) Gold is FCC and has a lattice parameter of 0.40788 nm. Calculate the 3) Below are shown three different crystallographic planes for a unit cell of Due 07/02/1 Molybdenum, in nanometers. atomic radius of gold, in nanometers. some hypothetical metal; the circles represent atoms: (a) To what crystal system does the unit cell belong? Draw a unit cell. (b) What would this crystal structure be called? (c) If the density of this metal is 8.95 g/cm3, determine its atomic weight 0.30 nm 0.35 nm- (110) (001) (101) 4) Iron / Aluminum unit cell calculations (a) Draw the unit cell for Fe and label all the atom positions. (b) Compute the density of the Fe crystal based on the atomic radius and (c) Compute the atomic packing fraction for Fe (volume of atoms/volume (d, e, f Do (a), (b) and (c) again for an aluminum (Al) crystal. (print and use pg. 3) the unit cell structure. Compare your answer to a literature. of unit cell). For the Fe crystal structure above determine the following: (a) The angle between [111] and [101] (b) The angle between (112) and (001) (c) The linear atomic density along [111 (d) The planar atomic density of (110) 5) Note: keep the units in atoms Inm or atoms InmExplanation / Answer
Answer (1):-
We know Molybdenum has radius of 0.143 nm
For BCC -
?3 * a0 = 4r , thus
a0 =4r/?3 = 4 * 0.143 nm /?3 = 0.33 nm
Answer (2) :-
The unit cell of an fcc lattice has right angles, and each face has gold atoms touching each other along the diagonal.
Thus, one can draw a right triangle whose legs both have length of a = 0.40788 nm and whose hypoteneuse is 4r, where r is the radius of a gold atom. By the Pythagorean theorem: a2 + a2 = (4r)2 2a2 = 16r2 r = (21/2/4)a Substituting in a = 0.40788 nm, r = 0.14421 nm