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Answers_to_assn_1 - CHEM 123: Assignment # 1 page 1 of 10...

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CHEM 123: Assignment # 1 page 1 of 10 1. Simple cubic packing. The crystal lattic and unit cell for simple cubic packing are shown below. a) # spheres = () ( 1 8 sphere/corner 8 corners ) = 1 sphere The coordination number of each sphere is 6. To see this, focus on an “interior” sphere such as sphere “A” in the figure on the left. That sphere is in contact with 6 other spheres – the one above it, the one below, the one in front, the one behind, the one to its left and the one to its right. b ) If the radius of each sphere is “ R ” then the edge length of the unit cell is a = 2 R c) The volume of the unit cell is V cell = a 3 = 8 R 3 There is one full sphere contained within the unit cell so the volume of spheres contained within the unit cell is 3 spheres 4 3 VR = π . 3 4 3 3 0.5236 6 8 spheres cell V R Packing Efficiency V R π π == = = 2. Show that the following data are consistent with the fact that silver metal crystallizes in a face-centred cubic lattice: Edge length of unit cell, a = 408 pm Density of silver, 10.6 g cm 3 Atomic weight of silver, 107.9 g mol 1 , Avogadro's number, N A = 6.022 × 10 23 If silver crystallizes in a face-centred cubic lattice, then 1 23 1 3 10 3 107.9 g mol 6.022 10 atoms mol (4atoms) 10.6 g cm ( 408 10 cm) cell cell m V d × × ⎛⎞ ⎜⎟ ⎝⎠ = A The calculated density agrees well with the observed density.
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CHEM 123: Assignment 1 page 2 of 10 2 3 . A tetrahedral site ican be generated by placing four spheres of radius R at alternate corners of a cube. Since the spheres are in contact at the centre of each cube face, the length of the face diagonal of this cube is equal to 2 R . Since the centre of the tetrahedral site is located at the centre of the cube, the radius of the tetrahedral hole ( r t ) is equal to the difference between half the body diagonal and R . What is the radius of the tetrahedral hole? At left is an “exploded view” of a small sphere in a tetrahderal hole. In reality the larger spheres (of radius R ) are in contact along the cube faces. Let a be edge length, FD the length of the face-diagonal and BD the length of the body- diagonal of the cube above. The following right-angled triangles can be identified: FD = 2 BD a a FD = 2 From the first triangle: 22 2 2 2 (2 ) 2 4 2 aa R a R a R += = ⇒= Using the second triangle: ( ) 2 2 2 2 2( 2 ) 6 6 B DaF D R R R B D R =+ = + = = The distance from the center of the small sphere (at the centre of the cube) to the centre of the larger sphere (at the corner of the cube) is 1 2 6 2 BD R = . This distance is also equal to r t + R where r t
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Answers_to_assn_1 - CHEM 123: Assignment # 1 page 1 of 10...

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