HW9 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 09 12.14 Magnesium oxide has the rock salt crystal structure and a density of 3.58 g/cm 3 . (a) Determine the unit cell edge length. (b) How does this result compare with the edge length as determined from the radii in Table 12.3, assuming that the Mg 2+ and O 2- ions just touch each other along the edges? Solution (a) This part of the problem calls for us to determine the unit cell edge length for MgO. The density of MgO is 3.58 g/cm 3 and the crystal structure is rock salt. From Equation 12.1 n ( A Mg + A O ) V C N A n ( A Mg A O ) a 3 N A Or, solving for a a n ( A Mg A O ) N A 1/3 Inasmuch as there are 4 formula units per unit cell for the rock salt crystal structure, and the atomic weights of magnesium and oxygen are 24.31 and 16.00 g/mol, respectively, when we solve for a from the above equation a (4 formula units/unit cell)(24.31 g/mol + 16.00 g/mol) ( 3.58 g/cm 3 )( 6.022 10 23 formula units/mol ) = 4.21 10 -8 cm = 0.421 nm (b) The edge length is determined from the Mg 2+ and O 2- radii for this portion of the problem. Now for the rock salt crystal structure a = 2 r Mg 2+ + 2 r O 2- From Table 12.3 a = 2(0.072 nm) + 2(0.140 nm) = 0.424 nm
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
12.15 Compute the theoretical density of diamond given that the C—C distance and bond angle are 0.154 nm and 109.5°, respectively. How does this value compare with the measured density? Solution This problem asks that we compute the theoretical density of diamond given that the C—C distance and bond angle are 0.154 nm and 109.5 , respectively. The first thing we need do is to determine the unit cell edge length from the given C—C distance. The drawing below shows the cubic unit cell with those carbon atoms that bond to one another in one-quarter of the unit cell. From this figure, is one-half of the bond angle or = 109.5 /2 = 54.75 , which means that =9 0  54.75 = 35.25 since the triangle shown is a right triangle. Also, y = 0.154 nm, the carbon-carbon bond distance. Furthermore, x = a /4, and therefore, x = a 4 y sin Or a = 4 y sin = (4)(0.154 nm)(sin 35.25 ) = 0.356 nm = 3.56 10 -8 cm The unit cell volume, V C is just a 3 , that is V C a 3 = (3.56 10 -8 cm) 3 4.51 10 23 cm 3
Background image of page 2
We must now utilize a modified Equation 12.1 since there is only one atom type. There are 8 equivalent atoms per unit cell, and therefore = nÕA C V C N A (8 atoms/unit cell)(12.01 g/g- atom) ( 4.51 10 -23 cm 3 /unit cell )( 6.022 10 23 atoms/g- atom ) = 3.54 g/cm 3 The measured density is 3.51 g/cm 3 . 12.16 Compute the theoretical density of ZnS given that the Zn—S distance and bond angle are 0.234 nm and 109.5°, respectively. How does this value compare with the measured density? Solution This problem asks that we compute the theoretical density of ZnS given that the Zn—S distance and bond angle are 0.234 nm and 109.5 , respectively. The first thing we need do is to determine the unit cell volume from the given Zn—S distance. From the previous problem, the unit cell volume V C is just a 3 , a being the unit cell edge length, and V C (4 y sin ) 3 = (4)(0.234 nm)(sin 35.25 )  3 = 0.1576 nm 3 = 1.576 10 -22 cm 3 Now we must utilize Equation 12.1 with
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 19

HW9 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online