HW3 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 03 Question 4.1: Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy for vacancy formation of 0.55 eV/atom. Solution of 4.1: In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation 4.1. As stated in the problem, Q v = 0.55 eV/atom. Thus, N v N = exp Q v kT = exp 0.55 eV/atom (8.62 10 5 eV/atom- K )(600 K) = 2.41 10 -5
Background image of page 1

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

View Full DocumentRight Arrow Icon
Question 4.2: Calculate the number of vacancies per cubic meter in iron at 850 C. The energy for vacancy formation is 1.08 eV/atom. Furthermore, the density and atomic weight for Fe are 7.65 g/cm 3 and 55.85 g/mol, respectively. Solution of 4.2: Determination of the number of vacancies per cubic meter in iron at 850 C (1123 K) requires the utilization of Equations 4.1 and 4.2 as follows: N v = N exp Q v kT = N A Fe A Fe exp Q v kT And incorporation of values of the parameters provided in the problem statement into the above equation leads to N v = ( 6.022 10 23 atoms /mol )( 7.65 g /cm 3 ) 55.85 g/mol exp 1.08 eV/atom (8.62 10 5 eV/atom K )(850 C + 273 K) = 1.18 10 18 cm -3 = 1.18 10 24 m -3
Background image of page 2
Question 4.5: For both FCC and BCC crystal structures, there are two different types of interstitial sites. In each case, one site is larger than the other, and is normally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0 1 2 1 4 positions—that is, lying on {100} faces, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystal structures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom. Solution of 4.5: In the drawing below is shown the atoms on the (100) face of an FCC unit cell; the interstitial site is at the center of the edge. The diameter of an atom that will just fit into this site (2 r ) is just the difference between that unit cell edge length ( a ) and the radii of the two host atoms that are located on either side of the site ( R ); that is 2 r = a – 2 R However, for FCC a is related to R according to Equation 3.1 as a 2 R 2 ; therefore, solving for r from the above equation gives r = a 2 R 2 = 2 R 2 2 R 2 = 0.41 R A (100) face of a BCC unit cell is shown below.
Background image of page 3

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

View Full DocumentRight Arrow Icon
The interstitial atom that just fits into this interstitial site is shown by the small circle. It is situated in the plane of
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/11/2011 for the course ENG 407 taught by Professor Tsakal during the Spring '11 term at Rutgers.

Page1 / 15

HW3 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

This preview shows document pages 1 - 5. Sign up to view the full document.

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