Essentials Of Materials Science And Engineering
Essentials Of Materials Science And Engineering
4th Edition
ISBN: 9781337385497
Author: WRIGHT, Wendelin J.
Publisher: Cengage,
Question
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Chapter 9, Problem 9.20P
Interpretation Introduction

(a)

Interpretation:

The critical radius of the nucleus in the liquid iron needs to be determined.

Concept Introduction:

Homogeneous nucleation is the process in which the nuclei that are formed randomly and spontaneously grows irreversibly and form into a new phase.

In this question, we need to calculate the critical radius of the nucleus.

The critical radius of the nuclei is the minimum size of the nucleus required for the formation of a new stable nucleus.

Expert Solution
Check Mark

Answer to Problem 9.20P

From the equation of the critical radius for the homogeneous equation, the value of the critical radius is 10.128×108cm.

Explanation of Solution

Given Information:

From the parameter of the iron, values of different properties are:

Latent heat of fusion, ΔHf=1737J/cm3

Surface free energy of solid-liquid phase, σsl=204×107J/cm3

Formula used:

The critical radius of homogeneous nucleation can be calculated by,

   rk=2σ slTmΔHfΔTmWhere,σsl=Surfacefreeenergyofsolid-liquidphase Tm=Equillibriumsolidificationtemperature ΔHf=Latentheatoffusion ΔTm=Temperaturedifferenceforundercooling

Calculation:

Temperature difference for undercooling, ΔT=420C0

Equilibrium solidification temperature, Tm=1538C0=1538+273=1811 K

The equation for the critical radius is,

  rk=2σslTmΔHfΔTm

Putting values in the equation,

  rk=2×204× 10 7×18111737×420rk=10.128×108cm

Interpretation Introduction

(b)

Interpretation:

The number of iron atoms in the nucleus needs to be determined.

Concept Introduction:

Homogeneous nucleation is the process in which the nuclei that are formed randomly and spontaneously grows irreversibly and form into a new phase.

In this section, we need to calculate a number of atoms in the iron nucleus.

The nucleus consists of a different number of neutrons, protons, and electrons.

Expert Solution
Check Mark

Answer to Problem 9.20P

From the equation of volume of the nucleus with a critical radius for the homogeneous equation, we get the value of a number of iron atoms in the nucleus is 350 atoms.

Explanation of Solution

Given Information:

The lattice parameter of solid BCC iron, a0=2.92A0=2.92nm

Formula Used:

The volume of the nucleus can be calculated by,

   V nucleus = 4 3 π ( r k ) 3 where,rkisthecriticalradiusNumberofunitcell=V nucleusVwhere,Vnucleusisvolumeofnucleus Visvolumeofunitcell

Calculation:

The volume of the unit cell,

  V=(a0)3=(2.92)3=24.897×1024cm3

The radius of the nucleus,

  rk=10.128×108cm

The volume of the nucleus,

   V nucleus = 4 3 π ( r k ) 3 =43π(10.128×108)3 Vnucleus=4352×1024cm3Numberofunitcell=V nucleusV =4352× 10 2424.897× 10 24 =175

Atoms per nucleus = (Number of the unit cell) × (Number of atoms in the unit cell)

  = 1752                           =350 atoms

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Chapter 9 Solutions

Essentials Of Materials Science And Engineering

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