Foundations of Materials Science and Engineering
Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 16.13, Problem 65AAP
To determine

The average magnetic moment for cobalt.

Expert Solution & Answer
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Answer to Problem 65AAP

The average magnetic moment for cobalt is 1.7Bohr magnetons/atom.

Explanation of Solution

Write the expression to calculate average magnetic moment for cobalt (N).

  N=MsμB(Atomic density)                                                                                  (I)

Here, saturation magnetization is Ms and Bohr magneton is μB.

Write the expression to calculate atomic density of the cobalt.

  Atomic density=nV                                                                                             (II)

Here, number of atoms per unit cell is n and volume of the unit cell is V.

Write the expression to calculate volume of the unit cell for the cobalt (V).

  V=Area×height=(3a2sin60°)×c                                                                                            (III)

Here, lattice constant of the unit cell is a and height of the unit cell is c.

Conclusion:

Refer "Problem 16.64" in the textbook, obtain the values for lattice constant (a) and height (c) for cobalt.

  a=0.25071nmc=0.40686nm

Substitute 0.25071nm for a and 0.40686nm for c in Equation (III).

 V=(3(0.25071nm)2sin60°)×0.40689nm=(3(0.25071nm×1×109m1nm)2sin60°)×(0.40689nm×1×109m1nm)=6.644×1029m3

Substitute 6atoms/unit cell for n and 6.644×1029m3 for V in Equation (II).

 Atomic density=6atoms/unit cell6.644×1029m3=9.030×1028atoms/m3

The value of Bohr magneton (μB) is taken to be 9.27×1024Am2.

Substitute 1.42×106A/m for Ms, 9.030×1028atoms/m3 for Atomic density, and 9.27×1024Am2 for μB in Equation (I).

 N=1.42×106A/m(9.27×1024Am2)(9.030×1028atoms/m3)=1.7Bohr magnetons/atom

Thus, the average magnetic moment for cobalt is 1.7Bohr magnetons/atom.

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Chapter 16 Solutions

Foundations of Materials Science and Engineering

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