Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 2, Problem 2.7P
To determine

(a)

Thenumbers of atoms for FCC nickel (Ni).

Expert Solution
Check Mark

Answer to Problem 2.7P

The numbers of atoms for FCC nickel (Ni) is 91700atoms.

Explanation of Solution

Given:

The side of cube is 10nm

Lattice parameter for nickel is 0.352nm

Formula used:

Write the expression to find the number of atoms per unit volume.

na=Numberofatomsa3 …… (I)

Here, na is the number of atoms per unit volume and the Numberofatoms isthe numbers of atoms in a unit cell.

Write the expression to find the number of atom for the volume.

N=naa13 …… (II)

Here N is the number of atoms for the cube, a1 is the side of cube.

Calculation:

Nickel has Face Centered Cubic (FCC) structure with 4 number of atoms.

Substitute 4atoms for Numberofatoms and 0.352nm for a in equation (I) to find na.

na=4atoms ( 0.352nm )3=4atoms ( 0.352nm× 10 9 m 1nm )3=4atoms0.044× 10 27m3=91.70×1027atoms/m3

Substitute 91.70×1027atoms/m3 for na, and 10nm for a1 in equation (II) to find N.

N=91.70×1027atoms/m3(10nm)3=91.70×1027atoms/m3(10nm× 10 9 m 1nm)3=(91.70× 10 27atoms/ m 3)×( 103× 10 27m)=91700atoms

Conclusion:

Therefore, the numbers of atoms for FCC nickel (Ni) is 91700atoms.

To determine

(b)

The number of atoms by using the density of nickel.

Expert Solution
Check Mark

Answer to Problem 2.7P

The number of atoms by using the density of nickel is 91300atoms.

Explanation of Solution

Formula used:

Write the formula to find theatoms per unit volume.

NV=ργNi×NAMNi …… (III)

Here, NV is the atom per unit volume, ργNi is thedensity of Nickel, MNi is the molecular weight of nickel, and NA is the Avogadro’s number.

Write the expression to find the number of atom for the volume.

N=NVa13 ……. (IV)

Here, N is the number of atoms for the given volume.

Calculation:

The molecular weight of nickel is 58.71×103kg/mole.

The density of nickel is 8.902g/cm3.

Avogadro’s number is 6.02×1023atoms/mole.

Substitute 8.902g/cm3 for ργNi, 58.71×103kg/mole for MNi and 6.02×1023atoms/mole for NA in equation (II) to find NV.

NV=8.902g/ cm 3×6.02× 10 23atoms/mole58.71× 10 3kg/mole=8.902g/ cm 3×( 10 3 kg 1g )×( 1 cm 3 10 6 m 3 )×6.02× 10 23atoms/mole58.71× 10 3kg/mole=0.913×1029atoms/m3

Substitute 0.913×1029atoms/m3 for NV and 10nm for a1 in equation (IV) to find N.

N=0.913×1029atoms/m3(10nm)3=0.913×1029atoms/m3(10nm× 10 9 m 1nm)3=(91.3× 10 27atoms/ m 3)×( 103× 10 27m3)=91300atoms

Conclusion:

Therefore, the number of atoms by using the density of nickel is 91300atoms.

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Chapter 2 Solutions

Materials Science And Engineering Properties

Ch. 2 - Prob. 11CQCh. 2 - Prob. 12CQCh. 2 - Prob. 13CQCh. 2 - Prob. 14CQCh. 2 - Prob. 15CQCh. 2 - Prob. 16CQCh. 2 - Prob. 17CQCh. 2 - Prob. 18CQCh. 2 - Prob. 19CQCh. 2 - Prob. 20CQCh. 2 - Prob. 21CQCh. 2 - Prob. 22CQCh. 2 - Prob. 23CQCh. 2 - Prob. 24CQCh. 2 - Prob. 25CQCh. 2 - Prob. 26CQCh. 2 - Prob. 27CQCh. 2 - Prob. 28CQCh. 2 - Prob. 29CQCh. 2 - Prob. 30CQCh. 2 - Prob. 31CQCh. 2 - Prob. 32CQCh. 2 - Prob. 33CQCh. 2 - Prob. 34CQCh. 2 - Prob. 35CQCh. 2 - Prob. 36CQCh. 2 - Prob. 37CQCh. 2 - Prob. 38CQCh. 2 - Prob. 39CQCh. 2 - Prob. 40CQCh. 2 - Prob. 41CQCh. 2 - Prob. 42CQCh. 2 - Prob. 43CQCh. 2 - Prob. 44CQCh. 2 - Prob. 45CQCh. 2 - Prob. 46CQCh. 2 - Prob. 47CQCh. 2 - Prob. 48CQCh. 2 - Prob. 49CQCh. 2 - Prob. 50CQCh. 2 - Prob. 51CQCh. 2 - Prob. 52CQCh. 2 - Prob. 1ETSQCh. 2 - Prob. 2ETSQCh. 2 - Prob. 3ETSQCh. 2 - Prob. 4ETSQCh. 2 - Prob. 5ETSQCh. 2 - Prob. 6ETSQCh. 2 - Prob. 7ETSQCh. 2 - Prob. 8ETSQCh. 2 - Prob. 9ETSQCh. 2 - Prob. 10ETSQCh. 2 - Prob. 11ETSQCh. 2 - Prob. 12ETSQCh. 2 - Prob. 13ETSQCh. 2 - Prob. 1DRQCh. 2 - Prob. 2DRQCh. 2 - Prob. 3DRQCh. 2 - Prob. 4DRQCh. 2 - Prob. 5DRQCh. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10PCh. 2 - Prob. 2.11PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - Prob. 2.16PCh. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. 2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.26P
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Materials Science And Engineering Properties
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ISBN:9781111988609
Author:Charles Gilmore
Publisher:Cengage Learning