Manufacturing Engineering Questions & Answers – Crystallography-2

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This set of Manufacturing Engineering Interview Questions and Answers focuses on “Crystallography-2”.

1. What is the coordination number of a simple cubic (SC) unit cell?
a) 4
b) 6
c) 8
d) 2
View Answer

Answer: b
Explanation: There are six nearest neighbouring atoms for every atom in a simple cubic (SC) unit cell, in other words, every atom in a SC unit cell is surrounded by 6 other atoms, thus coordination number of SC unit cell is is 6.

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2. What is the coordination number of a face centered cubic (FCC) unit cell?
a) 4
b) 6
c) 8
d) 12
View Answer

Answer: d
Explanation: In an FCC structure, there are eight atoms, one atom each at the corner of the unit cell and one atom at the centre of each face. For any corner atom of the unit cell, the nearest atoms are face-centred atoms. Thus, the coordination number for an FCC structure = 4 centre atoms below the horizonal plane + 4 centre atoms above the horizontal plane + 4 centre atoms on the horizonal plane.
Hence, the coordination number for an FCC structure is 4 + 4 + 4 = 12.

3. What is the coordination number of body centered cubic unit cell?
a) 4
b) 6
c) 8
d) 2
View Answer

Answer: c
Explanation: For any corner atom of the BCC unit cell, the nearest atoms are the body centred atoms. There are eight-unit cells in neighbours with body-centered atoms. Hence, the coordination number for a BCC cubic unit cell is 8.

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4. Effective number of atoms in a simple cubic (SC) unit cell is equal to _________
a) 4
b) 1
c) 8
d) 2
View Answer

Answer: b
Explanation: Total number of atoms at corners = 8 and each corner atom is shared by total 8-unit cells. Thus, effective number of atoms in an SC unit cell: 8 × 8⁄8 = 1.

5. Effective number of atoms in a face centered cubic (FCC) unit cell is equal to ________
a) 4
b) 1
c) 8
d) 2
View Answer

Answer: a
Explanation: In an FCC unit cell, there are eight atoms: one at each corner of the cube and six face centered atoms of the six planes of the cube. As corner atoms are shared by eight adjacent cubes and the face centered atoms by two adjacent unit cells, total effective number of atoms in an FCC unit cell will be 4.
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6. Effective number of atoms in a body centered cubic (BCC) unit cell is equal to _____________
a) 4
b) 6
c) 1
d) 2
View Answer

Answer: d
Explanation: The unit cell of a cube contains eight atoms at the corners, which are shared by the eight adjoining cubes and one atom at the centre of the cube.
8 atoms at the corner: (8×18) = 1 atom + 1 centre atom in the unit cell, So, there are “2” effective number of atoms in a BCC unit cell.

7. The atomic packing fraction in a simple cubic unit cell is ________
a) 0.74
b) 0.52
c) 0.68
d) 0.66
View Answer

Answer: b
Explanation: a=r and APF = (volume of effective number of atoms/volume of unit cell).
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8. The atomic packing fraction in a body centered cubic unit is cell is ________
a) 0.74
b) 0.52
c) 0.68
d) 0.66
View Answer

Answer: c
Explanation: r=(√3)/4×a and APF = (volume of effective number of atoms/volume of unit cell).

9. If the radius of a copper atom is given as 1.27 Ao, its density (in kg/m3) will be?
a) 100.01
b) 86.25
c) 8979
d) 7968
View Answer

Answer: c
Explanation: Formula to calculate the density of a cubic metal:
ρ (kg/m3) = \(\frac{n × A.W}{a^3}\) × 1.66 × 10-27
[where, ρ = density of metal, n = effective number of atoms per unit cell, A.W = Atomic weight of the metal in amu and a = lattice parameter in meter] Given: radius of copper = 1.27 Ao = 1.27×10-10 m
We know that atomic weight of copper = 63.5 amu
Lattice parameter to atomic radius relation for cubic structures are as follows:

Crystal Structure Effective Number of Atoms per Unit Cell Effective Number of Atoms per Unit Cell
Simple Cubic (SC) a = 2r 1
Body Centered Cubic (BCC) a = 4r/√3 2
Face Centered Cubic (FCC) a = 4r/√2 4
Hexagonal Close Packed Cubic (HCP) a = 2r 6
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We know that copper has FCC crystal structure, so it has ‘4’ effective number of atoms per unit cell and given its atomic radius = 1.27×10-10 m
Therefore, lattice parameter of copper (a) = (4×1.27×10-10)/√2 = 3.59×10-10 m
Therefore, density of copper = \(\frac{4 × 63.5}{(3.59 × 10^{-10})^3}\) × 1.66 × 10-27 ≈ 8979 kg/m3.

10. The atomic packing fraction in a face centered cubic unit is?
a) 0.74
b) 0.52
c) 0.68
d) 0.66
View Answer

Answer: a
Explanation: a=2√2×r and APF = (volume of effective number of atoms/volume of unit cell).

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter