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Consider the atomic packing factor (APF) of the following crystal structures :
P. Simple cubic
Q. Body centred cubic
R. Face centred cubic
S. Diamond T. Hexagonal close packed
Q. Which two of the above structure have equal APF?
  • a)
    R and S
  • b)
    R and T
  • c)
    S and T
  • d)
    P and Q
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider the atomic packing factor (APF) of the following crystal stru...
simple cubic → 52%, bcc → 68%, fcc → 74%, diamond → 34%, hcp 74%
The correct answer is: R and T
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Consider the atomic packing factor (APF) of the following crystal stru...
Explanation:

The atomic packing factor (APF) is a measure of how efficiently atoms are arranged in a crystal structure. It is defined as the ratio of the total volume occupied by atoms to the total volume of the unit cell.

To determine which two crystal structures have equal APF, we need to calculate the APF for each structure and compare the values.

1. Simple Cubic (P)
In a simple cubic structure, the atoms are located only at the corners of the unit cell. Each corner atom contributes 1/8th of its volume to the unit cell.

- Number of atoms per unit cell (N) = 1
- APF = (Number of atoms per unit cell * Volume of one atom) / Volume of unit cell
- APF = (1 * (4/3)πr^3) / a^3
- APF = (4πr^3) / (3a^3)

2. Body Centered Cubic (Q)
In a body-centered cubic structure, the atoms are located at the corners and at the center of the unit cell. Each corner atom contributes 1/8th of its volume, and the center atom contributes its entire volume to the unit cell.

- Number of atoms per unit cell (N) = 2
- APF = (Number of atoms per unit cell * Volume of one atom) / Volume of unit cell
- APF = (2 * (4/3)πr^3) / a^3
- APF = (8πr^3) / (3a^3)

3. Face Centered Cubic (R)
In a face-centered cubic structure, the atoms are located at the corners and at the center of each face of the unit cell. Each corner atom contributes 1/8th of its volume, and each face atom contributes 1/2 of its volume to the unit cell.

- Number of atoms per unit cell (N) = 4
- APF = (Number of atoms per unit cell * Volume of one atom) / Volume of unit cell
- APF = (4 * (4/3)πr^3 + 6 * (1/2) * (4/3)πr^3) / a^3
- APF = (16πr^3 + 12πr^3) / (3a^3)
- APF = (32πr^3) / (3a^3)

4. Diamond (S)
In a diamond structure, each atom is surrounded by four equally spaced nearest neighbor atoms in a tetrahedral arrangement. Each atom contributes its entire volume to the unit cell.

- Number of atoms per unit cell (N) = 8
- APF = (Number of atoms per unit cell * Volume of one atom) / Volume of unit cell
- APF = (8 * (4/3)πr^3) / a^3
- APF = (32πr^3) / (3a^3)

5. Hexagonal Close Packed (T)
In a hexagonal close packed structure, the atoms are arranged in a closely packed hexagonal lattice with additional layers stacked on top. Each atom contributes its entire volume to
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Consider the atomic packing factor (APF) of the following crystal structures :P. Simple cubicQ. Body centred cubicR. Face centred cubicS. Diamond T. Hexagonal close packedQ.Which two of the above structure have equal APF?a)R and Sb)R and Tc)S and Td)P and QCorrect answer is option 'B'. Can you explain this answer?
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