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In a BCC unit cell, if half of the atoms...

In a BCC unit cell, if half of the atoms per unit cell are removed, then percentage void is

A

`68%`

B

`32%`

C

`34%`

D

`66%`

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The correct Answer is:
To solve the problem of finding the percentage void in a BCC unit cell after removing half of the atoms, we can follow these steps: ### Step 1: Determine the number of atoms in a BCC unit cell In a Body-Centered Cubic (BCC) unit cell, there are: - 1 atom at the center - 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell. Thus, the total number of atoms per unit cell is: \[ \text{Total atoms} = 1 + 8 \times \frac{1}{8} = 1 + 1 = 2 \text{ atoms} \] ### Step 2: Calculate the number of atoms remaining after removal According to the problem, half of the atoms are removed from the BCC unit cell: \[ \text{Remaining atoms} = \frac{2}{2} = 1 \text{ atom} \] ### Step 3: Calculate the volume of the unit cell Let \( A \) be the edge length of the unit cell. The volume of the unit cell is given by: \[ \text{Volume of unit cell} = A^3 \] ### Step 4: Calculate the volume of one atom The radius \( r \) of the atom in a BCC lattice can be expressed in terms of the edge length \( A \): \[ \sqrt{3}A = 4r \implies r = \frac{\sqrt{3}A}{4} \] The volume of one atom is given by the formula for the volume of a sphere: \[ \text{Volume of one atom} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{\sqrt{3}A}{4}\right)^3 \] Calculating this gives: \[ \text{Volume of one atom} = \frac{4}{3} \pi \left(\frac{3\sqrt{3}A^3}{64}\right) = \frac{\sqrt{3}\pi A^3}{16} \] ### Step 5: Calculate the packing efficiency The packing efficiency is defined as the ratio of the total volume of the atoms to the volume of the unit cell: \[ \text{Packing efficiency} = \frac{\text{Number of atoms} \times \text{Volume of one atom}}{\text{Volume of unit cell}} \times 100 \] Substituting the values: \[ \text{Packing efficiency} = \frac{1 \times \frac{\sqrt{3}\pi A^3}{16}}{A^3} \times 100 = \frac{\sqrt{3}\pi}{16} \times 100 \] Calculating this gives approximately: \[ \text{Packing efficiency} \approx 33.99\% \approx 34\% \] ### Step 6: Calculate the percentage void The percentage void is calculated as: \[ \text{Percentage void} = 100\% - \text{Packing efficiency} \] Substituting the packing efficiency: \[ \text{Percentage void} = 100\% - 34\% = 66\% \] ### Final Answer The percentage void in the BCC unit cell after removing half of the atoms is **66%**. ---
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AAKASH INSTITUTE ENGLISH-THE SOLID STATE -Assignment (SECTION - B) (OBJECTIVE TYPE QUESTION)
  1. In a unit cell, atoms A, B, C and D are present at corners, face - cen...

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  2. In a CsCl structure, if edge length is a, then distance between one Cs...

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  3. The correct statement about, CCP structure is

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  4. In a NaCl structure, if positions of Na atoms and Cl - atoms are inter...

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  5. If radius of a metal atom (A) is 5pm and radius of an electronegative ...

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  6. A metal can be crystallized in both BCC and FCC unit cells whose edge ...

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  7. In a unit cell containing X^(2+), Y^(3+) and Z^(2-) where X^(2+) occu...

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  8. On rising temperature and decreasing pressure in CsCl solid

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  9. In a ccp type structure, if half of atoms are removed from face cente...

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  10. In a BCC unit cell, if half of the atoms per unit cell are removed, th...

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  11. Calculate the number of atoms in a cube based unit cell having one ato...

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  12. Number of unit cells in 10 g NaCl

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  13. Some of the molecular solid upon heating produces small amount of elec...

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  14. NaCl becomes paramagnetic at high temperature due to

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  15. How many Cs^+ ions occupy the second nearest neighbour location of a C...

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  16. The ratio of number of rectangular plane and diagonal plane in a cubic...

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  17. In the calcium fluoride structure, the coodination number of the catio...

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  18. which of the following defects decrase the density decrease the d...

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  19. Glass is a

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  20. The packing efficiency of the 2D square unit cell shown below is

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