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Solid State Test - 31

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Solid State Test - 31
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  • Question 1
    1 / -0
    If the co-ordination number of an element in its crystal lattice is 8, then packing is :
    Solution

    The co-ordination numbers of an element in fcc, hcp and bcc crystal lattice are 1212, 1212 and 88 respectively. Hence, if the coordination number of an element in its crystal lattice is 88then packing is bcc.

  • Question 2
    1 / -0
    In a simple cubic cell, each point on a corner is shared by:
    Solution
    In a simple cubic cell, each point on a corner is shared by 88 unit cells.
    Thus, each point on a corner contributes one-eight to the unit cell.

    Hence, the correct option is C\text{C}

  • Question 3
    1 / -0
    The available space occupied by spheres of equal size in three dimensions in both hcp and ccp arrangement is:
    Solution
     Solution:(z= no. of atoms per unit cell) \text { Solution:}-\quad(z=\text { no. of atoms per unit cell) } \\
     [Percentage of occupied volume =z× Volume of sphere  volume of cube /hexagon]\text { [Percentage of occupied volume }=z \times \dfrac{\text { Volume of sphere }}{\text { volume of cube /hexagon}}]

    z=4,z=4,
     4π=a24 \pi=a \sqrt{2} \\
    r=a2r=a \sqrt{2} \\
     percentage =4×4/3πγ3a3×100%\text { percentage }=4\times \dfrac{4 / 3 \pi \gamma^{3}}{a^{3}} \times 100 \%
    =163×πγ3a3=\dfrac{16}{3} \times \dfrac{\pi \gamma^{3}}{a^{3}} \\
    =π32×100=\dfrac{\pi}{3 \sqrt{2}} \times 100 \\
    $$\text { (%of occupied Vol. } =74 \% \text { ) }$$

     for hcp, \text { for hcp, }

    z=62r=az=6 \quad 2 r=a \\
    r=9/2r=9 / 2 \\
     percentage =6×4/3πγ332a3\text { percentage }=6 \times \dfrac{4 / 3 \pi \gamma^{3}}{3 \sqrt{2} a^{3}}
    $$
     
     (Volume of hexagon =32a3)\text { (Volume of hexagon } \left.=3 \sqrt{2} \mathrm{a}^{3}\right) \\
    =πγ332a3×100\quad=\dfrac{\pi \gamma^{3}}{3 \sqrt{2} a^{3}} \times 100 \\
    =π32×100=\dfrac{\pi}{3 \sqrt{2}} \times 100
     
    $$\text { (%of occupied Vol. } =74 \% \text { ) }$$
     
     Hence correct ans is (A)\text { Hence correct ans is }(A)
  • Question 4
    1 / -0
    For an orthorhombic system, axial ratios are abca\neq b\neq c and the axial angles are:
    Solution
    For orthorhombic system axial ratios are abca\neq b\neq c and the axial angles are   α=β=γ=90\;\alpha=\beta=\gamma=90^{\circ}. Thus, all the three edge lengths are unequal but all the angles are equal. They are equal to 90090^0.
  • Question 5
    1 / -0
    An element crystallizes in a structure having a fcc unit cell with edge length 200200 pm. Calculate its density (in g cm3^{-3}) if 200200 g of this element contain 24×102324\times10^{23} atoms.
    Solution
    Weight of 24×102324\times10^{23} atoms =200=200 g   
    \therefore\, Weight of 44 atoms =20024×1023×4=\displaystyle\frac{200}{24\times10^{23}}\times4 g   
    The fcc structure contains 44 atoms in a unit cell.
    Density =Mass of unit cellVolume of unit cell=\displaystyle\frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} =200×424×1023×(200×1010)3=\displaystyle\frac{200\times4}{24\times10^{23}\times(200\times10^{-10})^3} =41.6=41.6 g cm3^{-3} 
    Hence, the density of the unit cell is 41.641.6 g cm3^{-3}.
  • Question 6
    1 / -0
    Which of the following is a primitive unit cell?
  • Question 7
    1 / -0
    The more efficient mode of packing of identical atoms in one layer is :
    Solution
    The more efficient mode of packing of identical atoms in one layer is hexagonal close packing pattern.
    This is because the empty space in hexagonal closed packing is much less than that in square close packing.
  • Question 8
    1 / -0
    A tetrahedral void in a crystal implies that:
    Solution
    A tetrahedral void in a crystal implies that the void is surrounded tetrahedrally by four spheres. The four spheres are located at the four corners of the tetrahedron.
  • Question 9
    1 / -0
    A metallic crystal crystallizes into a lattice containing a sequence of layers AB AB AB.... . Any packing of spheres leaves out voids in the lattice. What percentage of the volume of this lattice is empty space?
    Solution

    AB AB AB describes the hexagonal close-packed arrangement of spheres.

    The available space occupied by spheres of equal size in three dimensions in the hexagonal close packed arrangement is 74%. 

    The vacant space is 10074=26100-74=26%.

    Hence, the correct option is B\text{B}

  • Question 10
    1 / -0
    The number of octahedral sites in a cubical close packed array of NN spheres is :
    Solution
    The ratio of spheres and octahedral void in ccp is 1:11:1. Therefore, for N spheres, the number of octahedral voids will be N. 
    The number of octahedral sites in a cubical close packed array of N spheres is N. 
    For example, a ccp lattice contains 44 atoms per unit cell. It has 44 octahedral voids and 88 tetrahedral voids.
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