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Solid State Test - 4

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Solid State Test - 4
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  • Question 1
    1 / -0.25

    Packing efficiency of body centred cubic unit cell is:

    Solution

    The packing efficiency of a body-centred cubic unit cell is 68%.

    Explanation: 

    The packing efficiency is the fraction of the crystal (or unit cell) actually occupied by the atoms. It is usually represented by a percentage or volume fraction.

    Mathematically Packing Efficiency is:

    Number of atoms ×volume occupied by 1 share ×100/Total volume of unit cell ×100

  • Question 2
    1 / -0.25

    Packing efficiency in a unit cell is never 100% because constituent particles are assumed to be:

    Solution

     Packing efficiency can never be 100% because in packing calculations all constituent particles filling up the cubical unit cell are assumed to be spheres.

  • Question 3
    1 / -0.25

    The packing efficiency of the two-dimensional square unit cell shown in the given figure is:

    Solution

    Let us take the radius of each atom to be ‘r ’. So, from the structure we can say that the diagonal of the square unit is ‘4r ’. …….(1)

    Let the side of the square unit be ‘L ’. So, the diagonal of the square will be:

     

    Now since we have 2 values for the diagonal, let ’s equate them to each other.
    (1) = (2)

    Total area (A) of the square unit can be written as:
    A  = L2
    In the above equation put the value of L from equation (3):


    = 16r2 /2
    = 8r2  ……….(4)
    Total number of spheres inside the square unit = 1 (the one at the centre) + 4 (1/4) (from the 4 spheres at the corners, each contributing 1 / 4 to the square unit cell)
    = 1 + 4 ( 1/4 ) = 2
    Area of 1 sphere = πr2
    Since there are 2 spheres, the total area of spheres will be = 2(πr2 ) ………...(5)
    Packing fraction or packing efficiency of the square unit cell is defined as the ratio of area occupied by the spheres and total area of the square unit cell
    Use the values of the area occupied by the sphere from equation (5) and total area of square unit cell from equation (4).
    Packing efficiency = total area of spheres / total area of the square unit cell
    = 2(πr2 )/8r2
    = π/4  (π= 3.14)
    = 0. 785

    So, the percentage packing efficiency will be 78.5%.
    The correct option is: (D) 78.54%

  • Question 4
    1 / -0.25

    Packing efficiency in a unit cell is never 100% because constituent particles are assumed to be:

    Solution

    The constituent particles i.e. atoms, molecules and ions are assumed to be spheres.

  • Question 5
    1 / -0.25

    What is the dimensional formula of packing fraction?

    Solution

    Packing fraction is a dimensionless quantity which is the ratio of space occupied to total crystal space available. Since both the quantities have the same units the ratio renders dimensionless.

  • Question 6
    1 / -0.25

    Arrange the types of arrangement in terms of decreasing packing efficiency.

    Solution

    HCP and CCP have the highest packing efficiency of 74% followed by BCC which is 68%. The simple cubic structure has a packing efficiency of 54%.

  • Question 7
    1 / -0.25

    Which of the following shows maximum packing efficiency?

    Solution

    Packing efficiency is defined as the percentage of space occupied by constituent particles packed inside the lattice.

    Packing efficiency = Number of atoms ×volume occupied by one share ×100/total volume of the unit cell

    • Packing efficiency in fcc and hcp structures is 74%

    • The packing efficiency of the body-centred unit cell is 68%.

    • The packing efficiency of the simple unit cell is 52.4%

     

    Since HCP shows the maximum packing efficiency, the correct answer is A. 

  • Question 8
    1 / -0.25

    What are the percentages of free space in a CCP and simple cubic lattice?

    Solution

    The packing efficiency in CCP and simple cubic lattice are 74% and 52%, respectively. Hence the corresponding free spaces will be 100% –74% = 26% and 100% –52% = 48%.

  • Question 9
    1 / -0.25

    If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC fashion then what is the surface area of each unit cell?

    Solution

    ABCABC arrangement is found in CCP.
    In closed cubic packing, relation between edge length of unit cell, a, and radius of particle, r, is given as a=2 √2r.
    Surface area (S.A.) = 6a2
    From the relationship,
    a2  = 8r2
    S.A. = 6a2  = 48r2
    When r = 166 pm, S.A. = 48(166pm) = 1.32 x 106  pm2

  • Question 10
    1 / -0.25

    If copper, density = 9.0 g/cm3 and atomic mass 63.5, bears face-centered unit cells then what is the ratio of surface area to volume of each copper atom?

    Solution

    Density, d of unit cell is given by d
    Given,
    Density, d = 9.0 g/cm3
    Atomic mass, M = 63.5 g/mole
    Edge length = a
    NA  = Avogadro ’s number = 6.022 x 1023
    z = 4 atoms/cell
    On rearranging the equation for density we get a3  
    Substituting the given values:

    Therefore, a = 360.5 pm
    The relation of edge length ‘a ’and radius of particle ‘r ’for FCC packing i.e. a = 2 √2r.
    On substituting the value of ‘a ’in the given relation, 127.46 pm
    Now, for spherical particles volume, V = 4 πr3 /3 and surface area, S = 4 πr2
    Required ratio = S/V=4 πr2 /(4 πr3 /3) = 3/r (after simplifying)
    Thus, S/V = 3/127.46 = 0.0235.

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