The Solid State

  • Question 233
    CBSEENCH12005581

    Calculate the efficiency (percentage of volume occupied and unoccupied) of packing in case of a metal crystal for simple cubic.

    Solution

    Let ‘a’ be the edge length of the unit cell and r be the radius of sphere.

    As sphere are touching each other

    Therefore a = 2r

    No. of spheres per unit cell = 1/8 × 8 = 1 

    Volume of the sphere = 4/3 πr3 

    Volume of the cube = a3= (2r)3 = 8r3

    ∴ Fraction of the space occupied = 1/3πr3 / 8r3 = 0.524

    ∴ % occupied = 52.4 %



    Question 234
    CBSEENCH12005582

     (a) Determine the type of cubic lattice to which a given crystal belongs if it has edge length of 290 pm and density is 7.80 g cm–3. (Molecular mass.= 56 g mol–1)

    (b) Why does zinc oxide exhibit enhanced electrical conductivity on heating?

    Solution

    we know that

    d =ZMa3NA

    as we have given 
    d= 7.80 g cm–3.
    M=56 g mol–1
    a= 290 pm or a3 =2.43 x10-23
    putting all value in above equation we get,

    Z =a3 x d xNAMZ= 2.43 x10-23 x 7.80 x6.022 x102356 =2.03

    hence it belong to bcc crystal lattice.

    b) Zinc oxide is white in colour at room temperature. On heating it loses electron and turns yellow in colour. 


    ZnO Zn2+ +12O2 +2e- 

    The electron liberate after heating, can act as charge carrier and thus on heating zinc oxide it exhibit electrical conductivity.

    Question 235
    CBSEENCH12005583

    The compound CuCl has the ZnS (Cubic) structure. Its density is 3.04 g cm–3. What is the length of the edge of unit cell? (At. mass of Cu = 63.5, CI = 35.5) 

    Solution

    Formula mass of CuCl =63.5 +35.5

                                          =  99.0

    The number of formula units per cell of ZnS is 4. It has face centred cubic structure.

     we have Z  =4density (d) =3.4 g Cm-3N0 (Avogadro's number) =6.023 x1023Mass =99.0 gramTherefore using formula density =Mass x Zunit cell volume(Cm)3 xNavod.no.d =M x Za3 x N0a3 =99 x43.4 x6.023 x1023a3 =1.932 x10-22cm3unit cell length a= (1.932 x10-22)13takinig log both sidelog a =13log  1.932 x10-22       =13(0.2860-22) =13(-21.7140)       = -7.238 =0.762 x10-8taking antilog a =antilog 0.762 x10-8 =5.758 x10-8 cm = 5.758 x 10-8cm




    Question 236
    CBSEENCH12005584

    Potassium crystallises in a simple cubic unit cell. It has an atomic mass of 209 and its density is 91.5 g m–3. What is the edge length of its unit cell? 

    Solution

    we have given,
    Mass= 209 g
    Number of  atom per unit cell =1 (Simple cubic)
    density =91.5 g m-3
    NA =6.023 x1023
    edge length of unit cell =?
     
    By applying formula


    density = Mass x Number of atom per unit cellvolume of unit cell  x avogadro's numberd= M x Za3 x NA91.5 = 209 x 1a3 x 6.023 x 1023a3 =209 x 191.5 x 6.023 x 1023

    a= 15.59 x10-8 cm

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