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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 %
(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?
we know thatd =ZMa3NAas we have given d= 7.80 g cm–3.M=56 g mol–1a= 290 pm or a3 =2.43 x10-23putting all value in above equation we get,Z =a3 x d xNAMZ= 2.43 x10-23 x 7.80 x6.022 x102356 =2.03hence 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.
Formula mass of CuCl =63.5 +35.5
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
we have given,Mass= 209 gNumber of atom per unit cell =1 (Simple cubic)density =91.5 g m-3NA =6.023 x1023edge length of unit cell =? By applying formuladensity = 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 1023a= 15.59 x10-8 cm