solid state numerical 2016

Solid state

1d

d1

M

Log 186 = 2.2695Log 8.55 = 0.9320

Log 6.022 = 0.7797A.L (0.5578) = 3.613

Md

M

1

1

an = n loga1

1

2. Copper crystallizes into a fcc structure and the unit cell has length of edge 3.61 x 10-8cm. Calculate the density of Cu. Atomic mass of Cu is 63.5gmol-1. [2 Marks]Given- a = 3.61 x 10-8cm, d = 63.5gmol-1

,d= ?, Z= 4

d

1

1

M

3. Silver Crystallizes in FCC structure with edge length of unit cell, 4.07 x 10-8cm and if density of metallic silver is 10.5g cm-3. Calculate the molecular mass of silverGiven- a = 4.07 x 10-8cm , d = 10.5g cm-3, Molecular mass = ?

The no. of atoms per unit cell = Z = 4

d M

4. Determine the density of cesium chloride which crystallize in a bcc type structure with the edge length 412.1pm. The atomic masses of Cs and Cl are 133 and 35.5 respectively.

[2 Marks]Given- a = 412.1pm = 4.12 x 10-8cm, M = 135 + 35.5 =168g mol-1, d = ?

5. Unit Cell of iron crystal has edge length of 288pm and density of 7.86g cm-3. Determine The type of crystal lattice(Atomic mass of Fe = 56g mol-1).Given-

Solution

Formula

To findHence the unit

of crystal lattice is of BCC type.

6d

6

6

77

7

7

8. Atoms C and D form fcc crystalline structure. Atom C is present at the corners of the cube and D is at the faces of the cube. What is the formula of the compound.

Calculation- 1. As C is present at the 8 corners of the cube.The no. of atom of C in the unit cell = 1/8 x 8 = 12. As D is present at the faces corners of the 6 facc of cube.The no. of atom of D in the unit cell = 1/2 x 6 = 3Therefore, the formula of the compound is CD3.